Question 992554
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Show that if x^2 is odd, x is odd. Use proof by contradiction.
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We are given that  {{{x^2}}}  is odd.

We need to prove that  x  is odd. 


Let us assume that  x  is not odd.  Then  x  is even.  In other words,  x  is divisible by  2.  Hence,  x = 2n,  where  n  is integer.


Then  {{{x^2}}} = {{{(2n)^2}}} = {{{4n^2}}}  is even.


This contradicts to the fact that  {{{x^2&}}}  is odd,  which is given. 


The source of the contradiction is the assumption that  x  is not odd. 


Hence,  x  is odd.


The proof is completed.