Question

prove that
$\sqrt{2}$
irrational number


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Answer 1

Answer:

1. Assume that √2 is a rational number, meaning that there exists a pair of integers whose ratio is √2.

2. If the two integers have a common factor, it can be eliminated using the Euclidean algorithm.

3. Then √2 can be written as an irreducible fraction  a / b  such that a and b are coprime integers (having no common factor).

4. It follows that  a^2  /  b^2  = 2 and a^2 = 2b^2.   ( ( a / b )^n =  a^n / b^n   )  

( a^2 and b^2 are integers)  

5.Therefore, a^2 is even because it is equal to 2b^2. (2b^2 is necessarily even because it is 2 times another whole number and multiples of 2 are even.)

6. It follows that a must be even (as squares of odd integers are never even).

7. Because a is even, there exists an integer k that fulfills: a = 2k.

8. Substituting 2k from step 7 for a in the second equation of step 4:

2b^2 = (2k)^2 is equivalent to 2b^2 = 4k^2, which is equivalent to b^2 = 2k^2.

9. Because 2k^2 is divisible by two and therefore even, and because 2k^2 = b^2, it follows that b^2 is also even which means that b is even.

10. By steps 5 and 8 a and b are both even, which contradicts that  a / b

is irreducible as stated in step 3.

This means that √2 is not a rational number; i.e., √2 is irrational.