Proof by smallest counterexample and contradiction

[sonic cave]

I do not have much time to get this as it's due later today and I'm behind with readings. I do not understand proofs very well, but I think I can sort of do contradiction as well as counterexample. I mostly don't know how to read the math shown, is it saying "For all n greater than or equal to 1, n is the sum of y=0 in 3^y<3^n+1)"? And what exactly does that mean, like where do I start with proving it? When I am done proving it, the answer would be (besides showing my work) to say whether the statement is true or false?

[sonic cave]

Also, to make sure I get it, counterposition is finding a "not" or opposite example of something and seeing if that claim is true and if it is true then the original statement is true. And contradiction really seems to be the same, I do not get how they differ.

[thick tendon]

For any $n\ge 1$, $\sum_{y=0} 3^y$ is just a number

[limber arch]

SWR

[thick tendon]

For $n=1$, that number would be $3^0+3^1=1+3=4$

[limber arch]

SWR

[thick tendon]

For $n=2$, it would be $3^0+3^1+3^2=1+3+9=13$

[limber arch]

SWR

[thick tendon]

And so on

[sonic cave]

Ok one second sorry, trying to understand this

I am still confused, sorry. So the first part I see means all positive numbers

The middle part says?

Oh wait

It's two

[thick tendon]

Few min. I'll brb

[sonic cave]

Oki!

So it's saying for all positive numbers,
3^n+1 ?

What does the n sum of y=0 mean, like I am assuming you always put 0 in for y so that just gets rid of the 3y (because it's zero)

[thick tendon]

Are you familiar with summation notation?

[sonic cave]

No

This is assumed that we know, I did Calculus two years ago and remember nothing. I believe that was a symbol present in it

I'm watching a video on it now

Oh... this is so simple lol

https://www.youtube.com/watch?v=st-6vZESf8A&ab_channel=BrianMcLogan

Is this it, I am just making sure

3^(0) < 3^(3^(0+1))
3^(1) < 3^(3^(1+1))
3^(2) < 3^(3^(2+1))
3^(3) < 3^(3^(3+1))

So this would be for my homework (the problem I have shown), and it just goes to infinity but I stopped at 3 here. Is this correct? @thick tendon

Ok yeah that is not right

I am realizing that theres y and n

Alright so I think I get it now, 3^0 is 1 so it'd all be this instead?
1 < 3^(3^(0+1))
1 < 3^(3^(1+1))
1 < 3^(3^(2+1))
1 < 3^(3^(3+1))
Which is all true.

I just do not know how the proof by contradiction and counterexample work

Also brb 5 mins if you return in that time, sorry

Okay back

[thick tendon]

I'm bback too

[sonic cave]

Awesome

[thick tendon]

had to poof away for a bit

[sonic cave]

All good!

[thick tendon]

okay lemme read what you wrote

[sonic cave]

Thanks for your help so far, I had no idea of what sum notation was called so couldn't find anything online to explain

[sonic cave]

and tyt!

[thick tendon]

This isn't quite right

[sonic cave]

😦

[thick tendon]

So first, $\sum_{y=0}^n 3^y$ is just shorthand for writing
$$3^0+3^1+3^2+3^3+...+3^{n-2}+3^{n-1}+3^n$$

[limber arch]

SWR

[sonic cave]

I do not understand how this is so, how does the sigma work? Like each component of it

[thick tendon]

You are basically summing $3^y$ for every $y$ from $0$ to $n$

[limber arch]

SWR

[thick tendon]

The sigma is just a notational definition

By $\textbf{definition}$,
$$\sum_{y=0}^n 3^y=3^0+3^1+3^2+3^3+...+3^{n-2}+3^{n-1}+3^n$$

[limber arch]

SWR

[sonic cave]

So the addition replaces the < symbol?

[thick tendon]

The $<$ has no impact here

[limber arch]

SWR

[thick tendon]

$<$ is nowhere attached to $\Sigma$

[limber arch]

SWR

[thick tendon]

Read it as $\left[\sum_{y=0}^n 3^y\right]<\left[3^{n+1}\right]$

[limber arch]

SWR

[sonic cave]

Oh ok

Okay my second confusion is why the y changed to n-1 or n-2 ?

[thick tendon]

brb again

working 😦

[sonic cave]

All good, any help is nice!!

[thick tendon]

@sonic cave hello again. I should be free for ~45 minutes now

[sonic cave]

Oh okay!

[sonic cave]

Could you please explain this? @thick tendon

[thick tendon]

$n-2$ is the third to last, $n-1$ is the second to last, and $n$ is the last.

[limber arch]

SWR

[thick tendon]

What I was writing here is that we are summing all $3^y$ for all $y$ from $0$ to $n$

[limber arch]

SWR

[thick tendon]

Here's an example,
$$\sum_{y=0}^4 3^y=3^0+3^1+3^2+3^3+3^4$$

[limber arch]

SWR

[thick tendon]

The $y=0$ under the $\Sigma$ specifies your starting value. You calculate $3^y$ starting with $y=0$, then you increment $y$ to $1$ and add $3^1$ to your $3^0$. You repeat this process until $y=4$.

[limber arch]

SWR

[sonic cave]

Okay I am not sure I am going to get this lol

[thick tendon]

that's okay

We're gonna get there

[sonic cave]

It is okay! I am just going to try to do what I can and submit it because I am considering dropping out at this point haha

This was like the tipping point is all

[thick tendon]

I still think I can help you

but I won't force you

It's your choice

[sonic cave]

Yeah I am a bit tired is all and my head hurts so I do not think I can understand stuff right now. Sleep deprived

But thank you so much for your help!!

[thick tendon]

fair enough

sleep is important

[sonic cave]

Yeah and I wake early too, not a nice mic haha

mix*

[thick tendon]

whether or not this is due tomorrow, you should ask more about this tomorrow

I think it'll be good for you

[sonic cave]

I intend to drop in to office hours for some help before I completely give up, thank you for the advice though!

I made a list of all the stuff I'm not getting to ask

And like do some practice problems (available on course page)

But yeah thanks!

How do I close this thread? I am sorry I am unfamiliar with this server's commands

!solved

[thick tendon]

.solved