prove √5 is a irrational number????
Answers
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.
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.