# Prove by mathematical induction $n < 2n$

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.

• The argument looks fine but the presentation could use some improvement. – blub Mar 31 at 12:35
• 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"... – Bruno Reis Mar 31 at 12:37
• What is your definition of "$<$"? – Henning Makholm Mar 31 at 12:38
• when you say $k+1<k+k$, you should stipulate $k>1$ – J. W. Tanner Mar 31 at 12:39
• For $$k=1$$ the proof was given by the OP – Dr. Sonnhard Graubner Mar 31 at 12:42

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.