Trying to convince myself I'm making progress:

prove: 1+n = n+1

base case: 1+0 = 0+1

assume 1+n = n+1
prove 1+(n+1) = (n+1) + 1

1+(n+1) = (1+n) + 1 (associativity)
= (n+1) + 1 (induction hypothesis)

and we're done...yes?

Yes.

(Edit: Note that your "assume" step is "Assume that for some number n, 1+n= n+1" as opposed to "Assume that for all numbers n, 1+n = n+1," which is what you're trying to prove. Often it helps to keep those qualifiers in there.)

S = all natural numbers is important, because that's how you showed (by induction on 'n') that m+1 is in M if m is in M. It's a mini-induction within an induction proof.

When you're proving something like "addition is commutative," you should recognize that you're not allowed to assume pretty much anything. I mean, if you can't even assume that 3+5 = 5+3, what the hell can you assume!?!?

So you can only use axioms and the statements you've proved from the axioms; here, this means a+0 = 0+a (axiom) and associativity (which we proved).

In my opinion, which is worth little since I'm not teaching your class, when you're learning to write proofs, you should have to be extremely explicit with each step of the proof. If I were to do an inductive proof in my thesis, I would not say "Let M = {m in M : P(m)}. First, we show 0 in M. Then we show m in M => (m+1) in M. Therefore M = all naturals." I would say something like

"We prove this by induction on m. For the base case, m=0, the following is true (...). Now assume that the statement is true for some m. We show that it's true for (m+1). (...). This proves the theorem."

But as a student, I think it's good to be pedantic and extremely clear about every step. If you want to show that something is true for all natural numbers, I think it's good to construct the set of numbers for which that something is true and show that it's the naturals (by first showing 0 is in there, and then showing that if 'n' is in there, so is 'n+1').

At least this was the way that I learned inductive proof at first. Actually, the way I learned induction was that

Any set S with the properties
(1) 0 is in S, and
(2) If n is in S, then n+1 is in S.
must contain the natural numbers.

Don't get too hung up on it. Hopefully, though, it makes a bit more sense now.

If it were "assume for all numbers, n..." how would that change our proof?

Also, I thought I was doing this for all of the natural numbers (not just "any"), since I start at 0, and then induct on n...

Great! Now we're getting to the heart of this.

Remember our dominoes. To show that all dominoes get knocked down, we show first (base case) that the first domino gets knocked down. Then we show that assuming some domino gets knocked down (this is our inductive hypothesis), then the next one gets knocked down also. The conclusion is that all dominoes get knocked down, since

first one down => second one down => 3rd one down => ...

So in your proof above:
You've just knocked the '0' domino down.

You're assuming that there exists an 'n' such that the n_th domino gets knocked down.

Rephrase: Now you'll show that this implies that the (n+1)st domino gets knocked down also.

So we've shown the following two statements:
(1) 1+0 = 0+1, and
(2) if 1+n = n+1, then 1+(n+1) = (n+1)+1

This is enough to show that all dominoes are knocked down.
In my set language, if we let S = {n in naturals: 1+n = n+1}, then
(1): 0 is in S
(2): if a number 'n' is in S, then (n+1) is also in S.

These together imply that all naturals are in S (this is the "principle of mathematical induction").

EDIT2: if you "assume for all numbers n, 1+n = n+1" then you've assumed what you're trying to prove. That is bad.

Ok that makes a ton more sense.

Thanks a lot

Little help with probabilities please

For this question pretend you are drawing cards without replacement from the infamous "Iraq's Most Wanted" deck issued by the U.S. Military. If you are drawing from the full deck of 55 cards, what are the following probabilities:

You draw three cards, which end up being Saddam Hussein and his two sons (whose pictures were also in the deck of cards.)

3/55 x 2/54 x 1/53
is what im assuming

1/(55c3)?

Since there is only one of each card of interest, you can only draw the 3 cards one way. There are 55 choose 3 ways to draw 3 cards, so the probability is just the reciprocal of 55 choose 3.

Are you sure? im pretty sure you can draw Sadam, son 1 and son 2 in 6 different ways.

and my pretty sure i mean 100% sure. the guy above you had the correct answer.

Who are you responding to?

...Our answers are numerically the same?

Yeah but Rosie isn't just pretty sure, he's 100% sure.

pretty sure i got #1. I first just took the upper half of the unit circle and transformed it into the entire unit circle by squaring it. then found the map to go from the circle to the upper half plane and combined them.

#2 I'm not really sure how to construct the definition from scratch. i know that the conformal map needs to be analytic and not equal to zero, and i also know that the conformal map preserves angles.

#3 seems simple but i can't get it. I wrote down that analyticity implies continuity and then i tried to show that the limit at infinity equals infinity implies that there is some point where the limit is 0 but i can't do it.

#4 I tried using induction but i end up getting z=1 as the extra root i need but it has to be inside the unit circle so i don't think it works.

any ideas on any of these would be great.

if i am looking to calculate the percentage of the confidence intervals that i created at a 75% confidence level that contain a population mean of 50.... would i used the formula E=z ((st.dev)/route of n)
im working on an excel project and i have 40x35 numbers
im pretty sure thats right but im not sure about the part containing the mean of 50
i think thats just throwing me off

im looking for the percentage of the confidence levels...... i created a spread sheet 35 rows 40 columns of random numbers ... 40 variables, 35 random numbers, mean of 50 with a stdev of 4
normal distribution
then
i did the confidence interval alpha .25 stdev 4 size 35 and alpha .05
i have two sheets
one at 75 % and one at 95% confidence intervals

3. Assume that f(z) never vanishes. Then 1/f(z) is well-defined, analytic, and bounded (you need to use the limit condition for the last part to show that |f| has a minimum, and this minimum is >0). Apply Liouville's theorem.

4. Use Rouche. On |z|=1, |e^z|\leq e^{|z|}\leq e < \pi = \pi|z|^n.

What's the standard way to interpret notation for partial derivatives? (Google isn't helping for some reason.)

If I write f_x1x2x3, would you interpret that as taking x1 then x2 then x3? Or the reverse?

wow thanx. for #4 does it not work if i take the nth root of both sides? that gets me z=(e^x/pi)^(1/n)*e^(y/n+2kpi/n), k=0,1,...,n-1 gives me n roots.

What are x and y? I don't think you are going to be able to solve that algebraically.

z=x+iy so i rewrote e^z as e^x*e^iy

Yes. And it equals d^3 f/dx3 dx2 dx1

Yeah, in that case, taking n-th roots and arguing like you did doesn't get you anywhere. You can solve the equation e^z=1 without much difficulty, but something like the equation e^z=z is a lot harder. Just use Rouche's theorem like I suggested.

gotcha. for #1 i just noticed that since the region in the z plane is open, when i square it i don't actually get the entire unit disk. i am missing the real axis in the disk. i'm not really sure how to deal with this.

This isn't a homework problem, but I figure someone here can answer it.

I want to construct a cheap set of shelves using boards and cinder blocks.
Something like this.

The cinder blocks will be in groups of two.
Will one of these arrangements provide more support than the other or does it not matter?
A. blocks one on top of the other. B. blocks standing next to each other.