If is a real number, then the area of a circle of radius is .
If there is a line and a point not on , then there is exactly one line containing that is parallel to .
If is a triangle with sides of length and then
If is a continuous function on [a, b] and is an function such that , then...
If , then there is an integer q such that . Let q = n.
If , then there is an integer q such that . Let q = 1.
If , then there is an integer q such that . This implies , and so , and thus .
If n is an even integer, then for some integer k, .
If n is an odd integer, then for some integer k, .
If n is even, then . For integers j and k, let .
, so is even.
If n is odd, then . For integers j and k, let .
, so is odd.
If a|b, and b|bm then a|bm, implying aj = bm for some integer j.
Also, if a|c, and c|cn then a|cn, implying ai = cn for some integer i.
We let x = (j+i).
ax = aj+ai
ax = bm+cn
Which implies a|(bm+cn).
Another proof: Suppose that and . Hence there are integers and such that and . Define the integer by . Then
Because , it follows
implies that for some integer, x.
implies that for some integer, y.
for some integer, j.
Let , hence .
Suppose that . Hence there is an integer such that . if is a positive integer, define the integer by . Then
Because , it follows
Proof by contraposition:
Assume is not even, then and .
Let then , it follows that if is not even then is not even.
It is true that does not divide . Suppose that . This means there is an integer such that . Then, we have:
We may consider the integer . Therefore, we have that . Then , Contradiction!
Let a non-zero rational number, hence there are integers and both different from zero, such that . Let an irrational number.
Suppose that the product is a rational number, hence there are integers and different from zero such that , this is , it follows that .
The last equatily means that is a rational number, which is a contradiction because we supposed that was irrational. By contradiction, it follows that the product must be irrational.
Suppose that and , but does not divide . Hence there are integers and such that and . Suppose that the equation has a solution such that and are integers, then
Let , then , it follows that , which is a contradiction.