Question

ample 9 : Prove that
$\sqrt{3}$

is irrational.​

Answer 1

Answer:

Say 3–√ is rational. Then 3–√ can be represented as ab, where a and b have no common factors.

So 3=a2b2 and 3b2=a2. Now a2 must be divisible by 3, but then so must a (fundamental theorem of arithmetic). So we have 3b2=(3k)2 and 3b2=9k2 or even b2=3k2 and now we have a contradiction.

Answer 2

Answer:

hii dear...,..

Step-by-step explanation:

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