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.
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.