Random math questions

[QuantumJG]

This is my attempt at this problem:

Let



Then



Claim:



Proof:

N=1



Now assume this holds for N, want to show that



Now



Therefore



So

[#1procrastinator]

Sorry for the late reply, took me a while to (and attempt to) digest that 

For your claim and proof, are you proving that

is true?

Cause I think you used that in your proof in this part:



(looks like you subbed in )

[QuantumJG]

Sorry for the late reply, took me a while to (and attempt to) digest that 

For your claim and proof, are you proving that

is true?

Cause I think you used that in your proof in this part:



(looks like you subbed in )

What I did is called proof by induction. Proof by induction has three steps:

(i) Show that what you're trying to prove holds for the most basic case. This is called the base "case".

(ii) Assume that what you're trying to prove holds for all $n\le N$ where $N$ is some integer.

(iii) Using the assumption you made in (ii), show that it holds for $N+1$. If you can show that, then you have shown it holds for any value of $n$, since I can arbitrarily make $N$ as large as I want. It's a real, useful and fun proof technique. Specialist maths should include some of these proof techniques (including proof by contrapositive).

[#1procrastinator]

Claim:

So that's what we want to prove right?

Proof:

N=1

^ That's the base case?

Now assume this holds for N, want to show that

^ So if we can show that the above is true, then we can show what we're trying to prove is true?

Now

^ The there means the series up to the nth term, right?

This is where I get confused cause it I'm not sure what we're proving now. It looks like you put in


But isn't that what we're trying to prove?

[Jeggz]

What I did is called proof by induction. Proof by induction has three steps:

(i) Show that what you're trying to prove holds for the most basic case. This is called the base "case".

(ii) Assume that what you're trying to prove holds for all $n\le N$ where $N$ is some integer.

(iii) Using the assumption you made in (ii), show that it holds for $N+1$. If you can show that, then you have shown it holds for any value of $n$, since I can arbitrarily make $N$ as large as I want. It's a real, useful and fun proof technique. Specialist maths should include some of these proof techniques (including proof by contrapositive).

Thanks alot for that Quantum!
I needed help with that for uni maths 

[kamil9876]

In general it is an idea called telescoping (funnily a lecturer once used this to show that maths has applications to astronomy )

In general notice that where   is any sequence. To see this just reorder the terms as follows

[#1procrastinator]

In general it is an idea called telescoping (funnily a lecturer once used this to show that maths has applications to astronomy )

In general notice that where   is any sequence. To see this just reorder the terms as follows

Is that what QuantumJG was proving? (the claim part)

[kamil9876]

Yeah that's right, he proved exactly that by induction, usually the cleanest way of expressing it but not necessarily the most intuitive.

[#1procrastinator]

Hmmm...I'm still confused by this part:


Cause to me, it looks like he used what he was trying to prove
)

[Jeggz]

I don't know if this is the right thread to be posting in..
but can someone please help me with these two questions?

(1) Prove that the cube of an odd integer is odd.
(2) Prove that if n^2 is divisible by three then n is also divisible be 3. (For this question..I tried to do it the same way as if you were to do it for divisible by 2..but it didn't work  )

Thanks in advance!

[polar]

1. hint: write an odd integer as 2k+1

Spoiler

the first three terms must be even (or 0 when k=0), hence, adding 1 (an odd number) makes it's odd. hence, the cube of an odd integer is odd

[Jeggz]

1. hint: write an odd integer as 2k+1

Spoiler

the first three terms must be even (or 0 when k=0), hence, adding 1 (an odd number) makes it's odd. hence, the cube of an odd integer is odd

Ohh I get it now!!
Thanks for that 

[kamil9876]

Hmmm...I'm still confused by this part:


Cause to me, it looks like he used what he was trying to prove
)

He used induction, so yes he showed that IF it is true for then it is true for .

This is the general scheme for induction:

Show the following:

(1) True for N=1

(2) If true for N=k then true for N=k+1

(1) and (2) imply that it is true for all N as follows: Since by (1) it is true for N=1, then using (2) we get true for N=2. Then using (2) again we show it is true for N=3 etc... i.e it's like a domino thing.

[Jeggz]

1. hint: write an odd integer as 2k+1

Spoiler

the first three terms must be even (or 0 when k=0), hence, adding 1 (an odd number) makes it's odd. hence, the cube of an odd integer is odd

sorry again polar, but how would you say i communicate to my teacher that the first three terms must always be even..i understand it, but i don't know how to express it?

[polar]

sorry again polar, but how would you say i communicate to my teacher that the first three terms must always be even..i understand it, but i don't know how to express it?

an even number can be written in the form , if we can write each of first three terms in this form, then they are even: