2=0

[JackG]

Nice trick. Of course, anyone looking at the results for 1782^12 + 1841^12 and 1922^12 individually will think that it breaks Fermat's theorem, but entering (1782^12) + (1841^12) - (1922^12) shows that the two sides are not in fact equal. If you could just find values of a, b, and c so that their difference is greater than 12 decimal points away, however, I think that the calculator would fall for the trick because of rounding problems due to memory issues. At least it would work in Matlab, and I'm guessing the calculator stores data the same way.

So, anyone know some a, b, and c such that (c^n)/((a^n)+(b^n)-(c^n)) > 1*(10^13) ? Did I even do that right?

[AndrewJS]

If you are adding like this,
.5 + .5 + .5 + .5 = 2
Etc, then you are really doing a Sigma(2/n) for n times as n approaches infinity. Using the distributive property this sigma expression can be written:
n * (2 / n ) = 2
For all nonzero real numbers, this expression equals 2. At n = 0, there's a hole in the graph, yet its limit is 2.

The key about infinity and zero is that they are not well-defined. It depends HOW you're reaching infinity or zero, not only the fact that you are.

[Molten]

Originally Posted by AndrewJS

The key about infinity and zero is that they are not well-defined.

Both are just as well-defined as anything else is. Zero is no less defined then one. Understood that infinity is a sort of hard concept to grasp, but it is pretty well set even if you can't see it yourself in the real world.

Or in wikiwords [citation needed]

[AndrewJS]

Yeah that's true.

I was just trying to say that approaching 0 or inf from different curves/functions changes the behavior when you perform operations between them. Like how y = x and y = e^x both approach inf as x -> inf yet they do so at different "rates." e^x is much "faster" than x so, x/e^x has a limit of 0, whereas e^x/x has a limit of inf as x --> inf. This is related to L'Hopitals rule, and is sorta what his rule is doing by taking the derivatives of top and bottom of the fraction etc etc.

[ExTexan]

But from a ref's point of view, 1=2

b=a then "Multiply by (a)"

ab=a2 (can't make squared symbol) then "Subtract (b2)"

ab-b2=a2-b2 then "Factor"

b(a-b)=(a+b)(a-b) then "Divide by (a-b)"

b=(a+b) then "Say it ain’t so"

1=2

[Norman J]

Originally Posted by ExTexan

"Divide by (a-b)"

I know this one. But I won't give the answer away to those who don't.

[Mike Schreiber]

Originally Posted by Norman J

I know this one. But I won't give the answer away to those who don't.

It would of have been easier for me to figure out if he had stated that a=1 and b=1...I got it now though. Hint: Try substituting 1's in. Although you don't have to use 1's it is the only way to get 1=2 otherwise you get 2=4 or 4=8 etc...

[ExTexan]

I know this one. But I won't give the answer away to those who don't.

Thanks.....it made me scratch my head 40 years ago!

It would of have been easier for me to figure out if he had stated that a=1 and b=1

Ahh but making it easier wouldn't have been as much fun! and also....if 1=2 then doesn't 2=4? :-)

[biojae]

Originally Posted by ExTexan

also....if 1=2 then doesn't 2=4? :-)

So, then you are saying 1 = 4 then?

a = b; b = c; a = c;
a = 1; b = 2; c = 4;
1 = 2; 2 = 4; 1 = 4;

and if 4 = 8, then would 8 = 16... and so on?
and if so then 1 = inf

[DonRotolo]

Originally Posted by Molten

Don't always believe your calculator. Sometimes they lie.

Beat me to it

Originally Posted by ExTexan

Thanks.....it made me scratch my head 40 years ago!

Heh, when I had hair to scratch through...

If we're talking about odd equalities, how 'bout this one:
6 of one = 1/2 dozen of the other.

[ExTexan]

So, then you are saying 1 = 4 then?

Well I actually hadn't thought of it that way but I guess if the inital formulas are correct then that is a valid assumption! Good one!