Step-by-step explanation:

Let √3 be rational.

√3=p/q. p &q =Z where q is not equal to 0.

HCF(p,q)=1

√3q=p

Squaring on both side.

3q²=p²-----(1)

q²=p²/3

3 divides p² so it also divides p.

P=3k

Squaring on both side

p²=9k²

3q²=9k² (from (1)

q²/3=k²

3 divides q² so it also divides q.

p and q have at least 3 as common factor and are not coprime.

If p/q is not rational then √3 is also Irrational.(√3=p/q.)----(2)

Now, 5-√3 be rational.

5-√3=p/q

5-p/q=√3.

5q-p/q=√3.

√3 is irrational. (from (2)

So LHS is also irrational.

Hence, 5-√3 is irrational.

To prove: √3 in irrational

Proof:

Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p²/q² (Squaring on both the sides)

⇒ 3q² = p² →→→ (1)

It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p² = 9r² →→→ (2)

from equation (1) and (2)

⇒ 3q² = 9r²

⇒ q² = 3r²

Where q² is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

To prove:

Proof:5-√3 is irrational

Let us assume the given number be rational and we will write the given number in p/q form

⇒5 - √3 = p/q

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