Question

Prove that √3 is not a rational number


with clear steps

Answer 1

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HERE IS YOUR ANSWER!

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )


So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )


=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q


Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number



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Answer 2

↑ Here is your answer ↓
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Lets assume that √3 is rational

Then

$\sqrt 3 = \dfrac{p}{q}$

{ 'p' and 'q' are co - primes, q not equal to 0 }

$\sqrt 3 q = p$

Squaring both sides

$3q^2 = p^2$

'3' divides ' p² ' so '3' divides ' p ' (Fundamental theorem of arithmetic) [1]

Now, put ' p = 3m '

$3q^2 = (3m)^2$

$3q^2 = 9m^2$

$q^2 = 3m^2$

'3' divides ' q² ' so '3' divides ' q ' (Fundamental theorem of arithmetic) [2]

From (1) and (2), we come to know that 

'3' divides both 'p and q'

Hence, 'p and q' are not co - primes

Therefore, our assumption is wrong

=> √3 is irrational