I don't get how irrational numbers exist.
for any two fractions m/n > p/q, there is some fraction (m+p)/(n+q) in between the two. thus there are an infinite amount of fractions of rational numbers between any two fractions.
then if we take 3/2 and 5/4, since there are an infinite of rational fractions between them, there must be some fraction which equates to sqrt(2). how am I wrong?
Why must there be?
The set of rationals between two numbers doesn't have to include all numbers between the two
Oh, how simple minded you are.
>there must be some fraction which equates to sqrt(2)
Well, just tell us what it is and then we can all go home. You will be right and you will have shamed thousands of years of mathematics.
Just tell us what that fraction is. Shouldn't be too hard, after all, there are infinite fractions so one of them must be it!
It's harder to explain the incredibly large and specific numerator and denominator of sqrt(2) and pi than to simply presume there is no exact decimal representation and therefore no exact fraction. Occam's razor. A mathematician might be able to give a more convincing argument, but do consider that a lot of truths in mathematics are really the simplest possible explanation -- something that at first seems simpler turns out to require many more assumptions and be harder to explain.
Because there is a basic proof showing sqrt(2) is not rational
it includes an infinite amount of ratios, therefor all numbers between the two. every possible decimal you could think of must be represented by the infinite ratios for they are infinite.
>it includes an infinite amount of ratios, therefor all numbers between the two.
Non sequitir. Rationals are not complete. Take real analysis.
No
did you really need to relate your image to bootlicking to make that post OP?
>every possible decimal
Only repeating decimals
OP, what, if anything, do you find wrong with the related pic?
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.
Why is m^2 divisible by 4
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.
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)
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.
>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.
>using Occam's razor for math
holy fucking shit you're an imbecile
>Occam's razor.
Oh? Mathematics has axioms? I can presume there is absolutely no mathematics because no mathematics has less assumptions.
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.
>Hitler
This is a worksafe board you know.
Veeky Forums - Science and First-Grade School Math
kill yourself, mathematics doesn't come from the authority of "thousands of years of mathematics".
If that's your argument for the existence of irrationals, you're more retarded than he is
you have discovered a property called "density". However, if a space A is dense in B, that doesn't mean that A=B.
highschoolers on this board need to stop confusing any infinite collection of things and some canonical case of an infinite collection things.
I saw someone asking if since there is a sum of the natural -1/12 or whatever that bs is any infinite sequence of integers "sums" to this as well.
OP seems to be assuming that just bc there is an infinite collection of things the limit must be in the set, rarely the case.