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Message: <blockquote data-quote="Persius" data-source="post: 19476462" data-attributes="member: 477838">Prime Numbers, 2,3,5,7(my lucky number).......The list goes on ......... are the most beautiful,powerful numbers in  Mathematics.Getting divided by 1 and herself shows how hard to tame these beautiful ladies(even harder than our ladies ).But there were mathematicians who tried their best and this is where we meet Pierre De Fermat who is one of a kind part time mathematician who did some remarkable effort in the field and introduced  Fermat's little theorem.The following proof is based on induction.It is very easy.There may be more ways to do this,but I always thought of Induction.If P is a prime, then for any integer A,( A^P - A ) must be divisible by P.    ((P|(A to the power p )-A)      In here I assume that you  have some idea about induction.But it is not essential.But you should definitely have an idea about how <a href="http://www.mathsisfun.com/algebra/binomial-theorem.html" target="_blank">Binomial Theorem </a> works in order to understand this.Let's assume that above is true,P divides a^p-a  (p| a^p-a).Now the task is to show it holds true for a+1.All we have to do is to expand (a+1)^p using binomial and see for ourselves.The following simplification is solely done on binomial theorem.P!= P * P-1 *P-2*...........................1 .You may get that each term except a^p-a has P! term(a+1)^p-( a+1)=a^p-a + Addition of term  (P!/(P-r)!r!) a^P-r.1^r from r=1 to r=P-1 Permutation  is an integer . A thing we count which is denoted by P!/(P-r)! Combination  is an integer. A thing we can count using (P!/(P-r)!)/r!  It means (P-r)!,r! completely get cancelled, and divide P!.Every Binomial Coefficient is an integer.So since P is a prime .P and 1 are  the only factors of P  .There is no way that P can be cancelled in a general term of binomial coefficient like (P!/(P-r)!r!). So (P-1!/(P-r)!r!)  must be an integer where P-r)!,r! completely get cancelled, and divide P-1! leaving P >>>> So that means we can take P outside of the general term  (P!/(P-r)!r!) a^P-r.1^r and write as P.(P-1!/(P-r)!r!)a^P-r.1^r This is true for all the terms r=1 to r= P-1 .So addition of term  (P!/(P-r)!r!) a^P-r.1^r from r=1 to r=P-1  is also divisible by P.Since we have assumed a^p-a is divisible by P .The expression is true for a+1 if it is true for a.The proof may be difficult to understand if you don't have much mathematics in you,but  it gets easier the more you do it.You and I,we  are not born as geniuses. </blockquote>

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