Proposition: √2 is an irrational number.

Proof:
We Know, 1 < 2 < 4
∴ √1 < √2 < √4
or, 1 < √2 < 2

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Here, 12 = 1, (√2)2 = 2, 22 = 4, (√2)2 = 2
∴ Therefore, the value of √2 is greater than 1 and less than 2.
∴ √2 is not an integer.

∴ √2 is either a rational number or a irrational number. If √2 is a rational number
Let, √2 =p/q; where p and q are natural numbers and co-prime to each other and q > 1
or, 2 = p2/q2; squaring
or, 2 = 2p2/q; multiplying both sides by q.

Clearly 2q is an integer but p2/q is not an integer because p and q are co-prime natural numbers and q > 1
Clearly 2q is an integer but p2/q2 is not an integer because p and q are co-prime natural numbers and q > 1

∴ 2q and p2/q2 cannot be equal, i.e., 2q≠ p2/q
∴ Value of √2 cannot be equal to any number with the form p/q , i.e., √2≠ p/q

∴ √2 is an irrational number. (Proved)
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