Proof Question: Prove integer + 1/2 is not an integer

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I was in the middle of proving something when I reached a contradiction, that .5 + an integer = an integer. However, this cannot be true, and I'm curious if its acceptable to just say that by definition of integers .5 + an integer is not an integer, or do I have to prove it?
Furthermore, if I have to prove it, how would I go about this? I would say let x and y be integers, so x + .5 = y, right?
Since x and y are integers then x = x/1 and y = y/1, so x/1 + 1/2 = y/1.
2x/2 + 1/2 = y/1
so
(2x + 1/2)/2 = y/1
and then... If I said that 2x +1/2 was not a whole number so dividing it by two must give a fraction, and thus it can't be reduced to a whole number over 1... That doesn't sound like it works though becuase its just restating what I was trying to prove... Not to mention I'm not sure I can even say that a fraction divided by two doesn't give a whole number... Any ideas? Thanks.
 
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Are you saying that

\frac{(2x+\frac{1}{2})}{2}=\frac{y}{1}

shows that \frac{1}{2}+n1=n2?

n is an arbitrary integer.

(there's a problem with this post. unwanted spacing)
 
There is a MUCH quicker way.
supose 0.5+N=M where N and M are integers. Then 0.5=M-N. But if M and N are integers, M-N is an integer. But this implies 0.5 is an integer. This is a cointradiction. Done.
 
How do you figure Setting x=x/1 forces x to be an integer?
 
To show 1/2 is not an integer, use its definition:

1/2 is the number that satisfies the equation 2x=1.

Now, for any integer n, 2n is never 1 (why? because 2n is always even and 1 is odd, and no integer is both even and odd.) Hence, 1/2 cannot be an integer.

Of course, to argue that no integer is both even and odd uses the quotient-remainder theorem, which in turn relies on the well-ordering principle of the positive integers.
 
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