what's socratic induction?
113053
Induction in general is moving from a specific to a general axiom. For instance, that man can walk, therefore men walk. Socrates (as used as a character by Plato) used this method very often, and it is part of the Socratic method.
363975
Algebra, concepts of proof, logic, set theory (baby version), number theory, maybe some other stuff.
57681
1(1!) + 2(2!) + 3(3!) ... n(n!) = (n+1)!-1\n\nYou can prove it by induction:\nFor n=1, 1(1!) = 2!-1 = 1\nNow assume 1(1!) + 2(2!) + 3(3!) ... n(n!) = (n+1)!-1 for n. Then show that it is true for n+1 as well: \n\n1(1!) + 2(2!) + ... n(n!) = (n+1)!-1\nAdd (n+1)(n+1)! to both sides:\n1(1!) + 2(2!) + ... n(n!) + (n+1)(n+1)! \n= (n+1)!-1 + (n+1)(n+1)!\n= (n+1)!(1+(n+1)) - 1\n= (n+1)!(n+2) - 1\n= (n+2)! - 1\n\nSince the equation is true for n=1, and when it is true for n it is also true for n+1, it must be true for all positive integers.
546909
Definitions of the terms. The axioms for the natrual numbers are as follows:\n\n1) 0 is a natural number,\n2) The successor of a natural number is a natural number,\n3) No two numbers have the same successor\n4) 0 is not the successor of any natural number\n5) If P(x) is a property of natural numbers such that P(0) is true and such that P(n) true implies P(succ n) is true, then P(n) is true for all natural numbers (Mathematical Induction)\n\nUse S(n) to denote the successor of n and *define*\n1=S(0) and 2=S(1).\n\nNow *define* addition inductively by\nn+0=n and\nn+Sm=S(n+m).\n\nThen,\n1+1=1+S(0)\n=S(1+0)=S(1)=2.
895063
Google the word algebra help.  Purple math is great, but there are other sites also.  I am sorry I do not have the names at the moment.
782198
Deductive.\n\nWhen you know a rule and apply it to a situation, you are DEDUCING the solution in that situation.\n\nThe other type of reasoning is inductive.\nWhen you observe many examples of a particular behavior (or relationship, or whatever), you may conclude INDUCTIVELY that it will occur in ALL situations. (But of course you can't be SURE that it will. Your degree of confidence is based on the number of situations in which you have seen it happen, and the fact that you have never seen it NOT happen.)
282120
The pre-Socratic philosophers rejected traditional mythological explanations for the phenomena they saw around them in favor of more rational explanations. Many of them asked:\n\n* Where does everything come from?\n* What is it really made out of?\n* How do we explain the plurality of things found in nature?\n* How might we describe nature mathematically?\n\nOthers concentrated on defining problems and paradoxes that became the basis for later mathematical, scientific and philosophic study.
67383
1. A(N)<2 for any N:\n\na) A(0) = sqrt(2) < 2\nb) induction: A(N)<2 => A(N+1) = sqrt(2)^A(N) < sqrt(2)^2 = 2\n\n2. The sequence A(N) increases:\n\na) 1<sqrt(2) => sqrt(2)^1<sqrt(2)^sqrt(2) => A(0)<A(1)\nb) induction: A(N)<A(N+1) => sqrt(2)^A(N)<sqrt(2)^A(N+1) => A(N+1)<A(N+2)\n\nFrom 1 and 2 it follows that A(N) HAS a limit X, with sqrt(2)<X<=2\n\n3. sqrt(2)^X=X:\nX = lim(N->inf.)A(N) = lim(N->inf.)A(N+1) = lim(N->inf.)sqrt(2)^A(N) = sqrt(2)^X\n\n4. X=2 or X=4:\nWe already know that 2 and 4 are solutions for sqrt(2)^X=X. Let's show that they are the ONLY solutions:\nThe second derivative of f(X)=sqrt(2)^X-X is f"(X) = ln(sqrt(2))^2 * sqrt(2)^X > 0 for any X. Then f is a strictly convex function, so f(X)=0 can have AT MOST two solutions. We already found two, so they must be the only ones.\n\n5. X=2:\nWe showed that X<=2 and (X=2 or X=4), you draw the conclusion.
958370
We wish to show that:\n[k=1, n]∑kk! = (n+1)! - 1\n\nClearly, this is true for n=1, since 1*1! = 2! - 1 = 1. Suppose it is true for some n. Then for n+1:\n[k=1, n+1]∑kk!\n(n+1)*(n+1)! + [k=1, n]∑kk!\n(n+1)*(n+1)! + (n+1)! - 1\n(n+2)*(n+1)! - 1\n(n+2)! - 1\n\nWhich is what we would get if we substituted n+1 for n in our formula. So by induction, the formula holds for all n. Q.E.D.
938293
There's proof by deduction, proof by process of elimination, proof by contradiction and proof by induction.
987606
usually they'd use it in engineering\n\naddition:\nwell, they use in the alogarithm form.i don't know any 'simple' analogy but the easiest way to explain to a child might be, they use it in engineering to calculate the strength of structures and stuffs like that.\ntry http://en.wikipedia.org/wiki/Exponentiation
878633
we need to prove this by induction...\n\nfor n belonging to Z\n\nf(n)=n*(n+1)*(2n+1)/6\n\nfor n=1\n\nLHS:\n f(1)= 1^2 =1\nRHS\n1*(1+1)8(2*1+1)/6=6/6=1\n\nThus its true when n=1;\n\nAssume f(k) is true i.e. \n1^2+2^2+......+k^2=k(k+1)(2k+1).. Let this be equation 1\n\nNow we need to prove f(k+1) is true...\nie\nf(k+1)=(k+1)(k+2)(2(k+1)+1)/6\n          =(k+1)(k+2)(2k+3)/6\nL.h.s\n\nf(k+1)=1^2+2^2........+k^2+(k+1)^2\n          =k(k+1)(2K+1)/6 + (k+1)^2 from eq 1\n          =(k+1)(k(2k+1)/6 +k+1)\n          =(1/6)(k+1)(2k^2+k+6k+6)\n          =(1/6)(k+1)(2k^2+7k+6)\n          =(1/6)(k+1)(k+2)(2k+3)\n\nHence f(k+1) is true if f(k) is true....\n\nThus its true for all positive integers.......
141606
There is an application to convert an AC current to DC.\nAssume that f is a sigmoid function, then integration of\nf^2 over one period gives a constant number which is considered as flattened sigmoid and is analogous with converting AC to DC.
490503
Lots of useful info here....\n\nhttp://en.wikipedia.org/wiki/Socrates
354990
oh yeah, he was the guy that discovered that thing like ummm, yeah, ok hold on while I cut & paste please....\n\nHe began his professional life as a lawyer, but he has become best known as a philosophical advocate and defender of the scientific revolution. His works establish and popularize an inductive methodology for scientific inquiry, often called the Baconian method. Induction implies drawing knowledge from the natural world through experimentation, observation, and testing of hypotheses. In the context of his time, such methods were connected with the occult trends of hermeticism and alchemy.
1086054
TESLA IS THE UNIT FOR MAGNETIC INDUCTION,HE GAVE A LOT OF THEORIES ON  THE ABOVE TOPIC
367692
If p(1) is true and, for each integer k>1 the statements  p(1),p(2),....p(k) imply the statement p(k+1) then p(n) is true for all.\n\nIn other words if a statement is true for the integers 1..k\nand you can prove it from those for k+1, than it is true for all integers.\n\nIt is a stronger version of normal mathematical induction where one use P(k) to prove P(k+1).