Mathematical Proof/Methods of Proof

There are many different ways to prove things in mathematics. This chapter will introduce some of those methods.

Introduction edit

In this chapter, for every method of proof introduced, we will discuss it in this manner:

  1. Explaining why the method works, by considering its underlying logic.
  2. Introducing how to use the method.
  3. Giving examples of using the method (and possibly also some previous method introduced) to prove some results.

Before introducing the first proof method, let us go through the meanings of some frequently used terms in mathematics books (some are already used in previous chapters actually), which are used more frequently starting from this chapter.

  • Definition: an explanation of meaning of a term.
  • Theorem: an important and interesting true mathematical statement.
  • Proposition: a (relatively) less important theorem.
  • Lemma: a true mathematical statement that is useful in establishing the truth of other true statements, and is less important than a theorem.
  • Corollary: a true mathematical statement that can be deduced from a theorem (or proposition) simply.
  • Proof: an explanation of why a statement is true.
  • Axiom: a true mathematical statement whose truth is accepted without proof.
  • Conjecture: a statement that is believed to be true, but is not proven to be true.

Direct proof edit

Many mathematical theorems can be expressed in the form of ", in which and are open statements about elements in a set [1]. This expression means "If then " is true for every . However, in practice, we rarely include the phrase "is true", and usually just write the theorem in the form of "For every , if , then ." Sometimes, we write instead "Let . If , then .", which has the same meaning.

In some situations, the statement is not stated in such a form directly, but can be expressed in such a form. For example, the statement "The square of an even integer is even." can be expressed as "For every , is even is even.", or "For every , if is even, then is even.", or "Let . If is even, then is even.", etc.

Now, we will introduce the first proof method to prove statements in such a form, which is known as direct proof. As suggested by its name, this method is quite "direct", and is probably the simplest method among all methods discussed in this chapter.

Consider the statement "". We would like to prove that it is true. Suppose is false for some . Then, the conditional must be true for this by definition, regardless of the truth value of . Hence, in the proof, we do not need to consider those for which the hypothesis is false.

Because of this, to give a direct proof of "", we first assume is true (So, we are considering every for which is true.), and then proceed to show that is true for every such (or else the conditional will be false).

This shows that is true for those for which is true, and this is enough to prove is true for every (for which may or may not be true), since we have mentioned that the conditional must be true for those for which is false.

Example. Prove that for every , if is odd, then is even.

Proof. Assume is odd. Then, by definition we have for some . Thus,

where (this follows from the definition of integers). This means can be written as for some , and hence is even.

In the previous proof, we have used the definitions of odd and even integers:

  • an integer is even if for some integer .
  • an integer is odd if for some integer .

We can write the proof more concisely using logical notations, as follows:

Proof.

Both styles of proofs are acceptable, but for beginners, it is recommended to use the first type. Also, the first type can let the readers understand the proof easier.

Remark.

  • The property of an integer of whether it is even or odd is called parity.
  • When we use "" in the proof, it is implicitly assumed it follows the same meaning as in the given statement, i.e., is an integer. We can also let to be an integer explicitly in the proof to be more clear.
  • In some other places, the first line "Assume ..." is simply omitted for convenience, but the assumptions are still used in the proof.
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Exercise. Consider the statement "For every , if is odd and is even, then is even.". A student provides the following proof to the statement:

Proof. Assume is odd and is even. That is, and for some . It follows that , and thus is even.

Is the proof correct? If not, point out the mistake, and write a correct proof.

Solution

The proof is incorrect. The mistake is that one should not use the same for both and , since using the same implies that , which is not necessarily the case from the assumptions. The following proof is a correct one:

Proof. Assume is odd and is even. That is, and for some . It follows that , and thus is even.


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Exercise. Prove that for every , if is even, then is odd.

Proof

Proof. Assume is even. Then, for some . Thus,

Since , is odd.


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Exercise. Prove every of the following statements.

(a) The sum of two arbitrary even integers is even.

(b) The sum of two arbitrary odd integers is even.

(c) The sum of an arbitrary even integer and an arbitrary odd integer is odd.

(d) The product of two arbitrary even integers is even.

(e) The product of two arbitrary odd integers is odd.

(f) The product of an arbitrary even integer and an arbitrary odd integer is even.

(Hint: rewrite the statements in the form of "" first.)


Solution

(a) First, rewrite the statement as "For every , if and are even, then is even."

Proof. Assume and are even. Then, and for some (notice that we should not use the same for both and , since may be different from .). Thus, , and hence is even since .

(b) First, rewrite the statement as "For every , if and are odd, then is even."

Proof. Assume and are odd. Then, and for some . Thus, , and hence is even since .

(c) First, rewrite the statement as "For every , if is odd and is even, then is odd."

Proof. Assume is odd and is even. Then, and for some . Thus, , and hence is odd since .

(d) First, rewrite the statement as "For every , if is even and is even, then is even."

Proof. Assume and are even. Then, and for some . Thus, , and hence is even since .

(e) First, rewrite the statement as "For every , if is odd and is odd, then is odd."

Proof. Assume and are odd. Then, and for some . Thus, , and hence is odd since .

(f) First, rewrite the statement as "For every , if is odd and is even, then is even."

Proof. Assume is odd and is even. Then, and for some . Thus, , and hence is even since .


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Exercise. A real number is defined to be rational if there exist integers with such that . Prove every of the following statements.

(a) The sum of two arbitrary rational numbers is rational.

(b) The difference of two arbitrary rational numbers is rational.

(c) The product of two arbitrary rational numbers is rational.

(d) The quotient of an arbitrary rational number by an arbitrary nonzero rational number is rational.

Solution

(a) First, rewrite the statement as "For every , if , then .

Proof. Assume . Then, and for some with and . Thus,

Since with , .

(b) First, rewrite the statement as "For every , if , then .

Proof. Assume . Then, for some with . So, is rational since and . Thus, by (a), we have .

(c) First, rewrite the statement as "For every , if , then .

Proof. Assume . Then, and for some with and . Thus,

Since with , .

(d) First, rewrite the statement as "For every , if and then .

Proof. Assume and . Then, and for some with , , and . Thus,

Since with , .


Example. Prove that for every , if , then .

Proof. For every , . Thus, the hypothesis is false, and hence the statement must be true.


Remark.

  • Notice that we do not use direct proof here, since the false hypothesis makes the statement true. In this case, we call the proof as vacuous proof, and say that the result follows vacuously (the statement says nothing at all).

Example. Prove that for every , if , then .

Proof. Assume . Then,

Since is positive and is negative by assumption, we have , and thus


Remark.

  • Graphically, the function (we will discuss the concept of function in later chapter) looks like:

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Exercise. Prove that for every , if , then (this statement is the converse of the statement in above example). (Hint: The following property may be useful: for every , if , then either ( and ) or ( and ).) (This statement seems to be true by inspecting the above graph, but inspecting graph is not a valid way of proving this statement.)

Proof

Proof. Assume . Then, ("" is read "which implies" in this context). It follows that we have either

  • and , or
  • and .

Since no real number satisfies the first one, we must have the second one, which gives .


Combining the two results in the previous example and exercise, we get an "if and only if" result:

For every , we have if and only if .

In other words, using set language,

(Both sets represent the curve (strictly) under the horizontal line in the above graph.)

Letting and be the open statement "" and "" respectively, we can express the above result symbolically:

Recall that "" means "" and "". So, to give a direct proof to the statement in the form of "", a usual way is to break the proof into two parts:

  1. proving that (known as the "only if" part, or "" direction)
  2. proving that (known as the "if" part, or "" direction)

Example. Prove that for every , . (Hint: this inequality is equivalent to the inequality , obtained by multiplying both side by , which is a positive integer.)

Proof. Using the hint, it suffices to prove that for every , . But it follows from the fact that .

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Exercise.

We can strengthen the statement (i.e., make the statement stronger (recall the meaning of "stronger" in previous chapter)) in the example, by changing to a larger set. Determine which of the following (may be more than one) can be the larger set.

None of the above.

A student provides the following proof to the above statement:

Proof. Since is a positive integer, multiplying to both sides of the inequality yields . After rearranging, we get , which is always true.

Point out the mistake in the proof.

Solution

The mistake is that the student implicitly assumes , which is what we want to prove, at the beginning of the proof. Thus, this proof does not prove the result. (Notice that in the proof in the example, we do not assume . Instead, we start from the fact that .)


Example. Let . Prove that for every , if is prime, then is odd.

Proof. Assume is prime. Since the only primes in the set are 3,5,7 and 11, this means . Hence, is odd.

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Exercise. Prove that for every , if is even, then is prime.

Proof

Proof. Since the set only contains odd numbers, the hypothesis is false, and therefore the result follows vacuously.



Example. (A special case of AM-GM inequality) Prove that for every , if and are positive, then

Proof. Assume and are positive. Then,

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Exercise. Suggest a condition where the equality holds, i.e., .

Solution

The condition is .


Proofs related to congruence of integers edit

After introducing the method of direct proof, let us apply this method on proving some results relating to congruence of integers. Before this, let us introduce the concept of congruence of integers. We begin by a motivation for the definition of congruence of integers.

We know that an integer is either even or odd, i.e., can be expressed as or for some integer , according to whether the remainder is 0 or 1 when is divided by 2. Thus, if two integers and have the same remainder when divided by 2 (i.e., have the same parity), then the difference can be proved to be a multiple of 2 (it can be proved that the converse is also true). Similarly, an integer can be expressed as or for some integer , according to whether remainder is 0, 1 or 2 when is divided by 3. Hence, if two integers and have the same remainder when divided by 3, then the difference can be proved to be a multiple of 3 (the converse is also true).

Hence, "two integers have the same remainder when divided by some integer " is equivalent to "their difference is a multiple of ". This leads us to the following definition.

Definition. (Congruence of integers) Let and with . Then, the integer is congruent to modulo , denoted by , if is a multiple of , i.e., for some .

Remark.

  • Instead of " is a multiple of ", we can also say is divisible by , or divides , denoted by (in general, the notation means " divides " (and means " does not divide ").

Example. We have since , and since . But, since .

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Exercise.

1 Choose the integers that are congruent modulo 3.

5,8
-9,7
-9,18
0,37
None of the above.

2 Choose the integers that are congruent modulo 6.

5,8
-9,7
-9,18
0,37
None of the above.

3 Choose the integers that are congruent modulo 9.

5,8
-9,7
-9,18
0,37
None of the above.


Example. (Clock arithmetic) A familiar application of the concept of congruence of integers is clock arithmetic. For instance, when we want to know adding 5 hours to 8 o'clock gives what time, we first calculate . But . So, we now think about 13 is congruent to which integer between 1 and 12. The integer is 1. Hence, the time is 1 o'clock.

In general, we can get the resulting time by performing the modulo operation, which gives the remainder when dividing the sum of the two numbers by 12 (called modulus). The notation for " modulo " is , which gives the remainder when is divided by ( and are positive integers).

Now, let us apply direct proof to prove some results related to the congruence of integers.

Theorem. For every and for every with , if and , then

  • (compatibility with addition) .
  • (compatibility with translation) .
  • (compatibility with scaling) .
  • (compatibility with multiplication) .

(For the compatibility with translation and scaling, the assumption that is not needed.)

Proof. We will only prove the compatibility with addition and multiplication. The proof of other two compatibilities is left to the following exercise.

Compatibility with addition: Assume that and . Then, and for some . Thus,

Since , we have .

Compatibility with multiplication: Assume that and . Then, and for some . Thus,

Since , we have .

Remark.

  • We have applied the trick in the proof of compatibility with multiplication. One can also prove it without using this trick. The details are left to the following exercise.
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Exercise. Prove the compatibility with multiplication in the above theorem, without using the trick used in the above proof. (Hint: You may express in terms of and in terms of first.)

Proof

Assume that and . Then, and for some . That is, and . Hence,

Since , we have .

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Exercise. Prove the compatibility with translation and scaling in the above theorem.

Proof

One can of course use direct proof and the definition of congruence of integers to prove them, but we will just apply the compatibility with addition and multiplication (which are proven) to prove them (they can be regarded as corollaries of the compatibility with addition and multiplication).

Proof. Assume . Since , we have . Hence, by the compatibility with addition, we have . Also, by the compatibility with multiplication, we have .


In the above result, everything is with the same "modulo ". It is then natural to ask whether there are any results for different "modulo". The answer is yes, and to discuss a result, we need the concept of relatively prime integers.

Definition. (Relatively prime integers) Two integers are relatively prime if their greatest common divisor is 1.

Remark.

  • The integers are also called coprime or mutually prime.

Example.

  • 2 and 3 are relatively prime.
  • 4 and 12 are not relatively prime since their greatest common divisor is 4.
  • 9 and 66 are not relatively prime since their greatest common divisor is 3.

Theorem. Two integers and are relatively prime if and only if there exist integers and such that .

Proof. Omitted.

Using this result, we can deduce the following result:

Theorem. For every with , if , and are relatively prime, then .

Proof. Assume , , and are relatively prime. Then, we have and for some . This means . Also, for some . Multiplying both sides by the integer , we get

Putting , we get
Putting this into , we get
Since , we have .

Remark.

  • This theorem says in particular if an integer is a multiple of , and also a multiple of , and and are relatively prime, then the integer is a multiple of .
  • For example, if an inteeger is a multiple of 3 and also a multiple of 7, then the integer is a multiple of 21 since 3 and 7 are relatively prime.
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Exercise. Give an example of integers such that , , and are not relatively prime, and .

Solution

Take . Then, and . However, .

Example. (Reflexivity, symmetry, and transitivity of the congruence of integers) Prove that for every with ,

  • (reflexivity) for every , .
  • (symmetry) for every , if , then .
  • (transitivity) for every , if and , then .

(Since the congruence of integers satisfies these three properties, it is said to be an equivalence relation. We will discuss the concept of (equivalence) relation in a later chapter.)

Proof.

Reflexivity:

Since , we have .

Symmetry:

Assume . Then, for some . Thus, , and hence since .

Transitivity:

Assume and . Then, and for some . Hence, . Thus, since .


Remark.

  • Notice that not all relations satisfy all these three properties. For instance, "" does not satisfy the reflexivity and transitivity, "" does not satisfy the reflexivity and symmetry.

Example. Applying the compatibility with multiplication in the above theorem, we can get the following result:

Let be an integer, be a nonnegative integer, and with . If , then .

We can prove it (a bit informally) as follows:

Proof. Assume . First, . Second, we have . Applying this argument again, we get . Thus, we have for every nonnegative integer by applying the argument "again and again".

To prove the result formally, we need to use proof by mathematical induction, which will be discussed later.

Example.

(a) Consider the remainder of when divided by 4 (i.e., ) for . Do you observe any pattern? Hence, suggest a conjecture.

(b) Prove the conjecture.

Solution.

(a) We have

Notice that the remainder is 1 for , and 3 for . It is thus natural to expect that the same pattern continue for all other larger integers. Thus, a conjecture is

Let be a nonnegative integer. If is even, then . Also, if is odd, then .

(b) We will just prove "If is even, then .". The proof of the second part is left to the following exercise.

Proof. Assume is even. Then, for some nonnegative integer ( is a nonnegative integer). It follows that . Since , it follows that by the result in previous example. Hence, .

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Exercise. Prove the second part in the above conjecture.

Proof

Proof. Assume is odd. Then, for some nonnegative integer . It follows that . From the prove above, we have . Thus, we get .



The following lemma is quite useful for proving results about congruence of integers.

Lemma. (Euclid's division lemma) For every integer and with , there exists a unique pair of integers and such that

Proof. We separate the proof into existence part and uniqueness part.

Existence part:

Case 1: .

Then, we set and . After that, the equation can be rewritten equivalently as , and also the inequality can be rewritten equivalently as . Through this, we transform this case to case 2.

Case 2: .

Subcase 1: and .

Then, we set , , and . After that, the equation can be rewritten equivalently as (this is equivalent to ). Also, the inequality can be rewritten equivalently as . Through this, we transform this case to the subcase 2.

Subcase 2: and .

Set and . Then, we have .

Subsubcase 1: . Take and . Then, we have

and we are done.

Subsubcase 2: .

Set and . Then, we have with . Since there are exactly nonnegative integers less than , we need to repeat this process at most times (, so is at most the preceding integer of ) to get a such that [2] (and also a ).

So, we take and . Then, we have

and we are done.

Uniqueness part:

Assume there exists another pair of integers and (in addition to the pair of integers and ) such that

Since we have also
we get, by subtracting the two equations,

Also, by considering the above two inequalities, we get

Putting it in the above equation, we get .

Thus, we have and .

Remark.

  • This lemma is the basis of Euclidean division, as known as division with remainder.
  • is called the dividend, is called the divisor, is called the quotient, and is called the remainder.
  • We employ the method of proof by cases in the proof, which is not a "new" way of proof strictly speaking. We just consider different cases in the proof. However, when we use this method, we need to ensure that the cases covers all possibilities for the "for every", so that we actually prove the statement.
  • From this lemma, we know that every integer has remainder either 0 or 1 or 2 or ... or (from the inequality ) when divided by an integer with . In other words, for every integer , we have either or or ... or . For brevity, we can also write .

Example. Prove that for every integer , .

Proof. By Euclid's division lemma, we have . Thus, we have . But and . So, the result follows by the transitivity of the congruence of integers.

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Exercise. Propose a similar result for the congruence modulo 5, and prove it.

Solution

Proposition: For every integer , .

Proof. By Euclid's division lemma, we have . Thus, we have . But and . So, the result follows.




Proofs related to sets edit

The proofs related to sets are often in the form of

  1. A set is a subset of another set.
  2. A set equals another set.

To prove that a set is a subset of another set, we use the definition of subset:

A set is a subset of another set if for every element , if , then .

Thus, we can employ direct proof to prove that a set is a subset of another set.

For the equality of two sets, recall that:

A set equals another set if and only if and .

Thus, to prove the equality of two sets, we often need to separate the proof into two parts: (i) proving that ; (ii) proving that .

Example. (Transitivity of "") Prove that for every set and , if and , then .

Proof. Assume and . Then, for every ,

and hence .


Example. (Associative law) Prove that for every set and , .

Proof. First, we prove that . For every ,

Thus, we have .

Now, we prove the reverse subset inclusion, i.e., . For every ,

Thus, we have .

Notice that the proof for reverse subset inclusion is very similar to the proof for the first part. Indeed, it is just a reverse of the proof for the first part. Hence, we can actually simplify the proof as follows:

Proof. For every ,

Thus, we have .

Using this proof, we can prove the set equality directly (set equals set if for every , ). (However, we should be careful about whether we really have "".) But, in many cases, the proof for reverse subset inclusion is not just simply obtained by reversing the proof for the first part, and we have to separate the proof into proofs for two subset inclusions.

We can observe that to prove different laws for the sets, we just simply use the corresponding laws in logic to prove them. Hence, the proofs for other laws, e.g., commutative law and distributive law, are similar.

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Exercise. (De Morgan's law) Let and subsets of a universal set . Prove that .

Proof

Proof. For every ,


Example. Let and be subsets of a universal set . Prove that .

Proof. For every ,

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Exercise. Using this result, prove that . (Hint: you may use the laws in set theory in the proof.)

Proof

Here, we just use the laws of in set theory to prove the equality of sets:


Example. Prove that for every set and , and .

Proof. For the first one, for every ,

(no matter is true or false, we always have the first implication.)

For the second one, for every ,

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Exercise.

(a) Propose an assumption on the sets and such that we have both reverse subset inclusions, i.e., and . Prove them under this assumption.

(b) Propose an assumption on the sets and such that we have (the another reverse subset inclusion may or may not hold). Prove it under this assumption.

(c) Propose an assumption on the sets and such that we have (the another reverse subset inclusion may or may not hold). Prove it under this assumption.

(The assumptions proposed should be as weak (recall the meaning of weak in logic) as possible, so that the result can apply in more contexts.)

Solution

(a) Assumption: .

Proof. Assume . For the first one, for every ,

For the second one, for every ,

(One can also use the some results in set theory (e.g. , , etc.) in the proof.)

(b) Assumption: .

Proof. Assume . Then, for every ,

(c) Assumption: .

Proof. Assume . Then, for every ,

(We have the first implication, since if , then we have , and also since )



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Exercise. Consider the statement: for every set and , . A student gives the following proof:

Proof. For every ,

Is the proof correct? If not, point out the mistake and give a correct proof.

Solution

The mistake is in this step:

We have "", but do not necessarily have "", since we generally do not have , and thus we cannot show "" by contrapositive.

The following is a correct proof:

Proof. For every ,

On the other hand, for every ,


Example. Prove that for every set and , if and , then .

Proof. For every ,


Example. Prove that for every set and , .

Proof. For every ,

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Exercise. Prove that for every set and ,

(a) .

(b) .

Solution
(a)

Proof. For every ,

(Notice that for the third "", we cannot change it to "" since .)

On the other hand, for every ,


(b)

Proof. For every ,



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Exercise. Consider the statement:

For every set and , if and , then .

A student provides the following proof:

Proof. Assume and . For every ,

Is the proof correct? If not, point out the mistake and give a correct proof.

Solution

The proof is correct.

Example. Prove that for every set and .

Proof. For every ,

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Exercise. Prove that for every set and . (Hint: you may find the property useful.)

Proof

Proof. For every ,


Remark.

  • The reverse subset inclusion does not hold. For instance, take and . Then, and . We can see that there is not reverse subset inclusion in this case.



Example. Let and be sets. Prove that if and only if .

Proof. "" direction: Assume . Then, for every set ,

Thus, we have .

"" direction: Assume . Then, for every ,

Thus, .

We can also prove the "" direction more simply:

Assume . Since for every set , , and thus , we have

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Exercise. Let and be sets.

(a) Prove that .

(b) Give an example of and such that .

(c) We will have under an assumption. Propose an assumption, and prove the equality under the assumption. (Hint: construct some simple examples to observe whether there are any patterns.)


Solution
(a)

Proof. Claim: .

Proof. "" direction: Assume . Then, for every , . So, .

"" direction: Assume . Then, for every , . Also, . Hence, and .

For every set ,

(b) Take and . Then, , while .

(c) An assumption is " or ".

Proof. Assume or .

Case 1: .

Then, . Hence, . On the other hand, by the above example, we have , and thus . Therefore, .

Case 2: .

The proof is similar (just interchange "" and "").


Proof by contrapositive edit

Recall that the contrapositive of a conditional is the statement , which is logically equivalent to the original conditional . Hence, if we can show that , then it follows that . This gives us another method of proof, namely proof by contrapositive.

A proof by contrapositive of the statement is a direct proof of . That is, we first assume that is true, and proceed to show that is true. In other words, we first assume that is false, and proceed to show that is false.

Generally, when we encounter a statement , and observe that the consequent is "simpler" than the hypothesis , using proof by contrapositive is usually more preferable.

Example. Prove that for every , if is even, then is odd.

Proof. Here, we use proof by contrapositive, that is we will prove that "for every , if is even, then is odd."

Assume is even. Then, for some . Thus, , and hence is odd since .

Remark.

  • Using proof by contrapositive here allows us to work with with initially, instead of the more complicated expression .


Example. Prove that for every , is even if and only if is even.

Proof. "" direction (or "only if" part):

Assume is even. Then, for some . Thus, . Hence, is even since .

"" direction (or "if" part):

We use proof by contrapositive, i.e., we will prove that "for every , if is odd, then is odd." Now, assume is odd. Then, for some . Hence, , and so is odd since .


Remark.

  • Using proof by contrapositive for the "" direction allows us to work with , instead of , initially.
  • The statement is logically equivalent to "for every , is odd if and only if is odd." (since ).

Example. Prove that for every , if and only if .

Proof.

"" direction:

We use proof by contrapositive. First, assume . Since by Euclid's division lemma, we have . Hence, . But . It follows that , and hence .

"" direction:

Assume . Then, we have . But . So, we have .

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Exercise. Prove that if and only if .

Proof

Proof.

"" direction:

We use proof by contrapositive. First, assume . Since by Euclid's division lemma, we have . Hence, . But . It follows that , and hence .

"" direction:

Assume . Then, we have . But . So, we have .

Notice that we can also apply the fact that (proved before) to conclude that . Then, we can see that this statement is just logically equivalent to the above statement (since ).


Example. Let and be sets. Prove that if and only if .

Proof.

"" direction:

We use proof by contrapositive. Assume , i.e., there exists an element such that . Since , it follows that . But . So, .

"" direction:

Assume . Since holds for every set and , it suffices to show that :

For every ,

So, .


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Exercise. Prove that for every set and , if , then . (Hint: means that there exists an element such that and .)

Proof

Proof. "" direction:

We use proof by contrapositive. Assume . Then, there exists an element such that and . So, we have and . This means . Thus, .

"" direction:

We use proof by contrapositive. Assume . Then, there exists such that and . It follows that and , meaning that , and hence .


Example. Prove that for every nonnegative real number , if for every positive real number , then .

Proof. We use proof by contrapositive, i.e., we will prove that "if , then for some positive real number ."

Assume . This means since is nonnegative. Then, we pick which is a positive real number, and we have .


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Exercise. Prove that for every , if , then or .

Proof

Proof. We use proof by contrapositive. Assume and . Then, we have (by the property of "").


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Exercise. An integer is defined to be a perfect square if for some . Prove that for every , if is a perfect square, then is even or is even. (Hint: consider a previous result about congruence of integers, that is related to perfect square.)

Proof

We use proof by contrapositive. Assume is odd and is odd. Then, and for some . Hence,

This means since . However, no perfect square is congruent to 2 modulo 4 (recall that we have proved that for every integer , ). Thus, is not a perfect square.

Proof by cases edit

In the previous sections, we have actually employed the method of proof by cases already. It is natural to use this method when we need to break down a problem into several cases, and then tackle every case individually ("divide and conquer"). Sometimes, we may even need to further divide a case into several subcases. The following are some typical cases:

  1. an integer is even or odd;
  2. a real number is less than 0, equal to 0, or greater than 0;
  3. for two nonzero real numbers and , either or .
  • For every of the cases, it can be divided into two subcases:
  1. : ( and ) or ( and )
  2. : ( and ) or ( and )

But, we should be aware that when we use proof by cases to prove that "", the cases should cover all possibilities, i.e., all .

Example. Prove that for every , is even.

Proof. We divide the proof into two cases: is even and is odd.

Case 1: is even.

Then, for some . Thus,

Since , is even.

Case 2: is odd. Then, for some . Thus,

Since , is even.

It follows that is even for every .


Example. (Comparing parity) Prove that for every , and are of the same parity (i.e., both even or both odd) if and only if is even.

Proof. "" direction: Assume and are of the same parity.

Case 1: and are both even.

Then, and for some . Thus, , and hence is even since .

Case 2: and are both odd.

Then, and for some . Thus, , and hence is even since .

"" direction: We use proof by contrapositive. Assume and are of different parity.

Case 1: is odd and is even.

Then, and for some . Thus, . Hence, is odd since .

Case 2: is even and is odd.

The proof is similar to case 1 (just exchange the role of and ).


Remark.

  • We sometimes use the phrase without loss of generality (WLOG) to indicate that the proofs of the two situations are similar, and thus the proof of only one of these situations is needed. For instance, for the "" direction in the above proof, we may write "Assume and are of different parity. WLOG, assume is odd and is even. ...".
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Exercise. Let and be integers. Prove that is even if and only if is even or is even.

Proof

Proof. "" direction: We use proof by contrapositive. Assume is odd and is odd. Then, and for some . Hence,

and so is odd since .

"" direction: Assume is even or is even. WLOG, assume is even. Then, for some . Thus, . Since , is even.


The following inequality is quite important in mathematics.

Theorem. (Triangle inequality) For every , .

Proof.

Case 1: and .

Then, , and so

Case 2: and .

Then, , and so

Case 3: one of and is nonnegative, and the other is negative.

WLOG, assume and .

Subcase 1: .

Then,

Subcase 2: .

Then,

So, we have the desired inequality for every .

Proof by contradiction edit

Another important method of proof is proof by contradiction.

Let be a statement. Now, suppose we want to prove that is true. If we can show that the conditional is true (i.e., ), then the truth table for conditional tells us that must be false, and hence must be true, as desired.

The truth table:

In words, the method of proof by contradiction is as follows: we first assume that the statement we want to prove to be true is false, and then proceed to show that this gives a contradiction, and therefore our assumption must be wrong. That is, it is impossible for the statement to be false since it leads to something "absurd". Hence, the statement is true.

The name of proof by contradiction comes from the fact that the assumption that the statement is false is later contradicted by some other fact. This is also known as reductio ad absurdum, which means reduction to absurdity.

In general, we often use the method of proof by contradiction when the statement we want to prove is negative sounding, e.g. "There is no ....", "There does not exist ...", etc.

Also, to indicate to the reader that we are using proof by contradiction, it is recommended that we write "assume to the contrary that ..." (or something similar) instead of just "assume ..." for the assumption that the statement is false.

Example. Prove that no odd number can be written as a sum of two odd numbers.

Proof. Assume to the contrary that the statement "no odd number can be written as a sum of two odd numbers" is false. That is, assume to the contrary that there is an odd number that can be written as a sum of two odd numbers and . Then, we have and for some . Hence, we have

which means is even since . But this contradicts to our assumption that is odd. So, the result follows.


Example. Prove that there is no smallest positive real number.

Proof. Assume to the contrary that there is a smallest positive real number . Now, consider the number . Since , it follows that . Also, we have . Hence, is a positive real number that is less than , contradicting to our assumption that is the smallest positive real number.

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Exercise.

Which of the following statement is/are true?

There is no smallest positive rational number.
There is no smallest positive integer.
There is no smallest nonnegative integer.

Prove that there is no greatest positive real number.

Proof

Proof. Assume to the contrary that there is a largest positive real number . Now, consider the number . Since , it follows that . Also, we have . Hence is a positive real number that is greater than , contradicting to our assumption that is the greatest positive real number.



Example. Prove that the sum of an arbitrary rational number and an arbitrary irrational number is irrational.

Although this statement appears to be not so negative sounding, "irrational" just means "not rational". In this sense, we can see that this statement is also quite negative sounding. Also, it is not easy to show that a number is irrational directly (while, on the other hand, showing that a number is rational is quite easy: just express it as a quotient of an integer by a nonzero integer).

Proof. Assume to the contrary that the sum of an arbitrary rational number and an arbitrary irrational number is a rational number . So, we have where and where with and . But we can then rewrite the equation as

Since and , this means is rational, contradicting to our assumption that is irrational.


The following is a typical example in introducing proof by contradiction:

Example. Prove that the number is irrational.

Proof. Assume to the contrary that is rational. Then, for some with . By dividing and by some common factor that is at least 2, if necessary, we can further assume that and have no common factor greater than or equal to 2, i.e., has been expressed in (or reduced to) the lowest terms.

Taking square for the both sides of the equation , we get . Since , this means is even. By a previous result, this implies that is even. Hence, we have for some .

Now substituting it into the equation , we get , which means is also even since . By a previous result again, this implies that is even.

Since both and are even, they have 2 as a common factor, contradicting to our assumption that has been expressed in (or reduced to) the lowest terms.


The following example is another typical example for demonstrating proof by contradiction. But before discussing it, we need to introduce the following theorem since it is used in the proof.

Theorem. (Fundamental theorem of arithmetic) Every integer can be expressed as a product of one or more (not necessarily distinct) primes uniquely. That is, for some prime numbers , uniquely.

Proof. We will prove this theorem after introducing proof by mathematical induction.

Remark.

  • Notice that the "uniqueness" in the theorem is in the sense of the combination of prime number(s) used is unique. Simply changing the order of the multiplication is not counted as another expression.
  • This theorem is also called unique prime factorization.

Example. (Euclid's theorem) Prove that there are infinitely many prime numbers.

Proof. Assume to the contrary that there are finitely many prime numbers, say prime numbers. We first list them in ascending order: . Now, consider the number

Since the smallest prime number is 2, we have [3], and hence . Applying the fundamental theorem of arithmetic, there is a unique prime factorization for , and thus there is a prime number dividing .

Since are all the primes by our assumption, it follows that one of them, say , divides , i.e., for some . In addition, we know that divides , i.e., for some .

Therefore, we have . This means divides 1, which is impossible (notice that 2 is the smallest prime, and only 1 divides itself). So, we are arriving at a contradiction.


Remark.

  • This proof is first suggested by Euclid.

We can also use proof by contradiction to prove that a statement "" is true. We first assume that the negation of the statement, i.e., "" is true, and proceed to arrive at a contradiction. (Recall that . So, .) Thus, to prove that by contradiction, we assume that there exists such that is true and is false, and then deduce a contradiction.

Example. Prove by contradiction that for every , if is odd, then is even.

Proof. Assume to the contrary that there exists such that is odd and is odd. Since is odd, we have for some . Then,

This means is even since , contradicting to our assumption that is odd.


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Exercise. Prove by contradiction that for every nonnegative real number , if for every positive real number , then .

Proof

Proof. Assume to the contrary that for every positive real number and . Since is nonnegative, this means . Now, consider the number . By assumption, we have , which implies that , arriving at a contradiction.


Example. Prove that is irrational.

Proof. Assume to the contrary that is rational, i.e., for some (). Notice that . So, we only have the following cases:

Case 1: .

Now, we have

But is even, while is odd (product of even (odd) numbers is even (odd)).

Case 2: and are both negative integers.

Then, we write . Now, we have

But is even, while is odd ( and are positive integers).

So, in either case, we arrive at a contradiction.

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Exercise. A student provides the following proof to the claim " is irrational.":

Proof. Assume to the contrary that is rational, i.e., for some (). Notice that . So, we only have the following cases:

Case 1: .

Now, we have

But is even, while is odd.

Case 2: and are both negative integers.

Then, we write . Now, we have

But is even, while is odd.

So, in either case, we arrive at a contradiction.

Is the proof correct? If not, point out the mistake.

Solution

The proof is incorrect. Indeed, is rational. The mistake is that the student states that and also in the proof, which are not true.


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Exercise. Prove that for every , if is rational, then both and are rational, or both and are irrational.

Proof

Proof. Assume to the contrary that there exists such that and exactly one of and is rational, and the other irrational. WLOG, assume and . Then, since , and and , we have , contradicting to our assumption that .


Example. (Sum and product of a rational and an irrational number) Prove that for every , (a) if and , then ;

(b) if and , then .

Solution.

(a)

Proof. Assume to the contrary that there exist such that and and . Then, , contradicting to our assumption that .

(b)

Proof. Assume to the contrary that there exist and and . Then, (), contradicting to our assumption that .


Sometimes we can prove a statement using multiple methods:

Example. Prove that for every positive real number and , if , then .

We can prove it using (i) direct proof; (ii) proof by contrapositive; (iii) proof by contradiction.

(i)

Proof. Assume . Then multiplying both sides by gives . Also, multiplying both sides by gives . Combining these two inequalities, we get .

(ii)

Proof. Assume . Rearranging the inequality gives . Now dividing both sides by the positive number preserves the direction of inequality, and thus gives .

(iii)

Proof. Assume to the contrary that and . Then, we have . Also, since and are positive, . So we have , contradicting to our assumption that .


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Exercise. Let be a positive real number. Prove that if , then by (i) direct proof; (ii) proof by contrapositive; (iii) proof by contradiction.

Solution
(i)

Proof. Assume . Multiplying both side by gives . Then, dividing both sides by the positive number gives .

(ii)

Proof. Assume . Then, . Also, ( is positive). Thus, . This implies

(iii)

Proof. Assume to the contrary that and . Then, . Also, . So, , and hence . This contradicts to our assumption that .


We can see that when proving a statement "", we may be able to use both proof by contrapositive and proof by contradiction. However, the two proofs are different, and we will compare them in the following table:

Proving that "."
Proof by contrapositive Proof by contradiction
Assumption
Goal

From this table, we can see that the proof by contradiction is more advantageous in terms of assumption, since it has one more "help" from (when we assume , we can use both and .) However, in terms of goal, the proof by contrapositive is more advantageous since the goal is more clear (), while for the proof by contradiction, it is not clear that what the form of the contradiction is. There can be many "ways" for arriving at a contradiction (compare the contradiction in the proof of " is irrational" vs. that of "there are infinitely many primes").

Existence proof edit

Consider the statement "". This statement is true if is true for at least one . Otherwise, it is false. Thus, to prove such kind of statement, it suffices to find one element such that is true, and this is known as an existence proof. In particular, we should verify that the choice of element is actually belonging to and is actually true for that choice.

Example. Prove that there exists positive integers such that (such integers are known as a Pythagorean triple).

Proof. Take , and (which are integers). Then, .


Remark.

  • The choice of is not unique. For instance, one can also take and . But, giving one example of and that satisfies the requirement is enough.

Example. Prove that there exists an integer such that .

Proof. Take . Then, we have .


The following example demonstrates a more advanced version of existence proof: non-constructive proof.

Example. Prove that there exist irrational numbers and such that is rational.

Proof. We have proved that is irrational. So, we can make use of this fact in this proof.

Consider the real number . This number is either rational or irrational. Now we consider the following cases:

Case 1: is rational. Then, we can take , and then is rational.

Case 2: is irrational. Then, we can take and . Then, which is rational.

Thus, in either case, we can find two irrational numbers and such that is rational.


Remark.

  • This type of proof is known as non-constructive proof since it does not actually construct an example and , and so we do not know which two irrational numbers and satisfy this requirement. However, this proof does prove the existence of such and .
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Exercise.

(a) Prove that there exist distinct irrational numbers and such that is rational. (You may use the fact that is irrational.)

(b) Prove that there exist a rational number and an irrational number such that is rational.

(c) Prove that there exist a rational number and an irrational number such that is irrational. (You may use the fact that is irrational.)

(d) Prove that there exist rational numbers and such that is irrational.

Solution
(a)

Proof. Consider the real number . It is either rational or irrational. Now, we consider the following cases:

Case 1: is rational. Then, we can take and , and is rational.

Case 2: is irrational. Then, we can take and . Then, , which is rational.

(b)

Proof. Take and . Recall that we have proved that is irrational. Then, we have , which is rational.

(c)

Proof. Consider the real number . Now, consider the following cases:

Case 1: is irrational. Then, take and , and is irrational.

Case 2: is rational. Then, take and . Then, , which is irrational.

(d)

Proof. Take and . Then, , which is irrational.


Disproof edit

Consider the statement ".". Suppose we somehow believe that the statement is false. Then, we want to prove that it is false, i.e., prove that " such that is false." An element for which is false is known as a counterexample of the statement, and a process of showing that the statement is false (in other words, disproving the statement) is called a disproof of the statement. One counterexample is enough to disprove the statement.

Example. Disprove that for every .

Disproof. Take . Then, .


Example. Disprove that the product of two arbitrary irrational numbers is irrational.

Disproof. To disprove this, we need to find such that .

Take . Then, .


To disprove the statement ", we show that there exists an element such that is false, i.e., is true and is false.

Example. Disprove that for every , if , then .

Disproof. Take . Then but .


To disprove the statement "", we need to show that there is no such . That is, we need to show that "" (the negation of statement) is true. In this case, we are not just constructing counterexamples. We have to prove that , instead of .

Example. Disprove that there exists such that is even.

Disproof. To disprove this, we need to prove that for every , is even. But this has been proved previously (in the section about proof by cases).


Often, before proving or disproving statements, the truth of falseness of them are not known, and we need to decide whether every of them is true or false. After that, we will prove it if it is true and disprove it otherwise. Of course, our decision is not perfectly accurate, and sometimes we make mistake, which leads us to do the opposite thing, i.e., proving a wrong statement or disproving a correct statement. Clearly, when we do such thing, we cannot succeed. However, the process of performing such wrong thing may give us some insights about what the correct direction is, and how to prove/disprove the statement.

To decide whether a statement is true or false, you may follow some tips below:

  • Follow your mathematical intuition. As you have learnt more about mathematics, you may have intuition and idea about whether a statement is true or false, before proving/disproving it.
  • Construct some simple examples. Such examples constructed may be a counterexample (for disproving "")/example (for proving "") if you are lucky. Even if they are not counterexamples/examples, you may observe some kind of patterns from it, which may give you some insights about how to prove/disprove the statement.

Example. Prove or disprove every of the following statements:

(a) There exists a real number satisfying the equation .

(b) Let and be sets. If , then .

(c) For every real number , .

(d) There exists a real number such that .

Solution.

(a)

Disproof. Since is a real number, we have and . Hence, , and thus for every .

(b)

Disproof. Take , and . Then, , but .

(c)

Disproof. Take . Then the LHS is undefined (the denominator is zero), but the RHS is -1.

(d)

Proof. Take . Then, and , and we have .


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Exercise. Prove or disprove every of the following statements:

(a) There exists a real number such that and are both rational.

(b) There exists with such that .

(c) There exist even numbers and such that .

(d) For every , if and , then .

(e) For every with , there exists such that .

Solution
(a)

Disproof. We want to prove the statement is false. We use proof by contradiction.

Assume to the contrary that the statement is true, i.e., there exists a real number such that and are both rational. Then, we have is rational, which contradicts to the fact that (product of a rational number and an irrational number) is irrational.

(b)

Proof. Take and . Then, .

(c)

Proof. Take . Then, .

(d)

Disproof. Take and . Then, and , but .

(e)

Proof. For every , choose (sum/product of two rational numbers is rational). Since , we have , as desired.


Proof by mathematical induction edit

The last proof method discussed in this chapter is proof by mathematical induction. This method is used to prove a statement in the form of "For every integer greater than or equal to an integer , is true.". That is, it is used to prove that a sequence of statements is true. Of course, we can prove that every of these statements is true one by one, but the principle of mathematical induction gives an alternative and more convenient method.

To prove the principle of mathematical induction, we need the well-ordering principle. Before introducing it, we need the definition of well-ordered.

Definition. (Well-ordered set) A nonempty set of real numbers is well-ordered if every nonempty subset of has a least element.

Example.

  • The set is well-ordered since its nonempty subsets are , and every of these subsets has a least element.
  • The sets are not well-ordered since none of these sets themselves have a least element (it can be proved that there is no smallest integer/rational number/real number).

Now, one may ask that whether the set is well-ordered. It may appear that it is well-ordered. Here, we will just regard this as an axiom:

Axiom. (Well-ordering principle) The set is well-ordered.

The well-ordering principle can then be used to prove (a less general version of) the principle of mathematical induction.

Theorem. (The principle of mathematical induction)

The principle of mathematical induction can be illustrated by the sequential effect of falling dominoes. " is true" means the first domino can be pushed down, and "for every integer , " means for every domino, if it falls down, then it will push the next domino down also. With these two requirements satisfied, every domino will fall down eventually (every statement involved is true).

For every , if are statements satisfying

(i) is true; and

(ii) for every integer , ,

then the statements are true.

This theorem is quite intuitive: with the conditions satisfied, we have an "infinite chain of implications":

  • is true (by (i)).
  • Thus, is true (by (ii)).
  • Thus, is true (by (ii)).
  • Thus, is true (by (ii)) ...

But strictly speaking, this is not counted as a proof. We will give a formal (partial) proof to this theorem below:

Proof. Here we only prove the case where .

Assume to the contrary that the theorem is false. Then, the conditions (i) and (ii) are satisfied by the statements, but there exist some positive integers for which is false. Now, let the set

Since is a nonempty [4] subset of , by the well-ordering principle, contains a least element. Let us assume to be the least element.

Since is true by the condition (i), . This means the least element cannot be . Hence, , which means .

Since is the least element of , we have , i.e., is a true statement. Then, by the condition (ii), is true (we have ). It follows that . But this contradicts to our assumption that is the least element of (in particular, ).

Remark.

  • This proof can be extended to the case where . But the proof needs to use the result that the set is well-ordered. Once we have proven this result, we can simply change the set in the proof to , and also modify some sentences in the proof slightly for the extension.
  • Notice that in the inductive hypothesis, we assume is true for an arbitrary integer with , but not for every integer . This is because if we assume the latter, then we are assuming what we want to prove, and so the proof is invalid. While for the former assumption, we just assume is true for an arbitrary (but not every) integer with , but this does not lose generality since is arbitrary.

Using the principle of mathematical induction, to prove that the statement is true for every integer , it suffices to prove two things:

(i) is true.

(ii) For every integer , .

Thus, proof in this form is a two-step process. We often call the proof of (i) as the basis step or base case, and the proof of (ii) is called the inductive step. In particular, when we prove (ii), we usually use the method of direct proof, and first assume is true for an arbitrary integer with . Such assumption is often called the inductive (or induction) hypothesis.

Example. Prove that for every ,

is true.

Proof. Here we use the principle of mathematical induction.

Basis Step: Since , the statement is true.

Inductive Hypothesis: Let be an arbitrary positive integer. Assume is true. That is, assume

is true.

Inductive Step: Since

it follows that is true. Hence, by the principle of mathematical induction, is true for every positive integer .


Remark.

  • One can also prove this statement using a trick:
First write the sum as .
Now also write the same sum, but in reverse order: .
After that, add the two sums together in the following way:

Now, we can see that two times the sum gives . Thus, the sum is .
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Exercise. For every , consider the sum

Guess a general formula for the sum and prove it. (Hint: you may compute the value of sum for some values of , and see whether there are any patterns.)

Solution

When , the sum is respectively. Notice that and . Hence, it is natural to guess that the formula is, for every ,

But we have proved that . So this formula can be alternatively expressed as

Proof. Let be the open statement "".

Basis Step: Since , is true.

Inductive Hypothesis: Assume is true for an arbitrary positive integer . That is, assume

Inductive Step: Since

is true. Hence, by the principle of mathematical induction, is true for every positive integer .

Remark.

  • The following diagram illustrates this statement:


We can use the proof by mathematical induction for proving some inequalities also:

Example. Prove that for every integer , .

Proof. Let be the open statement "".

Basis Step: Since , is true.

Inductive Hypothesis: Assume is true for an arbitrary integer with . That is, assume

Inductive Step: Since
it suffices to show that . Now, consider :
Thus, is true. Hence, by the principle of mathematical induction, is true for every integer .


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Exercise. Prove that for every integer , .

Proof

Let be the open statement "".

Basis Step: Since , is true.

Inductive Hypothesis: Assume is true for an arbitrary integer . That is, assume

Inductive Step: Since

(we have since ) is true. Hence, by the principle of mathematical induction, is true for every integer .

Example. (Sum of a geometric sequence) Prove that for every integer and for every real number , we have

Proof. Let be the open statement "".

Basis Step: Since , is true.

Inductive Hypothesis: Assume is true for an arbitrary nonnegative integer . That is, assume

Inductive Step: Since

is true. Hence, by the principle of mathematical induction, is true for every .


Example. (Cardinality of power set) Prove that for every set and for every integer , if is finite with , then .

Proof. Let be a finite set with cardinality and be the open statement .

Basis Step: Since , and , is true.

Inductive Hypothesis: Assume is true for an arbitrary nonnegative integer . That is, assume . In other words, there are subsets of .

Inductive Step: First, fix an element . Now let . Then, contains elements, and thus by inductive hypothesis, there are subsets of .

Let be a subset of . Since is a subset of , it follows that is a subset of . But, since . So, is also a subset of , that is distinct from . Thus, every subset of gives rise to two subsets of , namely and . As a result, there are subsets of without the element and subsets with the element .

So, we have subsets of altogether. Hence, , and thus is true. Hence, by the principle of mathematical induction, is true for every integer .


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Exercise. (Generalization of De Morgan's law) Prove that for every subset of a universal set and for every integer ,

(Hint: you may use the ordinary De Morgan's law, i.e., for every subset of a universal set , .)

Proof. Let be the open statement ".".

Basis Step: By the ordinary De Morgan's law, is true.

Inductive Hypothesis: Assume is true for an arbitrary integer . That is, assume

Inductive Step: Since
it follows that is true. Hence, by the principle of mathematical induction, is true for every integer .


The strong form of induction edit

In this section, we will discuss a stronger form of mathematical induction. Sometimes, we need this form of mathematical induction since merely assuming " is true ..." in the inductive hypothesis may not be enough to show that is true. We may need some more helps, particularly the fact that the statement is true for all cases from the basis step to the given . That is, if "" is 1, then we assume are all true.

Since "" is stronger than "", we have the name "the strong form of induction".

Theorem. (The strong form of induction) For every , if are statements satisfying

(i) is true; and

(ii) for every integer , ,

then the statements are true.

This theorem is also quite intuitive: with the conditions satisfied, we get an "infinite chain of implications":

  1. is true by (i).
  2. Because of 1., is true by (ii).
  3. Because of 1. and 2., is true by (ii).
  4. Because of 1., 2., and 3., is true by (ii) ...

Proof. The proof for the strong form of induction is similar to that of the principle of mathematical induction. We just need to modify some parts of the proof. The blue part is the modified part.

Here we only prove the case where .

Assume to the contrary that the theorem is false. Then, the conditions (i) and (ii) are satisfied by the statements, but there exist some positive integers for which is false. Now, let the set

Since is a nonempty subset of , by the well-ordering principle, contains a least element. Let us assume to be the least element.

Since is true by the condition (i), . This means the least element cannot be . Hence, , which means .

Since is the least element of , we have , i.e., is a true statement. Also, (they are all less than ). Hence, are also true statements. Then, by the condition (ii), is true (we have ). It follows that . But this contradicts to our assumption that is the least element of (in particular, ).

Hence, to prove that the statement is true for every by the strong form of induction, we need to prove

(i) the basis step: is true; and (ii) the inductive step: for every integer , if are true (the inductive hypothesis), then is true.

We usually use the strong form of induction to prove results that are related to recurrence relation.

Example. A sequence of numbers is defined by , and

(an equation in such form is called a recurrence relation). Prove that for every .

Proof. Let be the open statement "". Now, we will use the strong form of induction to prove that is true for every .

Basis Step: Since , is true.

Inductive Hypothesis: Let be an arbitrary positive integer. Assume that are true. That is, assume for every integer with .

Inductive Step: We want to show that is true, i.e., . Now, using the recurrence relation, we have

So, we have proved that is true for the case where . Now, it remains to prove that is true when , but it is easy to prove that: , so is true. Hence, by the strong form of induction, is true for every .


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Exercise. A sequence of numbers is defined by , and

Guess a formula for for every , and prove it.

Solution

When compute for according to the recurrence relation, we get and . This pattern suggests us to guess that for every .

Proof. Let be the open statement "". Now, we will use the strong form of induction to prove that is true for every .

Basis Step: Since .

Inductive Hypothesis: Let be an arbitrary positive integer. Assume that are true.

Inductive Step: We divide our proof into two cases:

Case 1: . Then, . So, is true.

Case 2: . Using recurrence relation, we have

Thus, is true.

Hence, is true for every by the strong form of induction.


Example. A sequence of numbers is defined by and

then for every .

Proof. Let be the open statement .

Basis Step: Since , is true.

Inductive Hypothesis: Let be an arbitrary integer. Assume are true.

Inductive Step:

Case 1: . Then, .

Case 2: . Then, .

Case 3: . Then,

Hence, is true.

By the strong form of induction, is true for every .


Remark.

  • It may appear that we consider quite "suddenly" in the proof above. Indeed, if we "think from the bottom" for case 3, then it is natural to consider it: we want , which is in the form of , to be less than . So, it is natural consider the "something".
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Exercise. Consider the sequence of Fibonacci numbers, defined by and

(a) Construct a table for comparing the value of against the value of for .

(b) Hence, make a guess in the form of " for every integer .", and prove the statement.

(c) By considering the even Fibonacci numbers in the table in (a), make a guess in the form of " is even if and only if , for every .", and prove the statement. (Hint: since there is "if and only if" in the statement, in the inductive step, we need to prove both "" direction and "" direction. But, we can prove both directions at once in this case.

(d) Prove that for every , where (golden ratio) and (conjugate of golden ratio). This gives a direct formula to compute th Fibonacci number, which is more efficient than using the recurrence relation in the definition for the computation. (Hint: The roots of the quadratic equation are and .)

Solution

(a)

(b) From (a), it appears that when , increases faster than . So, it is natural to guess that for every integer .

Proof. Let be the open statement .

Basis Step: Since , is true.

Inductive Hypothesis: Let be an arbitrary integer with . Assume that are true.

Inductive Step: Using recurrence relation, we have

Hence, is true, and by the strong form of induction, is true for every integer .

(c) From the table in (a), when (which are all multiples of 3), is even. This suggests us to guess that is even if and only if for every .

Proof. Let be the open statement is even if and only if .

Basis Step: Since is odd and , is true ().

Inductive Hypothesis: Let be a positive arbitrary integer. Assume that are true.

Inductive Step: By recurrence relation, . So,

(We have , since by the compatibility with translation, it can be shown directly that , and after showing this, we have the desired logical equivalence.)

Thus, is true. Hence, by the strong form of induction, is true for every .

(d)

Proof. Let be the statement .

Basis Step: Since , is true.

Inductive Hypothesis: Let be a positive arbitrary integer. Assume are true.

Inductive Step:

Case 1: . Then, . Hence, is true.

Case 2: . Then, apply the recurrence relation:

Hence, is true.

Thus, by the strong form of induction, is true for every .



  1. Mathematical theorems can also have more than one variable, e.g., .
  2. It is impossible for to be negative, since (the number just before getting in the desired range), and thus .
  3. Of course we can write it as (or even writing greater than or equal to some greater number if we know more prime numbers), but it is not necessary for applying the fundamental theorem of arithmetic on .
  4. This is because there exist some positive integer for which is false.