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I have to prove that there is no largest irrational number from the result of the a previous proof: Prove that if $x$ is rational and $y$ is irrational then, $x+y$ is irrational. I was able to prove this so I'll put it here.

Proof: Assume to the contrary that since $x$ is rational and $y$ is irrational then $x+y$ is a rational number $z$. Thus $x+y=z$, where $x=\frac{a}{b}$ and $z=\frac{c}{d}$ for some integers $a,b,c,d \in \mathbb{Z}$ and $b,d \neq 0$. This implies that $y=\frac{c}{d}-\frac{a}{b}=\frac{bc-ad}{bd}$. Since $bc-ad$ and $bd$ are integers and $bd \neq 0$, it follows that $y$ is rational, which is a contradiction.

Not sure how to go about proving that there is no largest irrational number.

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    $\begingroup$ Assume there is one, then add $1$... $\endgroup$ – abiessu Apr 19 '16 at 18:31
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Assume there is one. Add 1. QED.

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  • $\begingroup$ Proof: Suppose to the contrary that there is a largest irrational number. We can find m in I and s in I so s is less than or equal to m. We can then say that s is less than or equal to m+1. Hence there is a largest irrational number, m in I, which is a contradiction. $\endgroup$ – Matt Apr 19 '16 at 20:37
  • $\begingroup$ I was wondering how this sounded. $\endgroup$ – Matt Apr 19 '16 at 20:37
  • $\begingroup$ @Matt: Where do $s$ and $m$ come from? There's no proof that $s \leq m+1$: you just assume it. Your "contradiction" is not a contradiction at all, just the negation of the statement you were trying to prove - the exact same thing you assumed at the start! If I were a teacher I wouldn't give this any marks: it doesn't prove anything, only assumes the opposite and plays around with unrelated variables. $\endgroup$ – Deusovi Apr 19 '16 at 20:41
  • $\begingroup$ So I assume there is a largest irrational number, which I'll call m, and I meant to say y from the previous proof. I was trying to say we could assume from the previous proof that y<=m, and then add 1 to m? $\endgroup$ – Matt Apr 19 '16 at 20:49
  • $\begingroup$ @Matt: Proofs should generally be kept entirely separate. For instance, $y$ is literally just "any irrational number"; there's no reason to refer to a variable from a previous proof when you don't have to. Also, you haven't completed the proof yet. Are you going for a proof by contradiction? If so, you actually have to derive a contradiction. $\endgroup$ – Deusovi Apr 19 '16 at 21:19
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Hint: $n+\sqrt 2$ is irrational for all $n \in \mathbb N$.

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A direct proof is probably simplest.

Given any irrational number $\alpha,$ we have by the previous result that $\alpha+1$ is also irrational. Since $\alpha<\alpha+1,$ then $\alpha$ cannot be the greatest irrational number. Since $\alpha$ was an arbitrary irrational number, then there is no such thing as "the greatest irrational number" at all.

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