A-level Mathematics/OCR/C1/Indices and Surds

The indices in this chapter heading refers to the power to which a number is raised. Thus $x^{2}$ is a number with an index of 2. This is never normally said as people prefer the phrase "x to the power of 2". Likewise, surds is a heading that is not often used. Occasionally you wil be asked to give an answer in surd form; this implies that you should give the answer in terms of constants, trig functions and square roots instead of working out an imprecise decimal approximation.

Indices

As you should know from GCSE, $x^{3}$ is a shortening of $x\times x\times x$. In the same way, any number to the power of n is that number multiplied by itself n - 1 times. To describe more detail, in the expression $x^{3}$, the $x$ is referred to as the base, and the 3 as the exponent.

By always reducing the expression back to its basic parts it is possible to come to conclusions about how to treat numbers raised to powers in algebra.

Multiplication

When you multiply indices you add the exponents together.

As you can see $x^{3}\times x^{2}=x^{3+2}=x^{5}$ .(law1)

Division

Division is, as expected, the opposite to multiplication. When you divide indices you subtract the exponents from each other.

${\frac {x^{4}}{x^{2}}}={\frac {x\times x\times x\times x}{x\times x}}={\frac {\not \!x\times \not \!x\times x\times x}{\not \!x\times \not \!x}}=x\times x=x^{2}$

Again this shows that ${\frac {x^{4}}{x^{2}}}=x^{4-2}=x^{2}$

Negative powers

Having done the above division, we begin to ask ourselves what happens when we have a fraction that has a higher exponent on the bottom than the top.

Using the trusted method of showing each x separately we obtain:

${\frac {x^{2}}{x^{3}}}={\frac {\not \!x\times \not \!x}{\not \!x\times \not \!x\times x}}={\frac {1}{x}}$

Using the just demonstrated method of subtracting the exponents:

${\frac {x^{2}}{x^{3}}}=x^{2-3}=x^{-1}$

This implies that x to the power of a negative number is one divided by x to the power of that number. The specific case of ${\frac {1}{x}}\;\left(x^{-1}\right)$ is referred to as the reciprocal of x or more often as the inverse of x.

Base raised to two powers

When you have a base raised to two powers you multiply the powers.

$\left(x^{2}\right)^{3}=\left(x\times x\right)\times \left(x\times x\right)\times \left(x\times x\right)=x\times x\times x\times x\times x\times x=x^{6}$.

As you can see that $\left(x^{2}\right)^{3}=x^{2\times 3}=x^{6}$.

Multiple bases to the same power

When you have two bases to the same power, you can raise both bases to the same power and multiply them.

For example: $\left(xy\right)^{3}=xy\times xy\times xy=x\times y\times x\times y\times x\times y=x\times x\times x\times y\times y\times y$ which is the same as $x^{3}y^{3}\,$. Here is an example with numbers: $\left(2\times 5\right)^{2}=\left(10\right)^{2}=100=4\times 25=2^{2}\times 5^{2}$. There is a similar situation with division: $\left({\frac {x}{y}}\right)^{2}={\frac {x}{y}}\times {\frac {x}{y}}={\frac {x\times x}{y\times y}}={\frac {x^{2}}{y^{2}}}$

Fractional powers

What if the power isn't even an integer? Suppose you wanted to find $x^{\frac {1}{2}}$, you could say that ${x^{\frac {1}{2}}}\times {x^{\frac {1}{2}}}=x^{1}$ (by law 3, addition of powers) which means that $x^{\frac {1}{2}}$ must be $\pm {\sqrt {x}}$. However it is customary to only use the positive root and so $x^{\frac {1}{2}}$ is defined as ${\sqrt {x}}$. You can use a similar argument for other such fractions, for example $\left(x^{\frac {1}{3}}\right)^{3}=x$ so $x^{\frac {1}{3}}={\sqrt[{3}]{x}}$. In cases when the numerator is not 1 we need to use other laws of indicies to prove the square definition, for example $x^{\frac {2}{3}}=\left(x^{2}\right)^{\frac {1}{3}}$ (using law 3), and $\left(x^{2}\right)^{\frac {1}{3}}={\sqrt[{3}]{x^{2}}}$ (using the definition above). It's useful to remember the general rule that $x^{\frac {a}{b}}={\sqrt[{b}]{x^{a}}}$.

The 0 and 1 Power

You may well have realised that ${x^{1}}=x\,$. You can prove this by ${\frac {x^{n}}{x^{n-1}}}$ which is clearly $x$ which is $x^{n-\left(n-1\right)}=x^{1}=x$ by law 2.
Also with $x^{0}\,$ we can prove that it is equivalent to 1, ${\frac {x^{n}}{x^{n}}}=1$, but by law 2 it is also equivalent to $x^{n-n}=x^{0}$

The Laws of Indices

The rules that have been suggested above are known as the laws of indices and can be written as:

1. $x^{a}x^{b}=x^{a+b}\,$

2. ${\frac {x^{a}}{x^{b}}}=x^{a-b}$

3. $x^{-n}={\frac {1}{x^{n}}}$

4. $\left(x^{a}\right)^{b}=x^{ab}$

5. $\left(xy\right)^{n}=x^{n}y^{n}$

6. $\left({\frac {x}{y}}\right)^{n}={\frac {x^{n}}{y^{n}}}$

7. $x^{\frac {a}{b}}={\sqrt[{b}]{x^{a}}}$

8. $x^{0}=1\,$

9. $x^{1}=x\,$

10. $(x^{m})^{\frac {1}{m}}=x\,$

Surds

In mathematics, a Surd is an expression containing a root term, with an irrational value, that can not be expressed exactly – for example, ${\sqrt {3}}=1.732050808...$ .

Sometimes it is useful to work in surds, rather than using an approximate decimal value. Surds can be manipulated just like algebraic expressions and sometimes it may be possible to eliminate the surd (called rationalising the expression), which may have not been possible if you tried to work with the approximate value. When asked to give the exact value, approximate decimal answers will not do and you will have to manipulate surds in order to give a final answer in simplified surd form.

Simplification of surds

Because surds can be manipulated like algebraic expressions, you can easily multiply out the terms and add the like terms. However, there are also a few rules that will be useful when simplifying surds.

Basic rule of surds

Because ${\sqrt {x}}\times {\sqrt {x}}=x$, it is useful to know that it can be rearranged to give ${\sqrt {x}}={\frac {x}{\sqrt {x}}}$ and ${\frac {1}{\sqrt {x}}}={\frac {\sqrt {x}}{x}}$.

Surds as indices

Because ${\sqrt[{n}]{x}}=x^{\frac {1}{n}}$ the laws of indices also apply to any n-th root. The most frequently used instances of this are laws 5 and 6 with surds:

• $\left(xy\right)^{n}=x^{n}y^{n}$ becomes ${\sqrt {xy}}={\sqrt {x}}\times {\sqrt {y}}$
• $\left({\frac {x}{y}}\right)^{n}={\frac {x^{n}}{y^{n}}}$ becomes ${\sqrt {\frac {x}{y}}}={\frac {\sqrt {x}}{\sqrt {y}}}$

The first of these points is often used to simplify a surd, for example ${\sqrt {200}}={\sqrt {100\times 2}}={\sqrt {100}}\times {\sqrt {2}}=10{\sqrt {2}}$. In an exam, you will be expected to write all surds with the smallest possible number inside the surd (i.e. the number inside the root shouldn't have any square factors).

Note: This rule only applies to non-negative numbers, attempts to involve negative numbers might give absurd results; $-1={\sqrt {-1}}\cdot {\sqrt {-1}}={\sqrt {(-1)(-1)}}={\sqrt {1}}=1$

Rationalising the denominator

Another technique to simplify expressions involving surds is to rationalise the denominator. This means removing a surd from a fraction. In the case of a fraction such as ${\frac {5}{\sqrt {3}}}$, both numerator and denominator can be multiplied by ${\sqrt {3}}$ to give ${\frac {5{\sqrt {3}}}{3}}$.

If the fraction is of the form${\frac {a}{b+{\sqrt {c}}}}$ the strategy used in the previous paragraph will work if it is slightly modified. This time you should multiply the numerator and denominator by ${b-{\sqrt {c}}}$. If you are familiar with the standard difference of two squares expansion you should already know what happens next:

${\frac {a}{b+{\sqrt {c}}}}\times {\frac {b-{\sqrt {c}}}{b-{\sqrt {c}}}}={\frac {ab-a{\sqrt {c}}}{b^{2}+b{\sqrt {c}}-b{\sqrt {c}}-{\sqrt {c}}^{2}}}={\frac {ab-a{\sqrt {c}}}{b^{2}-c}}$.

As you can see the denominator now does not contain any surds. For example: ${\frac {2}{{\sqrt {3}}-1}}\times {\frac {{\sqrt {3}}+1}{{\sqrt {3}}+1}}={\frac {2\left({\sqrt {3}}+1\right)}{3-1}}={\sqrt {3}}+1$

Common questions and mistakes

Splitting up roots

A common mistake is to split ${\sqrt {x+y}}$ into ${\sqrt {x}}+{\sqrt {y}}$ or $\left(x+y\right)^{2}$ into $x^{2}+y^{2}$, usually whilst moving it to the other side of the equals. Trying a few examples will quickly convince you that this is not possible:

• ${\sqrt {64}}\ \neq {\sqrt {32}}+{\sqrt {32}}$
• ${\sqrt {25}}\ \neq {\sqrt {12}}+{\sqrt {13}}$

And so on

What is the value of $0^{0}$?

The short answer is that for this course you don't need to know, and you can safely skip this section. If you're still interested then read on:

The question arises because $x^{0}=1$ for any $x$ and yet you would expect that $0^{y}=0$ for any $y$ as $0\times 0\times 0\dots =0$. It turns out using a value of 1 is quite useful (perhaps even necessary) in various parts of algebra, whereas making it zero doesn't help at all. Almost all mathematicians would therefore either say that $0^{0}$ is 1 or that it is undefined (that is, it can't be given a value). A more technical discussion can be found at http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/

This is part of the C1 (Core Mathematics 1) module of the A-level Mathematics text.