I am doing problems that involve inequalities. My understanding is that you through a string of inequalities show that one is less than the other. Kind of like the transitive property. For example:

$2n+1 \lt 2^n$ for $n=3,4,...$

Assume this is true for P(k):

Base case $k = 3$

$LHS: 7 \space \space \space RHS: 8$

$LHS \space \space \lt \space \space RHS$

This holds true for $(k)$

Prove for P(k+1):

$2(n+1)+1 \lt 2^{n+1}$

my understanding is that if I can find something that is greater than $2(n+1)+1$ and obviously below $2^{n+1}$ then the inequality holds true

$2k+3 \lt 2^{k+1}$

$(2k + 1) + 2 \lt 2^k2$

$(2k+1) + 2 \lt 2^k + 2$ by our hypothesis on $k$

It is very obvious that $2^k + 2 \lt 2^k2$ and doesn't need explanation so its safe to assume

$2k+3 = (2k+1)+2 \lt 2^k + 2 \lt 2^k2$

Thus $(2k+3) \lt 2^k2$

This seems perfectly logical to me since we are dealing with integers. These inequalities would not hold for $\mathbb{R}$.

I am having a hard time find a string of inequalities that proves $n^2 \leq 2^n+1$

Assume this holds true for $k$. $P(k) = k^2 \leq 2^k+1$ for $n = 1,2...$

Base case $P(1):$ $LHS: 1 \space \space \space RHS: 3$

$LHS \space \space \leq \space \space RHS$

holds true for $P(k)$


$(k+1)^2 \leq 2^k + 1$

$k^2 + 2k + 1 \leq 2^k2+1$

$k^2 + 2k + 1 \leq 2^k +1 + 2k +1 \leq 2^k2+1$

$k^2 + 2k + 1 \leq 2^k + 2k + 2 \leq 22^k+1$ by our hypothesis on $k$

I do not know where to go from here

  • 2
    $\begingroup$ You say, "It is very obvious that $2^k+2<2\cdot2^k$ and doesn't need explanation..." but I highly recommend including even trivial steps into your work. Many teachers would see a line like this and would question if the student knew how to prove it (sometimes the obvious things are the hardest to prove). $\endgroup$
    – Clayton
    Jul 29, 2019 at 19:33
  • $\begingroup$ Why did you bring up $\mathbb R$ when you said "This seems perfectly logical to me since we are dealing with integers. These inequalities would not hold for R"? Why bring up $\mathbb R$ at all? I also don't really understand what you mean by "string of inequalities". So I don't really understand what you are asking. $\endgroup$
    – fleablood
    Jul 29, 2019 at 22:10
  • $\begingroup$ Why are you trying to show $n^2<2^n+1$? Was that the question? What was the question actually? Anyway; you've proven $2^n > 2n+1$; if if $k^2 < 2^k +1$ then $(k+1)^2 = k^2 + 2k + 1 < (2^k +1)+(2k+1) < (2^k+1) + 2^k= 2^{k+1} + 1 $. $\endgroup$
    – fleablood
    Jul 29, 2019 at 22:49

3 Answers 3


You assume that $k^2\leq 2^k+1$ for some $k\geq 1$, and you want to prove that $(k+1)^2\leq 2^{k+1}+1$. Starting from the left-hand-side, you can proceed as follows:

\begin{align*} (k+1)^2 &= k^2+2k+1\\ &\leq (2^k+1) + 2k + 1, \end{align*} where the inequality follows from the induction hypothesis. If we can now show that

$$(2^k+1) + 2k + 1 \leq 2^{k+1},$$ then we we are done. By rearranging, this amounts to showing (for $k\geq 1$) that $$2k+2\leq 2^{k+1} - 2^k,$$ or

$$2(k+1)\leq 2^k(2-1),$$


$$2(k+1)\leq 2^k,$$


$$k+1\leq 2^{k-1}.$$

Oops! This last inequality is not true for all $k\geq 1$ like I hoped it would be. Don't worry about this, it happens. It doesn't mean that the original statement is wrong.

The last inequality fails for $k=1$ and $k=2$. But it does hold for all $k\geq 3$. I can remedy the proof by checking the base cases $k=1$, $k=2$, and $k=3$ in $k^2\leq 2^k+1$ by hand. Then I assume the induction hypothesis $k^2\leq 2^k+1$ for some $k\geq 3$, and I can proceed as above.

  • $\begingroup$ UGH I cant believe I didnt think to subtract it over!! Thanks! $\endgroup$
    – K. Gibson
    Jul 29, 2019 at 21:58

In the first case, the proof is quasi-automatic. You want to prove

$$2n+1<2^n\implies 2n+3<2^{n+1},$$ or $$(2n+1)+2<2\cdot2^n.$$

Using the LHS,

$$(2n+1)+2<2^n+2$$ and we need to wonder when $$2^n+2\le2\cdot2^n.$$ This simplifies as $$2\le2^n$$ or $$n\ge1.$$


Now regarding the second case, we want to show

$$n^2<2^n\implies (n+1)^2<2^{n+1}$$ or $$n^2+2n+1<2\cdot 2^n.$$

By hypothesis

$$n^2+2n+1<2^n+2n+1$$ and we want

$$2^n+2n+1\le2\cdot 2^n$$ or $$2n+1\le2^n.$$

By a simple modification of the first proof, we can establish this result for $n\ge2$.


You can not use induction on $\mathbb R$ as the induction step will prove if it is true for $x$ then it is true for $x+1$ but there is nothing that will assure that it is true for any $k; x < k < x+1$.

Are you trying to prove $2x + 1 < 2^x$ for all real $x\ge 1$? If so do the following.

Use induction to prove that for $n\in \mathbb N;n\ge 3$ that $2n+1 < 2^n$.

Then prove that for any $x \in \mathbb R$; that if $n < x < n+1$ then $2n+1 < 2x+1 < 2(n+1)+1$ and $2^n < 2^x < 2^{n+1}$ so then only way that $2^x \le 2x+1$ could be possible is if $2(n+1)> 2^n$.

Why can prove that is impossible by induction by proving that $2(n+1) \le 2^n; n\ge 3$ and $n\in \mathbb N$.


Claim 1: $2(n+1) \le 2^n$ if $n\in \mathbb N; n\ge 3$.

Proof by Induction:

Base case: $k=3$ then $2(3+1) = 2^3$.

Induction case: If $2(k+1) \le 2^k$ then

$2(k+2)= 2(k+1) + 2 \le 2^k+2 \le 2^k + 2^k = 2^{k+1}$.

Thus we have proven so for all natural $n> 3$ that $2(n+1)\le 2^n$

So as $2a+1 < 2(n+1)=2n + 2\le 2^n$ we have proven $2n+1 < 2^n$ for all natural $n \ge 3$.

Now if $x\in \mathbb R; x \ge 3$ then there is an $n$ so that $n < x < n+1$ then $2n+1 < 2x + 1 < 2n+2 \le 2^n$. And $2^x = 2^n*2^{x-n}$. As $x-n >0$ we know $2^{x-n} > 1$ so $2^n < 2^n*2^{x-n} = 2^x$. And so $2x+1 < 2^x$. And that's that.

Then you sweitch gears and ask how to prove $n^2 \le 2^n + 1;n \ge 3$. (Was that part of the question? A different question?

Well, base case is easy: $3^2 = 2^3 + 1$.

Induction follows:

If $k^2 \le 2^k + 1$ then

$(k+1)^2 = k^2 + 2k + 1 \le 2^k + 2k + 1 = 2^k + 2(k+1)$. Above we proved that $2(n+1) \le 2^k$ if $k \ge 3$ so $2^k + 2(k+1) \le 2^k + 2^k = 2^{k+1}$.

  • $\begingroup$ I have a question with regards to proof one. Do I really have to go beyond $2^n + 2 \lt 2^n2$? That is already enough to use the transitive property to show that $2(n+1)+1 \lt 2^(n+1)$ $\endgroup$
    – K. Gibson
    Jul 29, 2019 at 23:42
  • $\begingroup$ I'm probably not going to see connections like that on a test. This is introductory material for Real Analysis. $\endgroup$
    – K. Gibson
    Jul 29, 2019 at 23:44
  • $\begingroup$ You need to actually state your results. $\endgroup$
    – fleablood
    Jul 30, 2019 at 0:05

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