Question

4. Prove that
$5 \sqrt{3} \: is \: irrational , \: using \: the \: \\ fact \: that \sqrt{3 } \: is \: irrational.$

Standard:- 10

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Answer 1

❣️Hi there ❣️

Suppose 5√3 is rational number.

$5\sqrt{3} = \frac{p}{q}$

And q≠0 pand q are integers.

$\sqrt{3} = \frac{p}{5q} - - - - (1)$

Therefore , p and q are integers and q≠ 0

p and 5q are integers and 5q ≠ 0

$\frac{p}{5q} is \: a \: rational \: no$

It is contradicting with the fact that √3 is an irrational no.

Now , I have to prove that root 3 is an irrational no.

Suppose √3 is a rational no.

√3= p/q ,where p≠ q are integers -----(1)

p = √3q

p²= √3 q² ----(2)

3 is a factor of p²
3 is a factor of p---(3) too (if a prime is a factor of p² then it also a factor of p )

p = 3k

p²=3k

3q² = 9k ²

q²=3k²

3 is a factor of q²
3 is a factor of q too ---(4)

now (3) and (4)

3 is a common factor of p and q

p and q are not coprime

It is contradicing with eqn one

So that our assumption is wrong hence root 3 is an irrational no.




So that our assumption is False.

Hence 5√3 is an irrational no.

Thanks !