Is my process of proving inequalities using Mathematical Induction correct?

Question

$$ 2^n<(n+1)! \quad \text{for all integers} \quad n \ge 2$$

Base Case: $$ 2^2< (2+1)! = 4< 6 $$ Correct

Assumption: $$ P(k): 2^k<(k+1)!$$ $$P(k+1)= 2^{k+1}<(k+2)!$$

Now we start with the Assumption : $$ 2^k<(k+1)!$$ $$ \text{Multiply 2 on both sides} \quad 2^{k+1}<2(k+1)! $$

$$ \text{Now we can say} \quad (k+2)!> 2(k+1)! \quad \text{to prove this inequality}$$

$$ (k+2)(k+1)! > 2(k+1)! $$

So now all we must prove is that $$ k+2 > 2 \quad \text{Which is True for} \quad k\ge 2$$

Answer 1

Your proof is correct, but could be more clearly written. In general in a proof, you should start with what you know, and move to what you are trying to prove. In yours, in the middle you jump to what you are trying to prove, then proceed to work backwards. This is always harder to follow. (Also, you list what you are trying to prove under the title "Assumption". That is just plain bad, even though it isn't what you meant.) Rewrite the last part as :

Assumption: $$2^k<(k+1)!$$ To be proved: $$2^{k+1}<(k+2)!$$

Since $0 < 2 < k+2$ and $0 < 2^k < (k+1)!$, $$(2)(2^k) < (k+2)(k+1)!$$ $$2^{k+1} < (k+2)!$$

Hence by induction $2^n < (n+1)!$ for all $n \ge 2$

Answer 2

In addition to the good advice already in an answer, I recommend more care in how you use symbols, especially the $=$ symbol. Yes, it does mean the thing on the left is the same as the thing on the right, but sometimes one can disagree on exactly what each "thing" is.

For example, many (most?) people reading $2^2<(2+1)!=4<6$ would interpret it as \begin{array}{ccccc} 2^2<(2+1)! & \quad\text{and}\quad & (2+1)!=4 & \quad\text{and}\quad & 4<6 \\[1ex] \text{(true)} && \text{(oops!)} && \text{(true)} \end{array}

You obviously did not mean to write the equation with "oops!" below it, and I think most people will realize that, but they'll have to think about it harder than they should have to and they may hold that against you.

You could write something like "$2^2<(2+1)!$ because $4<6$", or if you want to keep it in one big math expression, $$2^2=4<6=(2+1)!.$$ People will still read the $<$ in the middle as applying only to the numeric expressions immediately to its left and right, but now that gives you $4 < 6$, which is the essence of this part of the proof. Also, people are used to seeing "chains" of inequalities like

$$ a \ \Box\ b \ \Box\ c \ \Box\ d \ \Box\ x \ \Box\ y \ \Box\ z$$

where each of the $\Box$ symbols is replaced by $=$, $<$, or $\leq$, and they know that this chain implies that $a \leq z$ (and if any of the relationships is $<$, then $a < z$).

For similar reasons, $P(k): 2^k<(k+1)!$ is fine but $P(k+1)= 2^{k+1}<(k+2)!$ is confusing. You could make an argument for writing $P(k+1)= \left(2^{k+1}<(k+2)!\right)$, but I still prefer "$P(k):$". Also, once you choose a style, use it consistently.