Math Is Fun Forum

I have used RealEstateBroker's "Lawyer at a Funeral" joke with friends. Thanks for providing me with such an enjoyable joke to share with the community.

I'll submit some:

-If you are stranded on a desert island with Adolph Hitler, Atilla the Hun, and a lawyer, and you have a gun with only two bullets, what do you do?
-Shoot the lawyer twice.

-How can you tell when a lawyer is lying?
-His lips are moving.

-What's the difference between a lawyer and a herd of buffalo?
-The lawyer charges more.

I found the description of it hilarious. Just imagine how a 1956 man's head would spin if he were transported to 2006 and saw how small and powerful hard drives have become.

Here's the 1956 man working with the pinnacle of his day's technology:

se7en, (Edit: Ricky didn't leave it out in his post, I just didn't see it. And yes, any real number a is a complex number with Im(a) = 0. By the way, I just want to be friendly and correct another statement of yours, to save you some confusion: you said 89 = 90 + i², which is true, but no, you cannot say Re(89) = 90 and Im(89) = i, because by definition a complex number is expressed in the form z = a + bi, where a and b are real numbers, and Re(z) = a and Im(z) = b. Since a and b are real you cannot let b = i.  This, along with the previous statement about how Re(z) and Im(z) must be real numbers, explains why Re(89) ≠ 25 and Im(89) ≠ 64i in your other problem (even though following your logic Im(89) should have been -64i, I assume it was just a simple mistake on your part with forgetting the subtraction though)) (refer to my post at the top of this page if you wish for another statement of the hypothesis). The thing is, it is conjectured that Re(s) = 1/2 for all non-trivial zeros s. As my post explained, ζ(s) = 0 for every negative even integer, and thus those zeros are considered trival.

In regard to the generalized Riemann hypothesis, no, it does not say that ζ(s) = 0 (or L(χ, s) = 0, since you mentioned the Dirichlet L-series and this series is really what makes this different than the plain old Riemann hypothesis) if Re(s) ≤ 1/2. What it is saying is that if ζ(s) = 0 or L(χ, s) = 0, then Re(s) ≤ 1/2 (basically it's the other way around from what you said). Proof of the Riemann hypothesis would help this hypothesis as the mathematical world knows that the trivial zeros of ζ(s) have the property Re(s) ≤ 1/2 since they are all negative even integers, and if one proves that all other zeros of ζ(s) (the non-trivial zeros) have the property Re(s) = 1/2, then all the bases are covered, we would be certain that Re(s) is either 1/2 or a negative even integer if ζ(s) = 0. Now L-functions are very similar to Riemann's zeta function (for example, L(χ, s) = ∑χ(n)/n[sup]s[/sup], and for χ(n) = 1, L(χ, s) = ζ(s)), and thus properties may easily be compared between the two, but this is only a fraction of the power of the Riemann hypothesis.

So, using the previous paragraph this question can be answered:

No, the conjecture says that it is only possible for a zero s to have the property Re(s) = 1/2 or s = -2, -4, -6... It cannot be 1/4, or 1/8 or anything other than a complex number with real part 1/2 or a negative even integer.

Correct, it remains to be proved.

Wow, this thread grew substantially overnight.

Ok se7en, this here isn't a question as to whether or not your thesis works, so I hope you may choose to reply to it: when do you think your algorithm will be sufficiently used by you so you'd feel comfortable publishing some proofs and thus releasing it to the mathematical community? Are you going to wait until you're done with Riemann, or do a few more problems, or wait until you actually present your master's thesis?

I found it interesting that Ricky brought up the Poincaré conjecture... I came here just to ask if you could do that, but I guess it's already been talked about.

Ok, I was assuming pretty darn good recitations of pi, like this one:

"Gaurav on Monday recited 10,980 digits, breaking the North American and U.S. record of 10,625, which had stood for 27 years. During an after-school session, he recited about two numbers per second for one hour, 14 minutes and 28 seconds."

That's what made me say hours . I guess I was imagining the musical idea as being a significant aid such that someone could set a record with it.

Haha, there's a man who recited 83,431 digits. It took nearly 24 hours. Great stamina there.

Alright, captain. Consider the set of all primes that divide a and/or b:

Then for some suitable nonnegative exponents

we have

Now p[sup]q[/sup] divides p[sup]r[/sup] if and only if q ≤ r, so

But we can also tell that for ab ≠ 0,

Now consider the product

Also, consider the product

Because of the fact that

we have

Thus we can conclude that the two products are equal, so

Is that good enough?

Well... seeing that there are 10 possible digits (0-9) and thus 10 notes/chords you would need to use, it'd be hard to create something remotely diatonic. Sure, you could maybe put in secondary dominants or something for the 3 extra notes... possibly even make them the tonic, supertonic, and mediant an octave higher, but with the way the digits go the music would be rather random and hard to memorize. Another negative aspect of using music is that once you get to a respectable amount of digits, it'll take several hours to recite them all, so you'd need to remember a pretty long piece. Maybe some guy who is a professial orchestral musician with an interest in mathematics could pull it off, who knows.

I'm not too sure what to say to help you other than calm down. Quickly check over your work to make sure you don't do something foolish. It's not worth submitting that answer faster if it's not the right answer.

I think you're thinking that his signature is his answer?

Hmm...

Types of music: Classical compositions, instrumental shred, classic rock, grunge.

A few favorite musicians/bands: Beethoven, Steve Vai, Joe Satriani, Pink Floyd, Led Zeppelin, Alice in Chains, Soundgarden, Pearl Jam.

Favorite Song: For the Love of God, by Steve Vai. This song controls my life.

He may not want to show the proof of Fermat's Last Theorem because he says his algorithm can solve all of these problems. So if someone sees the algorithm in the proof he posts for Fermat, they could take it and adapt it to the Goldbach Conjecture, Riemann Hypothesis, and other big problems.

I'm sorry, but I'm very reluctant to believe all of this. If you have such an algorithm, it must be extremely creative and original to be on a high school level (I assume it's there because you ask for the Riemann hypothesis to be explained in "terms a high school student can understand") and not have ever been caught by the countless mathematicians who have lived ever since our "high school level" mathematics have been around. I mean, just imagine how many mathematicians there have been who have tried to find a simple proof to Fermat's Last Theorem. If you've really managed to prove these great problems, big congratulations to you. I've been working on them this summer but I haven't gotten too far. I was hoping to solve mabye one, many moons from now... maybe if they do get solved, I can still prove them in another way. Man, if this is all true, I guess my chances for mathematical fame in this lifetime are gone, everyone in this era would instantly be downsized, and the greatest problems will have already been solved, so what could I do after that!?

Is Ricky's explaination good for you? He laid out the basic hypothesis for you, but we don't know how much you need to know for you to be able to use this algorithm.

(Hey Ricky, congratulations on post 1000)

Edit: He has 7 posts too, this is pretty insane. April 1st was a while ago though.