Revision:OCR Core 1 - Index Notation

Register

About Us | Help | Sign in

Create New Article | Editing Help

Advanced Search

Edit \| Hide
Welcome \| Study Help \| A Levels \| Applications, UCAS etc. \| University Discussion \| Careers \| General Discussion \| Have Your Say \| Health & Relationships \| Debate & Discussion \| Technology \| Sport \| Entertainment \| Chat \| Ask A Moderator

Revision:OCR Core 1 - Index Notation

From The Student Room

Jump to:navigation, search

TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 1 - Index Notation

1 7. Index Notation
2 Also See
3 Comments

7. Index Notation

Working with indices

Indices offer a useful way to express a long series of multiplications (where one is multiplying a number to itself several times in succession).

The fact that indices are a simplified notation makes them susceptible to erroneous deduction through poor understanding of the meaning of the notation. It is therefore useful (and vital if one is intending to use indices) that one is aware of the rules which exist (and also why they exist).

Let us define an index:

$a^{x} = a \times a \times a \times ... \times a$

Where there have been "x" values of "a" multiplied together. Hence, a numerical example is:

$2^{5} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 32$

One must know about the division and multiplication rules for indices.

Suppose one has the following:

$a^{x} \times a^{y}$

One can write this in the "long" manner, by writing "x" values of "a", all separated by multiplication symbols, and then another set of "a" values, however "y" of them, all separated by multiplication symbols. One has a multiplication symbol between these two groups of "a". This means that there are "x + y" values of "a", all separated by multiplication symbols, hence one can simplify this to a different power, giving:

$a^{x} \times a^{y} = a^{x + y}$ .

The division rule is similar. Suppose one has the following:

$a^{x} / a^{y}$

One can write this as:

$\frac{a \times a \times a \times ... \times a}{a \times a \times a \times ... \times a}$

Where the numerator has "x" values of "a" separated by multiplication signs, and the denominator has "y" values of "a" separated by multiplication symbols. One can therefore cancel this down, giving the division rule:

$a^{x} / a^{y} = a^{x - y}$

Sometimes one will be presented with a term such as:

$\left( a^{x} \right) ^{y}$

One should think of this in the logical manner. It is a collection of "y" terms of $a^{x}$ which are separated by multiplication symbols. These terms are, in turn, collections of "x" values of "a", all separated with multiplication symbols. This gives a total of $x \times y$ values of "a", all separated by multiplication symbols, giving the "power-on-power rule":

$\left( a^{x} \right) ^{y} = a^{xy}$

One must remember that one is dealing with terms of "a", which is a single value, it is not possible (without further arithmetic, or at all, depending on the values of the unknown terms) to simplify $a^{x} \times b^{y}$ for obvious reasons.

One must also be aware of the "factor rule", which states:

$(a \times b)^{x} = a^{x} \times b^{x}$ .

The reasoning for this is simply that there is a collection of "x" terms of $a \times b$ , and hence there are "x" values of "a", separated with multiplication symbols, and the same for "b".

Zero and negative indices

It is perfectly meaningful to have a negative, or zero index, and one must be aware of the meanings of both.

First, the zero index. One should consider the following:

$\frac{a^{x}}{a^{x}}$

One can deduce (from the division rule) that:

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

Also, one can deduce from basic number principles, that:

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

(This is because one knows that any number that is not zero, results in 1 when divided by itself).

Hence, one deduces that:

$a^{0} = 1$

The negative index:

Consider:

$\frac{a^{0}}{a^{x}}$

Through the division rule, this gives:

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

But, also, one can use the zero index rule, to give:

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

Hence, one can deduce that:

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

Fractional indices

It is perfectly logical for fractional indices to exist (and be defined) also.

Consider the following:

$\left( a^{\frac{1}{x}} \right) ^{x}$

One can use the "power-on-power rule", giving:

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

Hence, one can deduce that:

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

(This is due to the definition of a root).

This need not be limited to unit fractions, consider the following:

$\left( a^{\frac{y}{x}} \right) ^{x}$

Again, use the "power-on-power rule", giving:

$\left( a^{\frac{y}{x}} \right) ^{x} = a^{\frac{xy}{x}} = a^{y}$

Hence:

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

Equations with rational indices

Example

1. Find if:

$x^{\frac{3}{4}} = 27$

It is much easier to understand what the situation is if one is to put this into a more "humanly readable" form, thus:

$\sqrt[4]{x^{3}} = 27$

Hence:

$x^{3} = 27^{4}$

At this point one could calculate the $27^{4}$ , however this is somewhat of a tedious task (considering that the examination is a non-calculator one), hence, if one spots that there is a link with "3", one can continue using the "power-on-power rule", thus:

$x^{3} = \left( 3^{3} \right) ^{4} = 3^{12}$

Hence:

$x = \sqrt[3]{3^{12}} = 3^{4} = 81$

The general idea with equations of this type is to ensure that they are in a more understandable form (than the fractional index which they will begin in), after this one can work using the aforementioned rules, and the algebraic techniques which one will (hopefully) be honing.

Powers of negative bases

Consider the following:

a > 0

$(-a)^{x}$

Consider:

$x \equiv 0 \pmod{2}$

This will produce a positive result (if one is to consider that one will be multiplying a negative number to a negative number, and even number of times, it is clear that the result is positive).

Now consider:

$x \equiv 1 \pmod{2}$

This will cause the original statement to be a negative value (as one will be multiplying a negative to a negative an even number of times, giving a positive result, however one must then multiply again by a negative value, giving a negative result).

Consider now the fractional indices:

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