Question

prove √5 is a irrational number????​

Answer 1

Answer- The above question is from the chapter 'Real Numbers'.

Given question: Prove that √5 is an irrational number.

Solution: Let us suppose that √5 is a rational number.

√5 = p/q (where p,q are co-primes, q≠0)

Transposing q to L.H.S., we get,

√5q = p

Squaring both sides, we get,

5q² = p² --- (1)

5 is a factor of p².

Using Fundamental Theorem of Arithmetic, we get,

5 is also a factor of p.

Put p = 5c.

Put the value of p in equation 1, we get,

5q² = 25p²

⇒ q² = 5p²

5 is a factor of q².

Using Fundamental Theorem of Arithmetic, we get,

5 is also a factor of q.

⇒ √5  is a rational number.

This contradicts the statement that p and q are co-primes and √5 is a rational number.

∴ We arrive at a wrong result due to our wrong assumption that √5  is a rational number.

∴ √5 is not an irrational number.

Answer 2

Let us assume that √5 be a rational number .

So , √5 = p/q where q≠ 0 and p and q are co - prime nunbers .

Hence , √5 q = p

Squaring both the sides

(√5q)² = p²

5q² = p²

q² = p²/5

This shows that p is divisible by 5 .

Let p = 5r

q² = (5r)² /5

q² = 25r² /5

q² = 5r²

q² /5 = r²

This shows that q is divisible by 5 .

Hence , both p and q are divisible by 5. This contradicts our earlier supposition that p and q are co - prime numbers and our consideration of √5 as rational is wrong.

Hence √5 is irrational number.