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I'm in the 6th week of a well-known MOOC course created by Kevin Devlin, "Introduction to Mathematical Thinking." I enjoy the course & did well in the first weeks with conditionals and truth tables, etc.; however now that we are entering into proofs, I'm running into trouble with algebra required to understand the example proofs given in the lectures. My high school algebra was a long time ago! So this is a particular lecture example I'm stuck on. I should add that Devlin is not really involved in the course forums anymore, and mentor assistance is available, but spotty; hence I'm posting my question here as well as on the course forum.

Here is the example proof. I've taken it from the book that Devlin wrote for the course; it's identical to the lecture presentation. I have numbered the steps so that they can be commented on individually.

(0)

(1)

(2) Then ##A(1)## is the statement $$(1 + x)^2 > 1 + 2x$$

(3) which is true by virtue of the binomial expansion $$(1 + x)^2 = 1 + 2x + x^2$$

(4) together with the fact that ##x > 0##. The next step is to prove the statement $$(\forall n \in \mathcal{N}) \, [A(n) \Rightarrow A(n+1)]$$

(5) To do this we take an arbitrary (!) ##n## in ##\mathcal{N}## and prove the conditional $$A(n) \Rightarrow A(n+1)$$

(6) To do this we assume ##A(n)## and try to deduce ##A(n + 1)##.

(7)We have:

$$\begin{align*}

(1+x)^{n+2} \;&= \;(1 +x)^{n+1}(1+x) \\

&> \;(1 + (n + 1)x)\,(1 + x) \;\; [\text{by } A(n)] \\

&= \;1 + (n+1)x + x + (n+1)x^2 \\

&= \;1 + (n+2)x + (n+1)x^2 \\

&> \;1+(n+2)x \;\; [\text{since } x > 0]

\end{align*}$$

(8)This proves ##A(n + 1)##. ##\;\; \Box##

None except the above proof.

I've read this part of the book several times; have listened to this part of the lecture video several times; and have copied out the proof and walked myself through each line, asking myself if I understand it. What hangs me up is not the concept of the principle of mathematical induction; I believe I understand that & have no problems or questions about that. And I had no problems with the math for a previous natural number proof in the course that used a simpler formula. It's the algebraic manipulations here in step 7 where I am getting stuck & can't follow.

Here's how I have been describing step 7 to myself: We assume the antecedent in the conditional to be true - that is, ##A(n)## - and are now proceeding to show that the antecedent implies the consequent - that is, ##A(n+1)##. We do this step directly; that is, by deduction. More specifically, we change the existing ##n## on both sides of the inequality statement for ##A(n)## to ##n + 1## for the statement ##A(n +1)##.

This is all done in the block of 5 aligned statements. Here's how I have been explaining these statements to myself so far:

Line 1, ##\;\;(1+x)^{n+2} \; = \;(1 +x)^{n+1}(1+x)## - this replaces the ##n## in the lefthand side of the statement with ##n+1##

Line 2, ##\;\;> \;(1 + (n + 1)x)\,(1 + x) \;\; [\text{by } A(n)]## - this replaces the ##n## in the righthand side of the statement with ##n+1##. However I don't understand the reference to ##[\text{by } A(n)]##

Line 3, ##\;\;= \;1 + (n+1)x + x + (n+1)x^2## - this expands the righthand side

Line 4, ##\;\;= \;1 + (n+2)x + (n+1)x^2## - factoring the previous line, all but ##(n + 1)x^2##

Line 5, ##\;\;> \;1+(n+2)x \;\; [\text{since } x > 0]## - This is what I really don't get. We are now saying this line is less than the previous line. I don't understand how or why; and I don't understand the annotation ##[\text{since } x > 0]## - specifically, why ##x > 0## is suddenly important.

To sum up, I don't understand the annotations in lines 2 and 5 of this block of statements; and I don't understand what is being done in line 5 at all.

I suspect what is throwing me off is that the previous two example proofs I have looked at for induction (one from this course, one from Hammack's

1. The problem statement (example proof)

1. The problem statement (example proof)

Here is the example proof. I've taken it from the book that Devlin wrote for the course; it's identical to the lecture presentation. I have numbered the steps so that they can be commented on individually.

(0)

**Theorem:**If ##x > 0## then for any ##n \in \mathcal{N}##, $$(1 + x)^{n+1} > 1 + (n + 1) x$$(1)

*Proof:*Let A(n) be the statement ##(1 + x)^{n+1} > 1 + (n + 1) x##.(2) Then ##A(1)## is the statement $$(1 + x)^2 > 1 + 2x$$

(3) which is true by virtue of the binomial expansion $$(1 + x)^2 = 1 + 2x + x^2$$

(4) together with the fact that ##x > 0##. The next step is to prove the statement $$(\forall n \in \mathcal{N}) \, [A(n) \Rightarrow A(n+1)]$$

(5) To do this we take an arbitrary (!) ##n## in ##\mathcal{N}## and prove the conditional $$A(n) \Rightarrow A(n+1)$$

(6) To do this we assume ##A(n)## and try to deduce ##A(n + 1)##.

(7)We have:

$$\begin{align*}

(1+x)^{n+2} \;&= \;(1 +x)^{n+1}(1+x) \\

&> \;(1 + (n + 1)x)\,(1 + x) \;\; [\text{by } A(n)] \\

&= \;1 + (n+1)x + x + (n+1)x^2 \\

&= \;1 + (n+2)x + (n+1)x^2 \\

&> \;1+(n+2)x \;\; [\text{since } x > 0]

\end{align*}$$

(8)This proves ##A(n + 1)##. ##\;\; \Box##

## Homework Equations

None except the above proof.

**3. The attempt at understanding; and what's hanging me up**

I've read this part of the book several times; have listened to this part of the lecture video several times; and have copied out the proof and walked myself through each line, asking myself if I understand it. What hangs me up is not the concept of the principle of mathematical induction; I believe I understand that & have no problems or questions about that. And I had no problems with the math for a previous natural number proof in the course that used a simpler formula. It's the algebraic manipulations here in step 7 where I am getting stuck & can't follow.

Here's how I have been describing step 7 to myself: We assume the antecedent in the conditional to be true - that is, ##A(n)## - and are now proceeding to show that the antecedent implies the consequent - that is, ##A(n+1)##. We do this step directly; that is, by deduction. More specifically, we change the existing ##n## on both sides of the inequality statement for ##A(n)## to ##n + 1## for the statement ##A(n +1)##.

This is all done in the block of 5 aligned statements. Here's how I have been explaining these statements to myself so far:

Line 1, ##\;\;(1+x)^{n+2} \; = \;(1 +x)^{n+1}(1+x)## - this replaces the ##n## in the lefthand side of the statement with ##n+1##

Line 2, ##\;\;> \;(1 + (n + 1)x)\,(1 + x) \;\; [\text{by } A(n)]## - this replaces the ##n## in the righthand side of the statement with ##n+1##. However I don't understand the reference to ##[\text{by } A(n)]##

Line 3, ##\;\;= \;1 + (n+1)x + x + (n+1)x^2## - this expands the righthand side

Line 4, ##\;\;= \;1 + (n+2)x + (n+1)x^2## - factoring the previous line, all but ##(n + 1)x^2##

Line 5, ##\;\;> \;1+(n+2)x \;\; [\text{since } x > 0]## - This is what I really don't get. We are now saying this line is less than the previous line. I don't understand how or why; and I don't understand the annotation ##[\text{since } x > 0]## - specifically, why ##x > 0## is suddenly important.

To sum up, I don't understand the annotations in lines 2 and 5 of this block of statements; and I don't understand what is being done in line 5 at all.

I suspect what is throwing me off is that the previous two example proofs I have looked at for induction (one from this course, one from Hammack's

*The Book of Proof*) both had to do with equalities (identities??), not inequalities; and there is something about inequalities that requires or allows a different punch line. I get the feeling that the last two lines of that block of statements have to do with demonstrating an inequality that completes the proof; but although I have a tickle in my mind that says "You're almost there . . . " I can't quite close the gap yet.
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