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A Weird Math Problem ....
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| King_Mack |
Im trying to prove to my friend that what he's saying makes no sense. In any case, here's the argument...can someone refute it?
Read:- 'a2' below as "a squared," 'b2' below as "b squared," etc.
If a = b (so I say) [a = b]
And we multiply both sides by a
Then we'll see that a2 [a2 = ab]
When with ab compared
Are the same. Remove b2. OK? [a2-b2 = ab-b2]
Both sides we will factorize. See?
Now each side contains a - b. [(a+b)(a-b) = b(a-b)]
We'll divide through by a
Minus b and olé
a + b = b. Oh whoopee! [a+b = b]
But since I said a = b
b + b = b you'll agree? [b+b = b]
So if b = 1
Then this sum I have done [1+1 = 1]
Proves that 2 = 1. Q.E.D.
good luck :) I only argued on grounds that he keeps assuming hehe but that argument aint good enough.. |
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| mmilo |
if a=b,
then a-b=0
and you cannot divide by zero. |
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| King_Mack |
i took that approach, but he's not too impressed by that hrm...
is there anything else you can see that flaws from what he presented? I mean..from first glance, it seems mathematically correct(his induction)....but theres something in there that doesnt make sens. I cant put my finger on it though, unless the only thign is the division issue.. |
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| {b.s.e.} |
I think your problem is right here: Remove b2. OK? [a2-b2 = ab-b2]
I don't even know where b^2 comes from. We have a^2, as well as ab. Even if you had b^2 you cannot subtract that from ab since they are not like terms. I think you need to make substitutions so that your algerbra is suitable. But of course, once you make subs, everything will equal zero and will render your equation void. :D
I hate math, I'm going to play some chess. :disbelief |
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| Perfect_Cheezit |
| just wait for Noisician, he'll set things straight |
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| _Nut_ |
(original eq)
a = b
(both sides mult through by a)
a^2 = ab
When with ab compared
a = b
ab = b^2
Are the same. Remove b2. OK? [a2-b2 = ab-b2]
a^2 = ab = b^2
(division to remove the middle term)
a/b = 0 = b/a
where a/b = -b/a
then
a^2/b = -b
a^2 = -1
therefore
a = sqrt(i) [you cannot take the sqrt of negative nums so you have to bust out the imaginary numbers algebra]
you were coming up with some odd . saying things factored and all that. following through the begining this is what i computed. Are you trying make up some theorum of sorts? if you are you need to lay off the green. |
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| DasBrotBesser |
you're forgetting that ab=a^2=b^2 since a=b
hence it should be a^2=ab=B^2
remove b^2
therefore a^2-b^2=b^2-b^2=0
but even without that, a-b=0, so then it would be
(a+b)(0)=b(0)=0
so then your last equation would be 0=0 |
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| DrUg_Tit0 |
| quote: | Originally posted by King_Mack
Now each side contains a - b. [(a+b)(a-b) = b(a-b)]
We'll divide through by a
Minus b and olé |
As mmilo said earlier, the division by 0 is where the flaw lies. First you have (a+b)*0=b*0=0, and so far the argument is a correct one. But after the division, you basically get (a+b)*0/0=b*0/0 which is not defined because 0/0 can be equal to anything. |
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| Flyboy217 |
| quote: | Originally posted by King_Mack
i took that approach, but he's not too impressed by that hrm...
is there anything else you can see that flaws from what he presented? I mean..from first glance, it seems mathematically correct(his induction)....but theres something in there that doesnt make sens. I cant put my finger on it though, unless the only thign is the division issue.. |
This explanation is correct. Dividing by zero invalidates the equation. Your friend's "proof" belongs to a whole class of such spurious explanations of false equalities. |
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| whiskers |
| quote: | Originally posted by DrUg_Tit0
As mmilo said earlier, the division by 0 is where the flaw lies. First you have (a+b)*0=b*0=0, and so far the argument is a correct one. But after the division, you basically get (a+b)*0/0=b*0/0 which is not defined because 0/0 can be equal to anything. |
i confirm that too, there are a bunch of problems in computer & math textbooks that rely on this "division by 0" approach to have you find the error in the algorithm. |
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| zarathustra |
| Division by zero is illegal. |
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| astroboy |
| quote: | Originally posted by King_Mack
i took that approach, but he's not too impressed by that hrm... |
Impressed or not, you simply CAN NOT divide by zero. As soon as he does that his proof becomes invalid. |
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