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Yogita Ingle 5 years, 6 months ago
Let us consider that √5 is a “rational number”.
We were told that the rational numbers will be in the “form” of \frac {p}{q}form Where “p, q” are integers.
So, \sqrt { 5 } = \frac {p}{q}
p = \sqrt { 5 } \times q
we know that 'p' is a “rational number”. So 5 \times q should be normal as it is equal to p
But it did not happens with √5 because it is “not an integer”
Therefore, p ≠ √5q
This denies that √5 is an “irrational number”
So, our consideration is false and √5 is an “irrational number”.
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