Question

17] Prove that
$\sqrt{3 | } - \sqrt{2}$
is an Irrational number.​

Answer 1

Answer:

method of contradiction

let the given no be a rational no.

but root3 is an irrational number

hence our consumption is wrong

the given no. is an irrational no.

hence proved

Answer 2

Step-by-step explanation:

R.T.P: $\sqrt{3} - \sqrt{2}$

Proof: consider $\sqrt{2}$

assume $\sqrt{2}$ is rational i.e. it can be expressed in the form of p/q such that p and q are co-primes and q≠0

therefore, $\sqrt{2}$=p/q

squaring on both sides

2=p²/q²

2q²=p²

here, p² is divisible by 2 so p is also divisible by 2 (Gauss' theorem)

so, p can be expressed in the form of another integer r such that

p=2r

squaring on both sides

p²=4r²

sub p²=2q²

2q²=4r²

q²=2r²

here,q² is divisible by 2 so q is also divisible by 2 (Gauss' theorem)

p and q have a common factor so our assumption is false so, $\sqrt{2}$  is irrational

similarly, $\sqrt{3}$ is also irrational

irrational - irrational = irrational

therefore, $\sqrt{3} - \sqrt{2}$ is irrational

hence, proved