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As part of a proof I'm writing I want to prove that for any integer $n>0$, $\frac{2n+1}{n+1}$ is not an integer.

I'm stuck on how to go about proving this -- I understand why it's true in my head but struggle to put it into words. I'll give it a go, though:

$\frac{2n}{n}$ is clearly an integer, as any number $n$ doubled can obviously be divided by itself, always giving back $2$. If you look at the next $n$, this would be also true for $\frac{2(n+1)}{n+1}$ as the $+1$ part is also doubled and can therefore be divided safely. $\frac{2n+1}{n+1}$ gives you something in the middle of $1$ and $2$, and is therefore not an integer.

This seems clunky and badly formed. Also seems like possibly a proof by induction could be coming out of it? Although, that seems a little overkill.

Would greatly appreciate some pointers. Thanks.

As a side note: I know this may not be able to be answered generally as it probably depends on the context of the entire proof, but would something like this generally need to be proved? Or would it suffice to say "$\frac{2n+1}{n+1}$ is not an integer for any $n$" and leave it at that?

werner_b
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  • You should not leave anything without proof, even if it seems too trivial. As for the proof, your argument is correct. Just try to write it more precisely. – Aniruddha Deshmukh Oct 25 '21 at 06:34
  • @AniruddhaDeshmukh will do, thank you – werner_b Oct 25 '21 at 06:37
  • @SaucyO'Path do you have a counterexample? I can only think of $n=0$ which thankfully isn't really applicable for my greater proof. Should have specified $n > 0$ though. – werner_b Oct 25 '21 at 06:37
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    @werner_b The other counterexample was $n=-2$, which also doesn't apply anymore. Is there any reason you cannot simply write is as the following? $$\frac{2n+1}{n+1} = 2 - \frac{1}{n+1}$$ – Brian Moehring Oct 25 '21 at 06:41
  • @BrianMoehring there's no reason I couldn't, that does make it a lot clearer to see that you'll get some non-integer from that though. I'll try work it in, thanks. – werner_b Oct 25 '21 at 06:47

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I would recommend that if you are ever stuck trying to prove some statement, then you somehow make the statement more concrete. For instance, with this problem you can start by plugging in concrete values for $n$.

  • $n = 1$ gives $\frac{2n + 1}{n+1} = \frac{3}{2} = 1.5$
  • $n = 2$ gives $\frac{2n + 1}{n+1} = \frac{5}{3} = 1.666\ldots$
  • $n = 3$ gives $\frac{2n + 1}{n+1} = \frac{7}{4} = 1.75$
  • $n = 4$ gives $\frac{2n + 1}{n+1} = \frac{9}{5} = 1.8$
  • $n = 5$ gives $\frac{2n + 1}{n+1} = \frac{11}{6} = 1.8333\ldots$

These values seem to indicate that $1 < \frac{2n + 1}{n+1} < 2,$ which if true, would imply that $\frac{2n + 1}{n+1}$ is not an integer. The next step is to see if you can prove this new claim. To start, check if $\frac{2n + 1}{n+1} < 2$ is true: $$\frac{2n + 1}{n+1} < 2 \iff 2n + 1 < 2(n+1) \iff 2n + 1 < 2n + 2$$ where the right hand side (and so also the left hand side) is true, which proves that $\frac{2n + 1}{n+1} < 2.$ Then check that $1 < \frac{2n + 1}{n+1}$: $$1 < \frac{2n + 1}{n+1} \iff n + 1 < 2n + 1$$ where the right hand side is true.

In more abstract fields like topology or abstract algebra, you might be able to instead substitute some specific concrete object. You can find concrete examples in textbooks or online (e.g. this Wikipedia list for topologies). For example:

  • In topology, consider replacing a general topological space with $\mathbb{R}^n$, the unit circle, the Cantor set, etc.
  • In group theory, consider replacing some general group with the dihedral group of order 8.
  • In linear algebra/functional analysis, consider replacing an arbitrary linear map with rotation by $90^{\circ}$ around the $z$-axis in $\mathbb{R}^3.$ Or replace an arbitrary linear functional with the map $(x, y) \mapsto a x + b y$ from a $2$-dimensional space, where you can optionally make this even more concrete by replacing $a$ and $b$ with $1$ and $1$ (or with $1$ and $0$, or with $-2$ and $1$, etc.).
mgkrupa
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