### Archive

Archive for the ‘surds’ Category

## Rational and Irrational Numbers

All real numbers are either rational or irrational.

A number x is rational if it can be written in the form a/b where a and b are integers and b≠0 and a number which cant be written in this form is irrational.

Most of the time we deal with rational numbers, for example all the integers are rational as they can be written in the form of themselves over 1, all fractions are obviously rational and also are terminating or repeating decimals. When an irrational number is written out it will appear as an infinite non repeating decimal. Examples of irrational numbers are π (pi) and √2, in fact many square roots are irrational numbers.

Numbers can usually be proved to be irrational by contradiction, by assuming that they are rational and writing them in the form p/q where p and q are co-prime. Then a contradiction is usually produced by showing that they both have common factor other than 1. For example when proving √2 is irrational you can show that p and q must both have a common factor of 2.

There infinitely many rational and irrational numbers, however whilst it is possible to use a method to count all the rational numbers, if you had an infinite amount of time, this cannot be done for the irrational numbers.

Showing a repeating decimal is rational

A repeating decimal can be shown to be rational using the following method:

Consider the repeating decimal x,

Multiply x by 10 n where n is the number of repeating digits
Now subtract x from x10 n to get (10n-1)x
This number will be a finite decimal as all the repeating terms after the decimal point should cancel.
divide through by (10n-1)x to get an expression for x in the form of a fraction.

eg let x = 1/7 = 0.142857142857142857………
then 1000000x = 142857.142857142857………
so 999999x = 142857.0
x=142857/999999
so 1/7 = 142857/999999 therefore 1/7 is rational

if you have any questions, comment or corrections please leave them as a comment below

By David Woodford

Categories: maths, surds

## How to use Surds – Add, Multiply and Rationalise

Surds are roots to some power (usually square roots) that are left in their “root” form as opposed to being calculated as a decimal. In a way they are the power equivalent of fractions and are used when decimal value of a root isn’t needed, such as in the middle of a calculation, so it is better to leave the result more precisely as a surd.

Surds usually take the form a+b√c where b and c are non-zero

eg) √2
4+2√3

Simplifying Surds

It is easier to deal with surds if you can get the value in the square root to its lowest possible value by taking any square factors outside the root.

For example √50 = √(25×2) = 5√2

To do this I looked for any factors of 50 which are square, eg 25=5×5, can took these outside the square root – when taking a number outside the square root you obviously have to take the square root of the number so when you take out 25 you get a 5 left outside.

So in general

√(a2b) =a√b

This has the benefit of when multiple surds are in a calculation al the surds will be in their lowest terms so they will be easier to group

Surds can easily be added just by considering then as another variable, so just add up all the coefficients.

eg) 2√3 +6√3 – 3√3 = 5√3

Note– only roots of the same number can be added and subracted in this way, roots of different numbers must be left as they are. This makes it important to simplify all surds as it means you will be able to spot all the surds that can be added or subracted.

eg) 2√3 + 4√5 cannot by simplified

but

3√2 +2√50 = 3√2 +2×5√2=13√2

Multiplying and Dividing Surds

To multiply or divide surds just multiply or divide the values within the roots with each other and the coefficents of the roots with each other. When surds contain roots and non root parts multiply as if they were brackets

eg) 3√2 x 4√5 = 12√10

3√2 ÷ 4√5 = ¾ √(2/5)

(2+4√3) x (3+2√5) = 6 +4√5+12√3+8√15

Note – you may be able to further simplify the surd after you multiply

eg) 2√6 x √3 = 2√18=2×3√2 = 6√2

In general

(a+b√c) x (d + e√f) = ad + ae√f + db√c + be√fc

Rationalising and Conjugates of Surds

When simplifying fractions all surds are usually removed from the bottom. In order to do this both the top and the bottom of the fraction are multiplied by the conjugate of the botom. If there is only a root on the bottom you can simply multiply the top and bottom by the root on the bottom.

The conjugate of a surd is simply the surd with the sign of the coefficient of the root reversed so

a+b√c becomes a-b√c

So to rationalise the fraction

(2+√3)/(3+√4)

we multiply the top and bottom by 3-√4 to get

(3-√4)(2+√3)/(3+√4)(3-√4)

which once the brackets are expanded becomes

(6+3√3-2√4-√12)/5

as the √4 x √4 becomes √16 = 4 and the bottom is the difference of two squares.

By David Woodford

Categories: maths, surds