Revision:Indices

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1 Positive powers
- 1.1 Basic identities
- 1.2 Common pitfalls
2 Zero and negative powers
3 Rational powers
4 Complex powers
5 Summary of Identities
6 Examples
7 Comments

Positive powers

A long time ago, someone noticed that multiplying a number by itself took up quite a lot of space, and decided to define a new operation called exponentiation. It's easiest to explain with an example:

$2^4 = 2 \times 2 \times 2 \times 2 = 16$

The basic idea is to multiply the number at the bottom (the base) by itself until you have a product of length equal to the number at the top (the index or exponent). should be read as "Two to the four", or "Two to the power of four" if you're feeling posh. In general:

$x^a = \underbrace{x \times x \times \ldots \times x}_{\text{a total x's in the product}}$

This isn't on the syllabus, but here's an equivalent definition you might find helpful if that's unclear:

$x^a = \left\{\begin{array}{ll} x \times x^{a-1} & a > 1 \\ x & a = 1 \end{array}\right.$

Basic identities

Now, three major identities hold. They're obvious when you think about them, but they're still pretty important.

$x^a \times x^b = x^{a+b}$

For example, $5^2 \times 5^3 = (5 \times 5) \times (5 \times 5 \times 5) = 5^5$ .

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

For example, $\frac{3^4}{3^2} = \frac{3 \times 3 \times 3 \times 3}{3 \times 3} = \frac{3 \times 3}{1} = 3^2$ .

$(x^a)^b = x^{ab}$

For example, $(5^2)^3 = 5^2 \times 5^2 \times 5^2 = 5^6$ .

Common pitfalls

Now, this isn't too difficult, but people often make two mistakes. The first one is to assume that because $x^a \times x^b$ and $\frac{x^a}{x^b}$ can be simplified easily, so can $x^a \times y^b$ and $\frac{x^a}{y^b}$ , where $x \ne y$ . Wrong. They can only be simplified if is a power of or vice versa. So, for example:

$3^4 \times 2^4$ can't be simplified.

$5^2 \times 25^2 = 5^2 \times (5^2)^2 = 5^2 \times 5^4 = 5^6$ can be simplified.

The second one is to assume that $x^{a^b} = (x^a)^b$ . They look equal, but they really aren't. For example:

$(5^3)^2 = 5^3 \times 5^3 = 5^6$

$5^{3^2} = 5^9$

See what I mean?

Zero and negative powers

Now, some bright spark noticed that the division identity above only worked for , and decided to define exponentiation for zero and negative exponents. Obviously, the key consideration when doing something like that is to make sure that zero and negative exponents behave as much like positive exponents as possible - that is, make them obey all three of the above identities. He eventually settled on the following definition for :

for all x.

I know, not an intuitive definition, right? But it's the only sensible definition under which all three identities hold:

$x^a \times x^0 = x^a \times 1 = x^a = x^{a+0}$

$\frac{x^a}{x^0} = \frac{x^a}{1} = x^a = x^{a-0}$

$(x^a)^0 = 1 = x^0 = x^{a \times 0}$

$(x^0)^a = 1^a = 1 = x^0 = x^{0 \times a}$

As for negative integers, it turns out from the division identity that the only sensible definition is:

$x^{-a} = \frac{1}{x^a}$

Again, all three identities hold:

$x^a \times x^{-b} = \frac{x^a}{x^b} = x^{a-b}$

$\frac{x^a}{x^{-b}} = x^a \times x^b = x^{a+b}$

$(x^a)^{-b} = \frac{1}{(x^a)^b} = \frac{1}{x^{ab}} = x^{-ab}$

$(x^{-a})^b = (\frac{1}{x^a})^b = \frac{1}{x^{ab}} = x^{-ab}$

This also renders the identity $\frac{x^a}{x^b} = x^{b-a}$ redundant, since dividing by is the same thing as multiplying by $x^{-b}$ . For this reason, sometimes $x^a \times x^b = x^{a+b}$ is known as the law of exponentiation.

Rational powers

But of course, this wasn't enough! Someone decided that integer powers were so good they might as well generalise to rational powers - in other words, fractions. As before, the key thing was to make sure that all the old identities worked. Assuming they do, then:

$(x^{\frac{1}{a}})^a = x^{a\times\frac{1}{a}} = x$

So the obvious choice for $x^{\frac{1}{a}}$ was:

$x^{\frac{1}{a}} = \sqrt[a]{x}$

Then we have:

$x^{\frac{a}{b}} = (x^{a})^{\frac{1}{b}} = \sqrt[b]{x^a}$

It turns out that both our identities hold, so we can take this as our definition. You might want to skip the proofs, though - they're a little messy, and they're not on the syllabus. Here they are anyway for completeness:

$\begin{eqnarray[*]} (x^{\frac{a}{b}})^{\frac{c}{d}} & = & \sqrt[b]{x^a}^{\frac{c}{d}} \\ & = & \sqrt[d]{(\sqrt[b]{x^a})^c} \\ & = & \sqrt[bd]{\sqrt[d]{(\sqrt[b]{x^a})^c}^{bd}} \\ & = & \sqrt[bd]{((\sqrt[b]{x^a})^c)^b} \\ & = & \sqrt[bd]{(\sqrt[b]{x^a})^{bc}} \\ & = & \sqrt[bd]{(x^a)^c} \\ & = & \sqrt[bd]{x^{ac}} \\ & = & x^{\frac{ac}{bd}} \\ \end{eqnarray[*]}$

$\begin{eqnarray[*]} x^{\frac{a}{b}} \times x^{\frac{c}{d}} & = & (x^{\frac{a}{b}})^1 \times (x^{\frac{c}{d}})^1 \\ & = & (x^{\frac{a}{b}})^{\frac{d}{d}} \times (x^{\frac{c}{d}})^{\frac{b}{b}} \\ & = & x^{\frac{ad}{bd}} \times x^{\frac{bc}{bd}} \\ & = & \sqrt[bd]{x^{ad}} \times \sqrt[bd]{x^{bc}} \\ & = & \sqrt[bd]{(\sqrt[bd]{x^{ad}} \times \sqrt[bd]{x^{bc}})^{bd}} \\ \end{eqnarray[*]}$ $\begin{eqnarray[*]} & = & \sqrt[bd]{(\sqrt[bd]{x^{ad}})^{bd} \times (\sqrt[bd]{x^{bc}})^{bd}} \\ & = & \sqrt[bd]{x^{ad} \times x^{bc}} \\ & = & \sqrt[bd]{x^{ad + bc}} \\ & = & x^{\frac{ad + bc}{bd}} \\ & = & x^{\frac{a}{b} + \frac{c}{d}} \\ \end{eqnarray[*]}$

Complex powers

It's possible to define exponentiation for irrational exponents like $\sqrt{2}$ , and complex exponents like $2 + 3\sqrt{-1}$ , but this is beyond the scope of A Level.

Summary of Identities

$x^a = \underbrace{x \times x \times \ldots \times x}_{\text{a total x's in the product}}$

$x^{-a} = \frac{1}{x^a}$

$x^{\frac{1}{a}} = \sqrt[a]{x}$

$x^{a} \times x^{b} = x^{a+b}$

$(x^a)^b = x^{ab}$

Examples

$2^{-2} = \frac{1}{4}$

$8^{\frac{1}{3}} = 2$

$32^{\frac{2}{5}} = 4$

$2^{2^{2^2}} = 2^{2^4} = 2^{16} = 65536$

Comments

This article is currently only suitable for A Level students and above. Perhaps as separate article for GCSE students (or maybe just foundation GCSE students is needed#) which starts more basically and has none of the latter stages.

Alternatively, it could be noted here that the latter sections are only needed for Higher GCSE/A Level maths.

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