OP, what, if anything, do you find wrong with the related pic?
Irrational numbers
Jordan Roberts
Asher Ortiz
Do you mean something like:
1) Take irrational number N
2) put it over 1 (N/1)
3) shift the decimal ((N*10)/10)
4) repeat step 3 indefinitely
You'd need a fraction with infinitely long 'whole number' numerator and denominator to describe an irrational number I'd think.
John Scott
Why is m^2 divisible by 4
Juan Wilson
If x is even, then we can write x = 2n. Then x^2 = 4n^2. On the other hand, if x is odd, we can write x = 2n+1, which implies x^2 = 4n^2 + 2n + 1. In other words, if x^2 is even if and only x is even. In the picture, (2) shows the m^2 is even, so we are able to conclude m is even.
Blake Lopez
because m is even
why is m even? because m^2 divides 2n^2 so 2 is a factor of m^2, then it's a factor of m (Z is a UFD and 2 is irreducible)
Camden Walker
I neglected to directly answer your question. Because m is even, we can write m=2k, so m^2 = 4k^2. Thus, m^2 is divisible by 4.
Andrew Jenkins
>presume there is no exact decimal representation
No, nothing is presumed. It's proven. If you're going to argue otherwise, simply point out a flaw in the proofs.
Colton Green
>using Occam's razor for math
holy fucking shit you're an imbecile
Daniel Robinson
>Occam's razor.
Oh? Mathematics has axioms? I can presume there is absolutely no mathematics because no mathematics has less assumptions.
Caleb Allen
OP, assuming you are not trolling...
I think I understand your original question: "If there are an infinite number of integers, then there are an infinite number of possible rational numbers (fractions), so doesn't this mean that ONE of them must be able to represent PI or sqrt(2) or some other irrational number?"
That is a good question, and it has been PROVEN FALSE for thousands of years, since the Greeks at least.
The proof is an example of reducto-ad-absurdum (you assume something is true and by trying to prove it is true you show that contradictions crop up, therefore it is false). Someone above posted this proof for sqrt(2), and that by attempting to prove it is a rational number, that it CANNOT be rational, since it leads to contradictions with other characteristics of rational numbers.
If you find any library books on mathematical logic, or math proofs, you will almost certainly find this proof there, along with other proofs, like there is an infinite number of prime numbers, etc.
Good question though-- the answer is NO, there really are irrational numbers, even though there are an infinite number of rational numbers. Don't let the haters get to you-- keep asking math questions.