Is my induction-step argument correct here?

[Sophdof1]

Hi I would really appreciate if someone could tell me if my induction step is correct here... thanks

 

[firemath]

Looks good!

One thing: Want to show P(n+1) holds, right?

 

[Sophdof1]

Looks good!

One thing: Want to show P(n+1) holds, right?

Yeah

 

[Steven G]

Hi I would really appreciate if someone could tell me if my induction step is correct here... thanks

IF p(n) holds for ALL n>1, then why show that p(n+1) is true? Instead, say that p(k) holds for some n=k. Then show that p(k+1) is true

 

[pka]

Hi I would really appreciate if someone could tell me if my induction step is correct here... thanks

Do you understand the logic of a proof by induction?
If we know that for $\displaystyle a>-1$ and we know that $\displaystyle (1+a)^1\ge 1+1\cdot a$.
Suppose we assume that $\displaystyle (1+a)^k\ge 1+k\cdot a$ is true for $\displaystyle k\in\mathbb{Z}^+$.
Here is what we do next:
$\displaystyle (1+a)^{k+1}=(1+a)(1+a)^k\\(1+a)^{k+1}\ge(1+a)( 1+k\cdot a)\\(1+a)^{k+1}\ge( 1+k\cdot a+a+k\cdot a^2)\\\\(1+a)^{k+1}\ge 1+(k+1)\cdot a$
So now: we know if true for $\displaystyle k$ then it is also true for $\displaystyle k+1$.
We started off with it being true for $\displaystyle k=1$ so it is true $\displaystyle k=1+1=2$.
Well then it must be true for $\displaystyle 2+1=3+1=4+1,\cdots.$
Thus $\displaystyle \forall n\in\mathbb{Z}^+$ & $\displaystyle a\ge-1$ then $\displaystyle (1+a)^n\ge1+n\cdot a$

BTW: this is very famous and known as Bernoulli's inequality.

 

[HallsofIvy]

At the end you have $\displaystyle (1+ a)^{n+1}\ge 1+ a+ na+ na^2\ge 1+ a+ na$ and then argue that since a>-1, 1+ a> 0. But you want it greater than 1, not just greater than 0. Instead, write it as $\displaystyle (1+ a)^{n+1}\ge 1+ na+ (a+ na^2)$ and show that $\displaystyle a+ na^2$ is greater than 0.

 

[lookagain]

At the end you have $\displaystyle (1+ a)^{n+1}\ge 1+ a+ na+ na^2\ge 1+ a+ na$ and then argue that since a>-1, 1+ a> 0. But you want it greater than 1, not just greater than 0. Instead, write it as $\displaystyle (1+ a)^{n+1}\ge 1+ na+ (a+ na^2)$ and show that $\displaystyle a+ na^2$ is greater than 0.

How can $\displaystyle \ a + na^2 \$ be shown to necessarily be greater than 0 if a = 0 is a
possibility with a > -1?

 

[firemath]

IF p(n) holds for ALL n>1, then why show that p(n+1) is true? Instead, say that p(k) holds for some n=k. Then show that p(k+1) is true

But you're trying to show that p(n+1) holds, correct?

I hate to be the one asking the obvious, but why not p(n+1)?

 

[Steven G]

But you're trying to show that p(n+1) holds, correct?

I hate to be the one asking the obvious, but why not p(n+1)?

Of course you want to show that p(n+1) is true, but let's replace n with k. So you want to show that p(k+1) is true IF you assume that p(k) is true. The OP assumed that p(k) is true for all k.

 

[firemath]

Of course you want to show that p(n+1) is true, but let's replace n with k. So you want to show that p(k+1) is true IF you assume that p(k) is true. The OP assumed that p(k) is true for all k.

Ohhhhhh! I get it. So since p(k) was assumed, we will use p(k+1).

 

[pka]

How can $\displaystyle \ a + na^2 \$ be shown to necessarily be greater than 0 if a = 0 is a
possibility with a > -1?

Your's is good point. However, that actually does not come into the proof.
$\displaystyle 1)~(1+a)^{k+1}=(1+a)(1+a)^k\\2)~(1+a)^{k+1}\ge(1+a)( 1+k\cdot a)\\3)~(1+a)^{k+1}\ge( 1+k\cdot a+a+k\cdot a^2)\\4)~(1+a)^{k+1}\ge 1+(k+1)\cdot a$
Going from steps 3) to 4) all that is used is $\displaystyle k\cdot a^2\ge 0$