by Jay Johansen | Mar 18, 1996
Note: These are, of course, gag proofs. If you haven't seen them before, try to find the flaw.
|a2=ab||multiply both sides by a|
|a2-b2=ab-b2||subtract b2 from both sides|
|a+b=b||divide by (a-b)|
|b+b=b||as a=b, substitute|
|2=1||divide by b|
Let us consider a set of k numbers.
First, consider the case where k=1. Clearly, if there is only one number in the set, as it is equal to itself (reflexive property of equality), all the numbers in this set are equal.
Second, assume that we have proven the theorem true for some value k. (As we just did above for k=1.) Now consider a set with k+1 numbers. Let us remove any one number from the set, call it x. That leaves k numbers, and by our preceding assumption, these numbers are all equal. Now replace x and remove another number, call it y. Again, all the remaining numbers must be equal. But note that when we removed x from the set, y was still a member, and therefore equal to all the other members; and when we removed y from the set, we had replaced x, so it was a member equal to all the other members. Therefore x and y are both equal to all the other members of the set, so they must equal each other (transitive property of equality). Thus, given the assumption that the theorem is true for a set of k numbers, it must also be true for a set of k+1 number.
By the principle of induction, we have proven that the theorem is true for k=1, and that whenever the theorem is true for k, it is also true for k+1. Therefore it must be true for 2, 3, 4, ... infinity. So even in an infinite set of numbers, they must all be equal.
I once attended a mathematicians' convention. The convention lasted three days. On the first day, I went to a lecture in a certain room. The speaker presented a proof of a theorem he was working on. He wrote out the proof on the blackboard (there were blackboards covering three walls of the room), and at each step he would explain the justification, "divide both sides by x", "as q must be a prime ...", etc. Then he reached step 14, said, "And of course it's obvious that ..." and wrote step 15.
Someone in the audience interrupted and said, "Wait a minute, I don't think that's obvious at all."
The speaker began, "No, you see, because ..." He got about two sentences into his explanation and then he paused. He stared at the blackboard for what seemed a long time, and then turned and walked out of the room. We all waited, but his time ran out and the next speaker came in and he hadn't returned.
On the last day of the conference, I happened to be in that same room hearing another lecture. Suddenly, the speaker from that first day walked into the room. It was obvious he hadn't shaved or changed his clothes since he'd left two days before. Completely ignoring the present speaker, he walked up to the first blackboard and began writing equations in a long and convoluted proof. He began with the equation from step 14 of the other day, filled all three blackboards, and finally on the last board he wrote step 15, proving that it did indeed follow. He triumphantly slammed down the chalk and proclaimed, "There! It's obvious!".
© 1996 by Jay Johansen
Wendee Jul 23, 2014
We have corrected the case n=5 and now we have two porofs of Fermat's last theorem (FLT)for n=5 using the Identity of Fermat equation, x+y-z.A fox, i have no any suitable word to call this man, has sent an E-mail to the authorities of my University saying that CNMSEM is not legtmate. Would you please tell us the open access journal that you edit.We are looking for the second simple proof of FLT similar to n=7 case in CMNSEM with one of my colleagues.R.A.D.P
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