I hope you did your homework, Veeky Forums

I hope you did your homework, Veeky Forums.

>what is the collatz conjecture?

at least write the question down properly brainlet OP

a real number can't be even or odd, only integers

Kurisu is a dumb worthless whore, just like all females, real and fictional.

>a real number can't be even or odd
what?

>a real number can't be even or odd, only integers

Whole real numbers are integers.

Doesn't work for negative numbers, disproved

hm

>Take any whole, real number.
OK, I choose 0, which is a whole, real number. (Wholes include naturals and 0).

Zero is even, divide by 2, get 0. Repeat indefinitely. Will always be zero never get to 4, and therefore never get to a 4,2,1 loop.

Conjecture disproven. Man, that was so much easier than the Collatz conjecture.

But is zero really even?

>this took Veeky Forums 17 minutes

Lol, what a bunch of retards.

Yep.

Define all even numbers 2n
Define all odd numbers 2n+1
Set n as whole number
???
zero is even qed

I actually tried to write out a proof and spent a good 5 minutes before I realized it's a troll. Am I a brainlet?

Well obviously "nobody has solved it before" is no reason to not try when it comes to math.

We did this in my intro proof theory class. It's a pretty simple; just use induction.

I tried the basic discrete maths 101 inductive proof kind of thing. I wrote out different cases for odd numbers, numbers divisible by 4, and numbers divisible by 2 but not 4.

This math is just too advanced for our baboon minds to comprehend.

Clearly not you fucking brainlet.

14.
disproved.

you didn't think before posting did you?

The proof is trivial for numbers of the form 2^n

Proposition: Numbers that are not powers of 2 are not numbers.

And I don't mean proposition as in 'to be proven'. I mean it literally. Lets forget about those disgusting numbers.

Therefore, the proof is trivial. QED

>Repeat indefinitely. Will always be zero never get to 4
prove it.

In the sequence 4 appears after a 1 or an 8.

Suppose there exists a natural number n such that

0/(2^n) = 1 iff
0 = 2^n

Which is a contradiction, as there is no such n.

Now suppose there exists n such that

0/(2^n) = 8 iff
0 = 8(2^n) iff
0 = 2^n

Again, a contradiction.

Therefore, zero will never get to 4.

Mathematics may not be ready for such problems.

>look mum I put unsolved problems on a picture with an anime girl on it xdDdxDD

>integrals
don't tell me you went into grad school in math just to do calculus

You just can't see the light.

mathematical physics is some alien shit

...

where do you get these problems?

I make them up as I read texts and make sure that I can solve them myself. Helps with studying.

The answer is Tohsaka Rin pleases old men for money.

Mathematics is not preppares for this kind of problems

isn't that equivalent to Weyl's theorem?

Try Whitehead's lemma.

whole numbers are positive, retard

Natural numbers are positive. Whole numbers can be both negative and positive.