Prove that 7+3√2 is irrational
Answers
Answer:
This contradiction arise due to our wrong assumption. Hence, 7+3√2 cannot be a rational number, ∵ it is an irrational number.
Answer:
let us assume that 7+3√2 is rational
7+3√2=p/q(p and q are coprimes)
3√2=p/q-7
√2=p-7q/3q
for this we need to see if √2 is rational or not
let's assume that √2 is rational
√2 = p/q
Here p and q are coprime numbers and q ≠ 0
Solving
√2 = p/q
On squaring both the side we get,
=>2 = (p/q)2
=> 2q2 = p2……………………………..(1)
p2/2 = q2
So 2 divides p and p is a multiple of 2.
⇒ p = 2m
⇒ p² = 4m² ………………………………..(2)
From equations (1) and (2), we get,
2q² = 4m²
⇒ q² = 2m²
⇒ q² is a multiple of 2
⇒ q is a multiple of 2
Hence, p,q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
√2 is an irrational number.
hope this helps