What is wrong with my proof by contradiction? 
There exist no integers $a$ and $b$ for which $18a+6b=1$.

Proof: Assume that $18a+6b=1$. We find that
$$6(3a + b)=1$$
which leads to
$$3a+b=\frac16$$
We know that the sum of two integers can't produce a non-integer result, therefore a contradiction arises, as the proof demonstrates that two integers can produce a non-integer value. $\blacksquare$
My professor said that if one ends up with fractions in a proof, there is likely a problem. Can someone explain why this is the case?
 A: I'm assuming you're in a number theory class or abstract algebra. At this level of such a course, we haven't formally reintroduced $\Bbb Q $ so fractions don't formally exist yet.
We have multiplication and addition. And we have the integers.
The better proof is to show that $\gcd(18,6)=6$ and hence that the smallest positive linear combination of $18$ and $6$ we can make is $6$.
A: The core idea of your argument looks fine to me. 
I suspect what your professor had in mind, is that you should not write $3a + b = \frac{1}{6}$. You should simply say that 
$$
6(3a + b)=1
$$
is not possible in integers, since $6$ is not an invertible element in $\mathbb{Z}.$
A: $18a+6b=2\cdot(9a+3b)$ is an even number, so it cannot equal $1$ which is an odd number.
A: Well here I think your professor means that if you add two integers, you should never get a number that is not an integer. 
This is because the integers are "closed" under addition (and also multiplication). Mathematically speaking, any two elements of the integers say $a$ and $b$ can be combined as $a+b$ or $ab$ and still only give a result that is an integer.
Side note: why might the integers not be closed under division?
A: Well, all the reasoning is sound here, but, actually, one might ask why exactly $\frac 1 6$ is not integer? To properly answer this, one needs to know how rational numbers are defined. Well, let us assume that $\frac 1 6$ is integer. That means that there is integer $k$ such that $\frac 1 6 = \frac k 1$. From the definition of rationals, this happens if and only if $6k = 1$ (as you can see, we are back to your original question). To finish argumentation, one argues that $6$ does not divide $1$ in $\Bbb Z$ to get contradition, so $\frac 1 6$ can't be integer.
A: Try using the fact that sum , difference of even numbers produces even numbers only. $18a, 6b$ are even. I can't say exactly what he meant but I guess  he meant using only properties explained till that moment.
A: Your proof is correct.
However, it's a bit too complicated.
You could just show the left-hand side is even for all integer $a$ and $b$, while RHS is always odd, so they are never equal.
