Prove that √2 is an irrational number.


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Answer:


Step by Step Explanation:
  1. Let's assume that √2 is a rational number.
    i.e. √2 = a/b (a and b are the two integers, such that, b is not equal to zero(0), and a and b do not have common factor other than 1)
  2. a = √2b
    Squaring both side,
    a2 = 2b2
    Therefore, a2 is divisible by 2 and it can be said that a is divisible by 2.
    Let a = 2m, where m is an integer.
    Squaring both side,
    a2 = (2m)2
    Now,
    (2m)2 = 2b2
    b2 = 2m2
    This means that b2 is divisible by 2 and hence, b is divisible by 2.
    This implies that a and b have a common factor and this is contradiction to the fact that a and b are a co-prime.
  3. Therefore, √2 is an irrational number.

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