Question

Q.9 Show that 5 √3 is an irrational number.​

Answer 1

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

Let us assume that 5√3 be a rational and we will write the given number in p/q form.

=> 5√3 = p/q

=> √3 = p/5q

​

Here p/5q is rational number but √3 is irrational number (proved above).

Thus LHS is not equal to RHS.

We observe that LHS is irrational and RHS is rational, which is not possible.

This is contradiction.

Hence our assumption that given number is rational is false

⇒5√3 is irrational number.