Posted on 02 January 2017 - 10:27 PM
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Automatically DeletedPosted on 02 January 2017 - 11:04 PM
AlbertEinstein wrote
HIVLIK wrote...
I can't speak much on Beal's conjecture, but I can say that since it is a conjecture, there is no proof for it yet. It is some sense a generalization of Fermat's last theorem, which, of course, was proven by Andrew Wiles in the mid-90s.
I can speak a bit on Fermat's last theorem and its proof. FLT was a mess to prove, and quite frankly, you'd have to be a very well-rounded mathematician to even UNDERSTAND the proof of it.
The proof focuses on algebra, but not algebra that most people are familiar with. Higher math deals with algebra broadly described as "abstract algebra." This field describes various algebraic structures like groups, rings, fields (which are rings), and vector spaces (specifically, linear algebra for vector spaces). Often as undergraduates, math majors take three semesters of abstract algebra: linear algebra, algebra I (groups and maybe group representations), and algebra II (rings, fields, Galois theory).
But those classes are just building blocks for the stuff that Andrew Wiles had to use. Other relevant algebraic fields are commutative algebra (Here, "commutative" is a*b = b*a) and representation theory (taking your algebraic structure and putting it into its respective "linear" form to make life easier, basically). Wiles used algebraic geometry (algebra and curves) and, naturally, (algebraic) number theory. He also had to work with elliptic curves and modular forms.
The proof went along the lines of (very loosely–maybe even wrong since I'm no expert on this):
1.) Show that the solutions to FLT form an elliptic curve.
2.) Prove the Modularity Theorem (Taniyama-Shimura conjecture at the time), which shows that all rational elliptic curves are modular.
3.) A solution to FLT would not be modular, so it would violate the Modularity Theorem. Hence, FLT is proven.
I've looked at the proof of this a few times, and I recognize some notation, but other than that, I don't really know what's going on. I'd need a lot more math.
Yes, I heard a little bit about Andrew Wiles' very long proof to what Fermat said would be "trivial." I think I'll be coming back here quite often.
Posted on 03 January 2017 - 12:18 AM
Zaptelus wrote
EhhThing wrote...
That took 3 seconds to find .-.
Last edited on 04 January 2017 - 01:27 AM by *deleted-96553
Automatically DeletedPosted on 05 January 2017 - 08:58 AM
I was always curious about this, how did people calculate irrational logarithms before calculators existed?Posted on 05 January 2017 - 12:54 PM
Hobskey wrote
My best guess is guess and check or Taylor series.
Calculate ln(2).
1.) e = 2.7182818284590 to what power is 2?
-Good luck with guess and check.
-The Taylor series is ln(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ….
So ln(2) = 1 - .5 + .333 - .25 + .2 - … and just keep going until you have a bunch of decimal places.
Last edited on 13 January 2017 - 06:30 PM by whitpillow
How do you factorx^4-81 and a^4-b^8
Posted on 13 January 2017 - 06:44 PM
whitpillow wrote
x^4-81 and a^4-b^8
difference of squares… (a^2-b^2) = (a+b)(a-b)
Posted on 13 January 2017 - 06:55 PM
ThatOneCombo wrote
whitpillow wrote...
difference of squares… (a^2-b^2) = (a+b)(a-b)
I dont understand :(
Posted on 13 January 2017 - 07:07 PM
whitpillow wrote
ThatOneCombo wrote...
I dont understand :(
Posted on 13 January 2017 - 07:41 PM
crustacean wrote
whitpillow wrote...
math!
Posted on 14 January 2017 - 12:12 AM
whitpillow wrote
x^4-81 and a^4-b^8
Yes, these are hidden differences of two squares.
x^4 = (x^2)^2, 81 = 9^2
(x^2)^2 - 9^2 = (x^2 + 9)(x^2 - 9)
You can factor that again using the same technique.
x^2 - 9 = (x+3)(x-3)
–> x^4 - 81 = (x^2 + 9)(x + 3)(x - 3).
For the next one, a^4 = (a^2)^2, (b^8) = (b^4)^2
–> = (a^2 + b^4)(a^2 - b^4).
Factor again since b^4 = (b^2)^2:
= (a^2 + b^4)(a + b^2)(a - b^2).
I mean, you could factor more, but it might not be over integers anymore.
Last edited on 19 January 2017 - 07:36 PM by *deleted-96553
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