You should be able to do this

[Bonus] Use a proof other than contradiction

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en.wikipedia.org/wiki/Eisenstein's_criterion
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>babbys first proof by contradiction
Trivial.

Take your pedophile cartoons back to

Youre so big and strong user can you please take my virginity

>>>/fucku/

Uh... r = p/q in reduced form. 2 = r^2 = p^2/q^2 must also be in reduced form. 2 in reduced form is 2/1 so p^2 = 2 and q^2 = 1. 1^2 < p^2 < 2^2 so 1 < +/-p < 2. There is no such integer.

if you dont want to use contradiction you can build infinitely many rational approximations a/b to sqrt(2) using continued fractions. each of these satisfy |sqrt(2)-a/b|

I have witnesses who testify that it's not true. Therefore, in a court of law, it is not true.

[math]\frac{\sqrt{2}}{1}[/math]

but how do you prove the theorem you're using without proof by contradiction?

>[Bonus] Use a proof other than contradiction
1. Write sqrt(2) as a continued fraction
2. What does an infinite continued fraction tell us about a number?
3. ????
4. PROFIT

1 + 1/2 + 1/4 + 1/8 + 1/16....

Good news. I just proved sqrt(2) is irrational because apparently 2 is also irrational.

...

root 2 is not a fucking integer you moron.

>"You should be able to do this"-thread
>page 1 rudin
>freshman/HS math

man, step it up op, atleast pick a non highschool problem.

plus your phrasing of the problem was actually quite poor.

and
>1 + 1/2 + 1/4 + 1/8 + 1/16....
is not an fucking infinite continued fraction

So it IS true that people with autism have a difficult time understanding sarcasm.

You wanna try my problem instead?

would paste "hurr durr just pretending" image, but I'll leave that conclusion as an exercise to the reader.

sheeeit, I'm really only familiar with very elementary ricci calculus, of the sort you'd encounter in sorta middle of the road physics, that's the extent of my diff-geo/tensorial knowledge.

This is an easy proof by contradiction but now I'm wondering if a constructive proof exists. My intuition says no but I'm honestly not quite sure.

I think I once remember seeing a ridiculous proof of this that wasn't really constructive but also didn't use contradiction.

I only started learning about tensors this summer.

that's nice, regrettably I am quite busy with drinking and the such like.

1) That's not what "continued fraction" means
2) You can have infinite sums converging to rational numbers

...

Sarcasm is almost always indicated by tone of voice. Expecting people to pick up on sarcasm over the internet is just idiotic.

Prove it

You should be able to solve this.

3.

is it 3 ?

This notation is messing me up. Is the [A, .] supposed to be the Christoffel symbol?

Commutator. [math]D\varphi = d\varphi + [A,\varphi] [/math] for [math] \varphi \in \Lambda^1(M)[/math].

Use Atiyah-Singer Index Theorem.

Also, what's a "principal vector bundle"? In a principal bundle the model fiber is the structure group itself, but if the structure group is a vector space then the bundle is trivial, because you can use a point as the classifying space of a contractible topological group.

Also T*M is the co-tangent bundle, not the "tangent space"

Moreover I guess you're assuming E is a complex vector bundle in order to even define its Chern character

I'm glad I wasn't the only one thinking that.

Ah, right. Then I believe the answer would be 3.

>Use Atiyah-Singer Index Theorem.
That is the Atiyah-Singer index theorem.

Then the application will be trivial.

Nah. It's just an application of the Atiyah-Singer Index Theorem as said. I haven't gotten into much of K-Theory, but I can sorta recognize the calculations.

That is indeed the case, since [math]P \vdash P[/math] for all [math]P[/math].

Well, algebraic topology of any kind, really.

the theorem is that for any real number c and any natural number N, there exists a and b with 1

>there exists a and b with 1

Instead of the square root, consider the n-th root for n>2.

Suppose there are integers p, q so that [math](\frac{p}{q})^n =2[/math]. Then
[math]p^n = q^n + q^n[/math]
Since n>2, this contradicts Fermat's Last Theorem, so the nth root of 2 is irrational.

I don't think this works, does it? Doesnt it have to be a^n = b^n + c^n where a,b,c are all different? So from this example p^n = q^n + q^n, you cant use fermats last theorm to validate anything since b = c ( q = q )

dont beat me up, im asking

>Fermat can't even show sqrt(2) is irrational

It does work, just read the statement of the theorem, it's for any a,b,c (all positive)

ok thank you was unsure

This has to be easy for Veeky Forums

(-1)^(n+1) * n?

Right? What a pleeb.

yes. Although (-1)^(n-1)*n makes a bit more sense

It's literally the same thing.

why?

One way to solve it is by subtracting columns until you get to a rectangular matrix.
There you will get (n-1) times a -1 and one time a n on the diagonal.
Its obviously the exact same result.

>proof without contradiction
Polynomial $x^2-2=0$ satisfies Eisenstein's criterion for p=2 => it is irreducible in Q => all its roots don't belong to Q.

en.wikipedia.org/wiki/Eisenstein's_criterion

> proof by contradiction
No thx

Get your un-excluded middle the fuck out of the math board.

>un-excluded middle
So you mean that the law of excluded middle is either true or not?

No, I mean that standard mathematics assumes the law of excluded middle as a fundamental tenant, allowing for proofs by contradiction. If you don't assume it then a lot of mathematics becomes inaccessible. Your choice, I guess.

No, I mean that standard mathematics assumes the law of excluded middle as a fundamental tenet, allowing for proofs by contradiction. If you don't assume it then a lot of mathematics becomes inaccessible. Your choice, I guess.

Proof by contradiction != law of excluded middle

For example, intuitionistic logic (the most popular logic without excluded middle) accepts [math]A \wedge \neg A \Rightarrow \perp[/math] while forbidding [math]A \vee \neg A[/math].

Constructive mathematics does not forbid contradiction. Instead it forbids constructs that depend on propositions of the form "there does not exist". So (for example) we cannot construct irrational numbers using the standard definition as a number with no pairs (p,q), because we can't construct a definition of irrationality at all...
Or maybe we can, by appealing to another constructive but equivalent way of defining irrationality (several have been suggested in this thread already).

Take your pedophile cartoons back to .

Fucking degenerates.

Nice spam

take your autistic programmed response to