Prove that √2 is an irrational number.
Last Updated :
08 Mar, 2024
The √2 is an irrational number because it cannot be expressed as a fraction of two integers and does not terminate or repeat in its decimal representation.
The proof that √2 is an irrational number typically follows a proof by contradiction.
Assume, for the sake of contradiction, that √2 is rational.
So, we have . Squaring both sides, we get . Rearranging, we find a2=2b2.
This implies that a2 is even, and consequently, a must also be even (since the square of an odd number is odd). Let a = 2k, where k is another integer.
Substituting this back into a2 = 2b2, we get 4k2 = 2b2, and simplifying gives b2 = 2k2.
Now, this implies that b2 is also even, and therefore b must also be even.
However, if both a and b are even, they have a common factor of 2, which contradicts our initial assumption that a and b have no common factors (other than 1).
This contradiction arises from assuming that √2 is rational, leading to the conclusion that √2 must be irrational. Therefore, √2 cannot be expressed as a fraction of two integers, confirming its irrationality.
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