<p>Hey guys. Sorry for bothering you again, but I seem to have found a way to go past the “boundless series” crap in my proof of the Fermat’s Last Theorem. Please do inform me if you have a little time to spare to glance at my new (updated) proof. 
Much obliged.</p>
<p>bump10char…
At the risk of sounding irritably persistent, I think I would be highly obliged if someone could find the time to read a small modification in my proof of FLT. I think I have corrected the snag about integers-whole numbers and other stuff.
Please do tell me to shut up if I’m irritating you guys with my stupid requests.
I showed my maths teacher the modifications. He thinks it’s correct, but then goes on to give me bullcrap about I could not have solved it, I’m a (stupid) high school kid, blah blah blah… :)</p>
<p>If it doesn’t come out too unreadable, you could post the proof in text format directly into the forum so that more poeple could actually see it.</p>
<p>A PROOF OF FERMAT’S LAST THEOREM</p>
<p>The solution to Fermat’s last conjecture is given below:
Fermat’s last theorem states that
a^n+b^n≠c^n
for n, a, b and c belonging to the set of positive integers, and n is greater than 2.
Now, we know that an+bn<(a+b)n for n>=2.
The proof is based on reduction ad absurdum, or proof by contradiction.
Therefore, let
a^n+b^n=c^n
for n>2.
Cn<(a+b)n
Taking the nth root on both sides, we get
c<a+b this=“” clearly=“” gives=“” us=“” a=“” triangle=“” inequality.=“” with=“” any=“” three=“” positive=“” integers=“” in=“” such=“” relation,=“” we=“” can=“” consruct=“” triangle.=“” an=“” extension=“” to=“” further=“” affirm=“” the=“” mathematical=“” validity=“” of=“” my=“” statements:=“” cn=“an+bn” an+bn<(a+b)n=“” for=“” n=“”>=2 by binomial theorem
As a result, cn<(a+b)n
Taking the nth root on both sides, c<a+b
To prove the statement that this is a triangular inequality, we have to affirm that a<c+b, and b<a+c.
an= cn-bn
cn-bn<(c+b)n
As a result, an<(c+b)n
Taking the nth root on both sides, we get
a<c+b
We can repeat the same steps to arrive upon the relation that b<a+c.</a+b></p>
<p>Now, it is known, according to the cosine rule of a triangle, that
c^2=a^2+b^2-2ab<em>cosC…………(i)
Thus, keeping this relation in mind, cn can be written as 〖〖[c〗^2]〗^(n/2), which can again be written as
[a^2+b^2-2ab</em>cosC]n/2 , on substitution from (i).
We know that c^n is an integer. Thus, [a^2+b^2-2ab<em>cosC]n/2 should be an integer. It is known that ‘a’ and ‘b’ are integers. However, |cos〖C|<1〗. This is because C≠0⁰, which makes cos C=1. C cannot be equal to 0⁰ because it is known that a triangle is formed by the three integers a, b and c, and no angle in a triangle is 0⁰.
Hence, unless C= 90⁰, in which case cos C=0, 2ab</em>cosC will be a non-integer, as an integer times a non-integer is a non-integer. Subsequently, a^2+b^2-2ab<em>cosC will also be a non-integer, as an integer (a^2+b^2 is an integer) minus a non-integer is a non-integer.
It is also known that a non-integer raised to any rational power is a non-integer. Hence, [a^2+b^2-2ab</em>cosC]n/2 will also be a non-integer. This contradicts the assumption that cn is an integer. Hence, a contradiction is reached.
If cos C is taken to be equal to 0, then we have cn= 〖(a^2+b^2)〗^(n/2). However, the binomial expansion of 〖(a^2+b^2)〗^(n/2) will yield a sum that is greater than (a^n+b^n). Hence, the only suitable choice of n is 2, which leaves us with c^2=a^2+b^2 as the only possibility. Pythagorean triplets are the only solutions to Fermat’s Last Conjecture.</p>
<p>I’ve posted the proof. If you’re facing any problem interpreting the symbols, do tell me.
Really appreciate your time. :)</p>
<p>“as an integer times a non-integer is a non-integer”</p>
<p>(1/2) is not an integer. 2*(1/2) (which is an integer times a non-integer) is 1, which is an integer–this statement from your proof is incorrect.</p>
<p>I want to say this is true if you replace “non-integer” with “irrational”–that is, an integer times an irrational is an irrational (the sketch of a proof I’m imagining would go something like, suppose x is irrational and y is a nonzero integer, but x<em>y is an integer; then you could write x as the ratio of two integers (x</em>y)/y, which is rational by definition, contradiction on x’s irrationality). But…I don’t really see a clean proof that cosC is irrational instead of non-integer. I suspect (though I could very easily be wrong here, as I don’t know math) there isn’t a particularly easy way, since there are infinitely many values of C such that cosC is a non-integer, rational number between 0 and 1.</p>
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<p>The simplest counterexample is sqrt(2)^2. Or any number of the form [a^(p/q)]^(q/p)=a (a,p,q all positive integers) where [a^(p/q)] is your non-integer and (q/p) your rational power.</p>
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<p>Indeed, 1/2*2 = 1 is the desired counterexample, and I think the strongest version of the statement you’re thinking of is that the product of a rational and an irrational is irrational (identical proof to Quelloquialism’s, slightly stronger statement). </p>
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<p>I think you might be thinking of something like an <em>integer</em> raised to a “funky power” is not an integer in the first place – not sure if there’s anything useful like this. It’s in fact characteristic of non-integers, rather than integers, to become integers when raised to funky powers – i.e., take any strange root of an integer, and you have a billion examples.</p>
<p>Ya true. lol. However, the power to which the expression is raised is (n/2). Thus, the non-integer can only be of the form a^(2/n), which leaves us only with finite possibilities for a particular n.
Nevertheless, I’ve found a flaw in my proof…it kinda baffles me that I did not find it before.
c^2=a^2 + b^2-2abcosC. Hence, c^n=[a^2+b^2-2abcosC]^(n/2). First I have equated, and then I say that this is not true. It’s like saying a=b, and then saying a is not equal to b.
The statement that I have to actually prove is [a^2+b^2-2abcosC]^(n/2) cannot be equal to x^n+y^n for any integers a, b, x and y. Moreover, a and b can be assumed to be relatvely prime, and so can y and z.
As you might have guessed, I have acquired some knowledge of number theory. :)</p>
<p>Hey guys.
I’m terribly sorry to add to this rather boring thread again :D</p>
<p>I just got my SAT I report back.
I took it twice [once in jan and then again in October]/</p>
<p>Best scores:
Math= 770
Critical reading= 730
Writing= 710</p>
<p>The total comes to 2210.</p>
<p>I know it’s not great, and compared to Caltech’s standards it’s pretty pathetic.
Do you think I’ll land a decent chance of getting in?</p>
<p>I believe I have a pretty solid Math paper to give in. It’s about this branch of analysis that I’m developing…</p>
<p>Oh. and I forgot to add one thing.</p>
<p>I recently took part in the Senior and Open Sections of the Singapore Math Olympiad.
I got a bronze for one, and a special mention for the other.</p>
<p>In my school cohort, out of the 40 best Math students who took part in the event, I got the highest in one section and the second highest in the other.
I think that’s probably because our school is just two years old, and we received little and inappropriate training [ I really don’t intend to sound like a pathetic wheener here :D]
We don’t have any facility for ‘olympiad training’, and my Math teacher just gave the cohort a couple of questions to do a week before the contest, and called it a day…</p>
<p>Moreover, I’m pretty sure that I’ll be getting the ‘Best Math student’ award at the end of the year, which will be given to the student who has performed the best in Math over the past two years of the IB Diploma course.</p>
<p>As you can see, I’m pretty much a Math maniac :)</p>
<p>You are going to compete with other Singaporeans with international Olympiad medals so you will have quite a tough time getting in.</p>
<p>With regards to your papers in Mathematics and all that, I think you will need a better grounding in your fundamentals of mathematics - your multivariate calculus, differential equations (ordinary, partial), linear algebra+functional analysis and discrete mathematics before really trying to embark on any sort of research. I’ve met some of the most brilliant individuals in my and my junior batch w.r.t. Mathematics (i.e. SMO Open Champion, IMO Silver) and I don’t they are ready to do any sort of serious research in Math either.</p>
<p>Oh. Thanks for the reply :)</p>
<p>As I’ve stated earlier, my dismal performance in the olympiads when compared with national standards was mainly because I had no freaking idea what to expect. This was my first experience at any olympiad. However, I do understand that that offers very little in way of justification.</p>
<p>I don’t think I’ll be competing with the Singaporean cohort though because I’m an Indian national studying in Singapore on scholarship.</p>
<p>My research in Math is simple: we all know that when we graph functions, we plot the y-value vertically, and the x-value horizontally, which gives us a coordinate.
Now let’s take a coordinate plane, and divide the x-axis in tiny bits. Now construct isoceles triangles on these lines- take the two ends of the small lines on the x-axis, and connect these two ends to any point in the graph. These lines connecting the line to the points should be equal in length. These lines will make an angle at any point on the graph One should notice that the higher the graph [i.e. the larger the y-value], the smaller the angle that will be made.
Hence, instead of constructing a graph by graphing the value of the range against the value of the domain, one can plot the same graph by considering various value on the x-axis, and then the angles that these value will make with the graph.
This turns the whole of analysis [as far as my knowledge goes] on its head.</p>
<p>Moreover, I’ve also developed a sort of calculus for this [differentiation, not integration].
Some of the patterns displayed in such functions are quite neat…</p>
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<p>You will be competing with the Singaporean cohort because you will graduate from a Singaporean-based JC or other high school.</p>
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<p>Well yeah. They need proof that you are indeed good in what you say you are good in - and usually olympiads and other sort of science fairs are the best bet this will happen.</p>
<p>Have you tried submitting your research to SSEF or any of the other science fairs?</p>
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<p>If you divide the x-axis into tiny but finite bits, you will end up with a series of points and not a continuous function. If you divide the x-axis into infinitely tiny bits, then I think you will lose the concept of “triangle”. </p>
<p>Anyway, to turn a standard one-one y=f(x) function into your concept, I think it is just “angle of triangle with base at location x-dx to x+dx” = tan^-1(f(x)/dx) where dx is of course the “tiny bit”.</p>
<p>I do have some achievements in the field of Math, but I’ll elaborate on that in my application form rather than brag about them here :D</p>
<p>The nearest thing that I’ve come to winning an olympiad is that I was captain of the Math team that won the Geeta Memorial Math Cup in 2004, which is like an all-State Math competition based on the unexplored recesses of Math [mostly miscellaneous].</p>
<p>However, you’re right in saying that my dismal performance in the olympiad does not explicitly show my skill at math.
However, I’ll try and demonstrate that through my paper. The school award should also count for something. :)</p>
<p>As for submitting my paper to SSEF, I was made aware of them earlier this year, only to know that the registration deadline had passed.
Moreover, because my school is only 2 years old, there was no backing from the authorities. The school is only focused on the curriculum at the moment.
That is perhaps the reason for the non-existent math olympiad training, which I’m told is an elaborate procedure in other Singaporean schools.</p>
<p>However, my Math teachers has always encouraged me to pursue my interest much beyond curriculum, in stark and covert violation of the ‘stick to text’ policy enforced by our crappy Head of Staff. I think he has some good things to say about me in his letter of recommendation.</p>
<p>You can take the lengths of the bases of the triangle to be as long as you like.</p>
<p>Have you heard of sampling in data processing? What basically happens is that at regular intervals of time [like 0.2 seconds say], information is retrieved from analog data, and stored as digital data. </p>
<p>In the same way, if we take the lengths of the bases to be, say 2 cm, we will retrieve only certain values of the function…at regular intervals.
However, as the lengths of the triangles grow smaller, and ‘approach’ a point, a more accurate representation of the function can be approached.
Moreover, this gives rise to the idea of calculus. The lengths of the bases only approach the dimensions of a point. Hence, the concept of a triangle and hence an angle still exists.
In fact, it is this sentiment of ‘limits’ that gave rise to calculus. :)</p>
<p>Say, for example, I want to graph f(x)= sinx, as per my definition of function.</p>
<p>Then my graph will extend from infinity to a height in which an isoceles triangle will make an angle of 1 radian/degree. This is because the range of sin x is -1 to 1. Hence, as the value of sin x approaches 0 at x=0, the value of the angle which the triangles make will also approach 0. Hence, the graph will approach infnity. Hope this helps. :)</p>
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<p>heh. In the same post, even.</p>
<p>lol.
I was just referring to my ‘olympiad’ experience…although this was not an olympiad, it was still a math competition.
I have some other achievements in math too, which I’m sure you’ll object to if I list them here. :D</p>
<p>bump…?</p>