proof by induction help

[shelly89]

Let $\displaystyle *^{n}(a) = {a*.......*a }$ n times

thus

$\displaystyle *^{1} = a$

$\displaystyle *^{2} = a*a$

$\displaystyle *^{3} = a*a*a$ etc


prove by induction that

$\displaystyle *^{n}a = (a+1)^{n} -1$ for all n in N


am really having problems understand how to do this, please can someone show a step by step solution guide?


so far i get

n = 1

$\displaystyle *^{1}a = a$

$\displaystyle (a+1)^{1} -1 = a$

so true for n = 1

assume true for n = k

so

$\displaystyle *^{k}a = (a+1)^{k} - 1$


next step in indution, cant seem to do it.....

n= k+1

$\displaystyle *^{k+1}a = *^{k}a *a$

thats equal to

$\displaystyle (a+1)^{k}-1*a$

where do i go from here ?

 

[JeffM]

Let $\displaystyle *^{n}(a) = {a*.......*a }$ n times

thus

$\displaystyle *^{1} = a$

$\displaystyle *^{2} = a*a$

$\displaystyle *^{3} = a*a*a$ etc


prove by induction that

$\displaystyle *^{n}a = (a+1)^{n} -1$ for all n in N


am really having problems understand how to do this, please can someone show a step by step solution guide?


so far i get

n = 1

$\displaystyle *^{1}a = a$

$\displaystyle (a+1)^{1} -1 = a$

so true for n = 1

assume true for n = k

so

$\displaystyle *^{k}a = (a+1)^{k} - 1$


next step in indution, cant seem to do it.....

n= k+1

$\displaystyle *^{k+1}a = *^{k}a *a$

thats equal to

$\displaystyle (a+1)^{k}-1*a$

where do i go from here ?

If I understand what you are writing $\displaystyle *^n = a^n.$

This is a weird problem because it is not true for any number a. You first have to determine whether it is true for any number or numbers. If so, then you have to prove that it is true no matter what positive integer n is.

So $\displaystyle *^2 = a^2 = (a + 1)^2 - 1 = a^2 + 2a + 1 - 1 = a^2 + 2a \implies$

$\displaystyle a^2 = a^2 + 2a \implies 2a = 0 \implies a = 0.$

Now what?

 

[shelly89]

If I understand what you are writing $\displaystyle *^n = a^n.$

This is a weird problem because it is not true for any number a. You first have to determine whether it is true for any number or numbers. If so, then you have to prove that it is true no matter what positive integer n is.

So $\displaystyle *^2 = a^2 = (a + 1)^2 - 1 = a^2 + 2a + 1 - 1 = a^2 + 2a \implies$

$\displaystyle a^2 = a^2 + 2a \implies 2a = 0 \implies a = 0.$

Now what?

Am afraid i dont follow,

here is the solution and i cant understand it from the second line onwards.

 

[JeffM]

Am afraid i dont follow,

here is the solution and i cant understand it from the second line onwards.

You don't follow me because I misread the problem or made a typo. Maybe I need a beer or two to clear my head. However, I find the given proof hard to follow myself.

$\displaystyle *^n(a) = a^n \implies *^n = a^{(n - 1)}.$ Not sure what I was thinking (or if I made a typo) in my prior post. BUT

$\displaystyle *^2(a) = a^2 = (a + 1)^2 - 1 \implies a^2 = a^2 + 2a + 1 - 1 = a^2 + 2a = a^2 \implies 2a = 0 \implies a = 0.$

So we can simplify $\displaystyle *^n(a) = 0^{(n - 1)}(0) = 0.$

$\displaystyle *^1(0) = 0 = 0 + 1 - 1 = (0 + 1) - 1 = (0 + 1)^1 - 1.$

So there is at least one integer > 0, namely 1, for which this proposition is true. Let's take any one of them and see whether that one's successor is also included in the set.

$\displaystyle \exists\ k\ such\ that\ k \in \mathbb Z,\ k \ge 1,\ and\ *^k(0) = (0 + 1)^k - 1.$

$\displaystyle *^{(k + 1}(0) = 0 = 1 - 1 = 1^{(k + 1)} - 1 = (0 + 1)^{(k + 1)} - 1.$ Done.

I wonder if a step was missed in the write up.

Now let's hope that those beers straightened me out and I have not made some other faux pas.