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Maths-

General

Easy

Question

Show that $square root of 3$ is an irrational number

hint Hint:

Hint:
The real numbers which cannot be expressed in the form of p/q(where p and q are integers and q ≠ 0) are known as irrational numbers. In the given question we are asked to prove if $square root of 3$ is irrational number. To do so we use the method of contradiction.

The correct answer is: Is an irrational number.

Let's assume that $square root of 3$ is a rational number
¶Step 1 of 2:
¶If $square root of 3$ is rational, that means it can be written in the form of p/q, where p and q integers that have no common factor other than 1 i.e. they are coprime than 1 and q ≠ 0.
¶So, $table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell fraction numerator square root of 3 over denominator 1 end fraction equals p over q end cell row cell square root of 3 q equals p end cell end table$
¶Squaring both sides
¶ $3 q squared equals p squared$ .......(1)
¶ $q squared equals p squared over 3$
¶This means p² can be divided by 3
¶Step 2 of 2:
¶So if p2 can be divided by 3, then can also be divided by 3
¶ So, let’s say $table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell p over 3 equals r end cell row cell p equals 3 r end cell end table$
¶Squaring both sides
¶ $p squared equals 9 r squared$
¶Putting the value of p² in equation (1)
¶ $table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 3 q squared equals 9 r squared end cell row cell q squared equals 3 r squared end cell row cell q squared over 3 equals r squared end cell end table$
¶So, this means q² is divisible by 3 and so q is also divisible by 3.
¶Final Answer: p and q both are divisible 3 but we assumed that p and q are coprime and this contradiction has arisen because of our incorrect assumption that $square root of 3$ is a rational number. So, we conclude $square root of 3$ that is an irrational number.

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