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How would one go about proving that $$0<\frac{n}{n+1}<1$$ by mathematical induction?

If $p(n)$ is the statement as above, then I know we show $p(1)$, and assume $p(n)$, but in this particular case I am not sure know to show $p(n+1)$.

user138999
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  • That's not a very interesting thing to do by induction, basically $n$ is always less than $n+1$, QED... Where did the problem come from? – Gregory Grant Apr 30 '15 at 13:32
  • Just a problem my tutor gave me, I now see why it's so easy ... Surprised I was so confused by it. – user138999 Apr 30 '15 at 13:34

3 Answers3

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Hint:

Prove by induction that $0<n<n+1$ for each positive integer $n$.

drhab
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For the inductive step, we need to show that $$0<\frac{n}{n+1} < 1 \implies 0 < \frac{n+1}{n+2}<1$$ We do this in two parts:

  1. Since $n>0$, we trivially have $\frac{n+1}{n+2}>0$ as a ratio of two positive integers.

  2. We can rewrite the right-hand side of our inductive hypothesis as $$n < n+1$$ (As above, $n+1>0$ so we don't need to worry about signs.)

    It’s easy to get from here to $n+1 < n+2$, and then dividing by $n+2$ gives $\frac{n+1}{n+2}<1$.

We conclude that $$0<\frac{n+1}{n+2}<1,$$ as required.

alexwlchan
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One more way for the last step: notice that $\frac{n+1}{n+2} = \frac{n+2-1}{n+2} = 1-\frac{1}{n+2}<1 \ \forall n$ because the $\frac{1}{n+2}$ term is strictly positive.

Alex
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