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Prove the following by mathematical induction

$n < 2n$, for all positive integer $n$.

This is what I have done:

Step 1: $n=1$: $1 < 2$

Step 2: $k < 2k$

$n=k+1$: $(k+1) < 2(k+1)$

$k + 1 < 2k + 1 < 2k + 2 = 2(k+1)$

Hence $P(k+1)$ is true whenever $P(k)$ and since $P(1)$ is true.

I didn't write all necessary assumptions but can anyone help me to check if my method is correct or if it needs improvements. Thank you.

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  • $\begingroup$ The argument looks fine but the presentation could use some improvement. $\endgroup$ – blub Mar 31 at 12:35
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    $\begingroup$ It seems correct. Just try to be more explicit in your steps by saying. "this is our induction hypothesis"... "this is what we'll prove"... $\endgroup$ – Bruno Reis Mar 31 at 12:37
  • $\begingroup$ What is your definition of "$<$"? $\endgroup$ – Henning Makholm Mar 31 at 12:38
  • $\begingroup$ when you say $k+1<k+k$, you should stipulate $k>1$ $\endgroup$ – J. W. Tanner Mar 31 at 12:39
  • $\begingroup$ For $$k=1$$ the proof was given by the OP $\endgroup$ – Dr. Sonnhard Graubner Mar 31 at 12:42
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We have to prove that $$n+1<2(n+1)$$ , adding $1$ on both sides of $$n<2n$$ we get $$n+1<2n+1$$ and $$2n+1<2(n+1)$$ so the proof is finished.

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