prove that root2 is irrational
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Let √2 = --( a and b are co-prime & b ≠ 0 )
Then,
√2 =
⇒ 2 =
⇒ 2b² = a²
Since, a² is divisible by 2
Therefore,
Now,
Let a = 2c
Squaring both sides :
a² = 4c²
⇒ 2b² = 4c²
⇒ b² = 2c²
Since, b² is divisible by 2
Therefore,
That means,
‘a’ are ‘b’ are divisible by 2 .
But, ‘a’ and ‘b’ were co-prime!
This contradiction arised due to our wrong assumption that √2 is Rational
...
Cheers!
Answer:
here's Ur answer buddy
√2 = p/q
2 = (p/q)^2
2 = p^2/q^2
p^2 = 2q^2
thus p^2 is even. the only way this can be true is that p itself is even . but then p^2 is actually divisible by 4 . hence q^2 and therefore q must be even . so p and q both are even which is a contradiction to our assumption that they have no common factors. the square root of 2 cannot be rational.
hope it helps u
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