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I'm just wondering if there exists proofs that certain numbers are irrational that do not begin by saying some like along the lines of "assume $k=a/b$ for integers $a$ and $b$" and then deduce a contradiction. My question is, are there any irrationality proofs out there that somehow prove irrationality directly or do they all somehow utilize proof by contradiction?

Bart Michels
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ASKASK
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    You need a "direct definition" of irrationality for that. An irrational number is (by definition) a number which is not rational, so in order to prove that a number is irrational you must prove it's not rational. – Daniel Apr 04 '15 at 01:25
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    Another example is the proof by contradiction of Euclid's theorem that there are infinitely many primes, which deduces a contradiction from the hypothesis that there are finitely many primes. So an analogous question to yours is: can you prove that a set is infinite without a proof by contradiction by assuming that it is finite? If infinite $:=$ not finite, like irrational $ :=$ not rational, then you may face the same hurdles. – Bill Dubuque Apr 04 '15 at 02:14

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Consider the polynomial $x^2-2$, by Eisenstein's criterion this polynomial is irreducible over $\mathbb{Q}$. It follows that the extension degree of $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ is $2$, hence strict. This shows that $\sqrt{2}\notin \mathbb{Q}$.

You can go through the proofs of Eisenstein's criterion and the other theory involved, we nowhere use that $\sqrt{2}$ is irrational.

Mathematician 42
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Comment: I remember coming across a real neat generalized way of proving the irrationality of $\sqrt{a}$ in the American Mathematical Monthly (credit due to Amrik Singh Nimbran). Maybe the following will be something along the lines of what you are looking for (Daniel Escudero's comment makes a good point though):

Generalized proof: Let $a$ be a square-free positive integer and $[\sqrt{a}]=b$. where $[\;]$ denotes the greatest integer contained in the square root. Then $b^2<(\sqrt{a})^2<(b+1)^2$. Thus, $a$ lies between two consecutive squares; hence, $\sqrt{a}$ is not an integer. Suppose that $\sqrt{a}$ is rational, i.e., $\sqrt{a}=\frac{m}{n}, 1<n<m<(b+1)n; m,n\in\mathbb{N}$. Using a division algorithm, we may write the following: $m=bn+r, n>r, r\in\mathbb{N}$. Hence, $\sqrt{a}=\frac{bn+r}{n}$. Let $g=(n,r)$. So, $n=gs$ and $r=gt, s>t, (s,t)=1; g,s,t\in\mathbb{N}$. Then $\sqrt{a}=\frac{bs+t}{s}$. Hence $$ (a-b^2)s^2=t(2bs+t).\tag{1} $$ Since $t\mid(t(2bs+t))$ and $(s,t)=1$, either $(a-b^2)=tu, u\in\mathbb{N}$, or $t=1$. In the first case, we would have $s(us-2b)=t\Rightarrow s\mid t$, which is impossible for $s>t$. In the second case, that is, $t=1$, we obtain $$ s\{(a-b^2)s-2b\}=1.\tag{2} $$ Equation $(4)\Rightarrow s\mid 1\Rightarrow s=1$, which contradicts our supposition that $s>t=1$. Further, it leads to $a=(b+1)^2$, contradicting our supposition $a<(b+1)^2$. Hence, $\sqrt{a}$ is irrational. $\blacksquare$

Daniel W. Farlow
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    This is a nice elegant proof, but you still do say "suppose that $\sqrt(a)$ is irrational". I'm starting to think that what I'm looking for is inherently impossible – ASKASK Apr 04 '15 at 02:09
  • @ASKASK That is why I prefaced the proof with a comment--to let you know that the assumption/contradiction will still be used. Perhaps I could have been clearer, but I thought maybe it would still prove somewhat useful as it is somewhat different than how most proofs start. – Daniel W. Farlow Apr 04 '15 at 02:11
  • Yeah I don't think it's possible. If you explicitly state and explain this (rephrase the comment) then I'll give you the check – ASKASK Apr 04 '15 at 02:13
  • @ASKASK I can rephrase it, but it is really due to Daniel's comment. I simply wanted to provide a nice way of going about proving the irrationality of a number in general that has a bit of a different "flavor" than what you generally encounter. An answer check may not be appropriate since my answer really doesn't answer your question. – Daniel W. Farlow Apr 04 '15 at 02:16
  • That's not an answer to the question. There are much shorter ways to prove irrationality of sqrts using only the division algorithm, e.g. this one, which could be shortened further if everything was inlined (at the cost of obscuring conceptual structure). – Bill Dubuque Apr 04 '15 at 02:26
  • @ASKASK Here is a another shorter proof. $\tag*{}$ ${\bf Theorem}\ \ $ If $\ x = \sqrt{n}\ $ is is rational then it is an integer. – Bill Dubuque Apr 04 '15 at 04:23
  • $\begin{align}& {\bf Proof}\ \ {\rm Deny,\ ,so},\ \ \overbrace{ x = \frac{a}b}^{\Large\color{#c00}{bx, =, a}},\ \ {\rm and},\ \ \color{#b0f}{b\nmid a},\ {\rm by},\ x\not\in\Bbb Z. {\rm \ wlog},\ b\ ,{\rm is}\ ,\color{#0a0}{\rm least}\ {\rm denom}\ &\color{#c00}ax, =, (\color{#c00}{bx})x =, bn.\ \ a = qb+r,,\ \color{#b0f}{0 < r} < b\ \ \text{by the Division Algorithm}\ &rx! =! (a!-!q\color{#c00}b)\color{#c00}x! =! bn!-!q\color{#c00}a =: k,\Rightarrow, x = \frac{k}r,,\ \text{contra },b,\text{ is the $\color{#0a0}{\rm least}$ denom}\end{align}$ $\tag*{}$ – Bill Dubuque Apr 04 '15 at 04:23