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[mystic gust]

words are good
totally fine

[wheat isle]

Well I have {E exists in N |E= 2k for some number K} {O exists in N| O = 2j -1 for some number J} {F exists in N| Fibbanuachi numbers}-- or F = {1,1,2,3,5,..}

is this a good start to have written up

[mystic gust]

idk if those will help solve it but if youre trying to formalize E, O, F then sure

[wheat isle]

in proofs are you required to formalize?

[mystic gust]

the formal way would be $E = {x \in \mathbb{Z} | \exists k \in \mathbb{Z} : x = 2k}$

[placid obsidian]

cute rizzly bear (nom nom nom)

[wheat isle]

as i dont know what the teacher looks for in anwsers since this is the first homework

[mystic gust]

its good to be as specific as possible on why something is true to leave no room for doubt

however

words are still fine

[wheat isle]

what is Z?

is that natural number

or integers?

[mystic gust]

integers

[mystic gust]

this isnt necessary because everyone knows what the even numbers are

unless you are learning about set builder notation

oh it said separate N not Z mbmb

[wheat isle]

Also something like this would be good right: The Fibonacci sequence will always contain both odd and even numbers in its sequence ad infinitum, as each number is obtained by adding the two preceding numbers. That means there will always be an odd odd even pattern. Thereby O ∩ F is infinite set because the pattern always outputs an odd number and for the other one as well

[mystic gust]

yep, a bit more proof on the odd odd even thing might be good but thats what i thought of as well

[wheat isle]

how could I incoporate the {E exists in N |E= 2k for some number K} part into that

with proving that the fib set and E set intersection is infinite

{E exists in N |E= 2k for some number K} ∩ F = {1,1,2,3,5,..} = K = 1[E exists in N | E= 2(1)| = 2 ∩ F= {1,1,2,3,5,...)

[mystic gust]

idk

[wheat isle]

.

I still confused like how do I make this a mathematical proof

[mystic gust]

do you know about induction

[rotund snow]

To prove that O inter F is infinite for example, you could try proof by contradiction, ie you could suppose that O inter F is finite and use some properties of a finite set to find a contradiction

Like for example a finite set always has a maximum… which in this case could help in the proof I think

[wheat isle]

what do you mean use properties of a finite set to find a contradiction?

[rotund snow]

I meant use the fact that a finite set has a maximum

A maximum which here is a part of the Fibonacci sequence

You could call it Fm

And try finding a contradiction

[wheat isle]

Hmm I’ve never done that don’t no where I would start

[rotund snow]

Well this is the way I see how to do it you can do another method

I can try explaining the method by contradiction if you want

Or give the steps

[wheat isle]

yea can u give the steps

[rotund snow]

Sure:
.let E=O ∩ F, let’s suppose E is finite, that means that E has a maximum, this maximum is odd and a part of the Fibonacci sequence let’s call it Fm
.we then have Fm that’s odd now consider Fm+1=Fm+Fm-1 there are two possibilities:
.Either Fm-1 is even, then Fm+1 is odd, knowing that the Fibonacci sequence is increasing, this contradicts the statement that Fm is maximum because Fm+1>Fm.
.second possibility Fm-1 is odd then Fm+1 is even which means Fm+2=Fm+1+Fm is the sum of an even and odd number thus Fm+2 is odd, so Fm+2>Fm which contradicts the fact that Fm is the maximum

In both cases we have a contradiction

That means E is not finite

[rotund snow]

I don’t think it’s really rigorously written but that’s the gist of my idea

Also there may be an easier way to prove it

For E ∩ F it’s a similar reasoning I think

[somber spoke]

Like proving that there are infinite prime numbers

[rotund snow]

Yeah exactly

[somber spoke]

Don't think so, I mean induction prove must be easier but too much long

[rotund snow]

Is my proof generally correct ?

[somber spoke]

Yes

[rotund snow]

Okay thanks, I wasn’t completely sure lol

[silk socket]

@rotund snow has given 1 rep to @somber spoke

[somber spoke]

I think also about the proof of $card(\mathbb{N}) < card([0,1])$

[placid obsidian]

Yojda

[somber spoke]

Finding something more

Btw, read How to Prove It: A Structured Approach by Daniel Velleman if you want to write proofs

[wheat isle]

Why do you assume that the maximum is odd?

[mystic gust]

because it's a member of O ∩ F

rotor confusingly named that set E

[wheat isle]

so assume that O n F and E n f are finite and limited sets. there are only limited number of elements in the sets. If you look at the fib sequence , you havbe two odds follewed by an even O,O,E. The rule of the Fibonacci is that each number is formed by adding the previous two. Therefore, if you have O,O you will get even next. This rule contradicts that you can have a finite number of odd and even Fibonacci numbers in a set. If you assume there is a limited number of elements in the fibbo set you will reach a point or maximum where you cant find two odd fibbo numbers with an even one. For example, {O,O,E,O} in this set, the pattern stops and the fib rule is broken because there is not a second consecutive odd odd pattern since it reached a max due to its quality of being finite. This in all proves that the inital assumption of O n F and E n F being finite is contradicted by the nature of the fibbonachis pattern. Thereby, O n F and E n F in fact are infinite sets.

i made this

would that be a good proof?

[mystic gust]

you don't need to use both contradiction (assume they're finite) and the OOE patern

i would say pick one

also O n F and E n F can't both be finite otherwise F would be finite

[wheat isle]

wait wdym?

[wheat isle]

how could i write it with only 1 contradiction

[mystic gust]

suppose there are finite members of E n F, take the highest number in that set
then by definition there are 2 odd's after it in the fibonacci sequence
those 2 odds would add up to the next term, an even number, which is a contradiction
and
similarly for assuming there are finite members of O n F

[wheat isle]

how is there a contradiction if its an even next?

it aligns perfectly with the Fibonacci sequence's recursive definition

what contradiction is there if an even comes next?

[mystic gust]

because we assumed there are only a finite amount of evens and we went to the last one

but then we proved there is still one more

[wheat isle]

i get the and part being wrong but why is having two contradictions wrong?

[mystic gust]

it's not but it's good to have as little complication as possible

[rotund snow]

Can you prove this like this:
1/x is a bijection from [1,infinity[ to [0,1] so card([1,infinity[)=card([0,1]) but card(N)=card(N{0} U {3/2}) and N{0} U{3/2} C [1,infinity[ thus card(N)<card([1,infinity[)=>card(N)<card([0,1]) is this correct ?

[rotund snow]

Yeah my bad I should’ve named it better

[mystic gust]

subsets of a set can have the same cardinality as that set

but we shouldn't have this discussion in the person's help channel