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Discussion Starter · #1 ·
Note: could not do supercripts, the "2" in steps 3&4 is "squared".

a=b=1
a=b
ab=b2
ab-a2= b2-a2
a(b-a)=(b+a)(b-a)
a=b+a
a=1+1
1=2
 

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In between step 5 and 6 you divided by (b-a). because (b-a)=(1-1)=0 at this point both sides of the equation become undefined because you can't divide by 0.
 

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Discussion Starter · #3 ·
Nice job! I saw that today while dropping the kids off at the pool in one of the engineering buildings at Ohio State.:D
 

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TeamOGF
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Its been a lot of years since algebra but I found myself following this. You'd lose me around the next bend though. ;)
 

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one of my fav math jokes is to take the integral of e to the sub x equal to a function f of u to the sub n
 

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MOD SQUAD
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well,i guess it's true that if you don't use it,you lose it:D
almost 50 years is too long to retain what i learned in those classes,since i haven't had a use for it since then:rolleyes:
 

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Okay I thought I knew algebra pretty well but I was lost on what you were showing here. I don't know if that means I am like Misfit and I have lost it from not using it? :D

I read the first line (a=b=1) and thought, well that is pretty simple. Both a and b are 1 since they both equal 1. That would make the next 4 statements true when the values are substituted in. However, the next line (a=b + a) is an invalid statement unless b is 0. But since I thought b was was it seems invalid. Then the next statement (a=1+1) would only be true if a=2 and again I had a=1 earlier. (You are losing me at this point in case you couldn't tell :p) And then the final statement that 1=2 was the real kicker. Because I learned all through school that they were indeed two different numbers altogether. So I guess they reinvented math since the day I took it. :D:D
 

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However, the next line (a=b + a) is an invalid statement unless b is 0.
this line "works" because both sides are being divided by the quantity (b - a), which in and of itself is legal (before taking into account the values of the variables). however, since (b - a) = (1 - 1) as per the first line, division by zero occurs, which cannot be defined, as someone else pointed out ;)
 

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Powderfinger
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For some reason as I was doing your math in my head I kept thicking this:

god is love
love is blind
Ray Charles is blind
Ray Charles is god.
 

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Ummmmm I think I'd rather be out fishin' !!!! YEP !!!
 

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this line "works" because both sides are being divided by the quantity (b - a), which in and of itself is legal (before taking into account the values of the variables). however, since (b - a) = (1 - 1) as per the first line, division by zero occurs, which cannot be defined, as someone else pointed out ;)
Actually that statement does not "work" because if it were true the b would equal 0 which the first line of the problem already stated that b=1...which is what a equaled as well.;)

I guess I just don't see the merit to the solving of the equation since the values are clearly stated in the first line. That and the fact that 1 can not equal 2.:D
 

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Actually that statement does not "work" because if it were true the b would equal 0 which the first line of the problem already stated that b=1...which is what a equaled as well.;)
taken out of context from the rest of the problem, it does "work", which is what I said:
which in and of itself is legal (before taking into account the values of the variables).
 

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alright seethe your equation has had me stumped, (probably because I haven't used calc in a while.) integral e^x = e^x +c I follow that but the rest is a total blur to me. please enlighten.
 

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alright seethe your equation has had me stumped, (probably because I haven't used calc in a while.) integral e^x = e^x +c I follow that but the rest is a total blur to me. please enlighten.
hehe

its really not solvable, it is just how you write the original equation. if you use the integral symbol and rewrite the equation it looks sort of like this:

Se^x = f(u) n

:D

quite sophomoric. my calc teacher told us this last quarter.
 

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nice, Ill have to remeber that one
 

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I guess I just don't see the merit to the solving of the equation since the values are clearly stated in the first line.
Amen...

That's sorta like an architect sketching blueprints for houses when he has never picked up a hammer nor seen a 2x4.

Sure...it might work...but in all reality it would not make sense to ever have it happen.
 
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