Mathmeticians: Do you believe 1 = 0
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Mathmeticians: Do you believe 1 = 0
http://www.diabloworld.com/forums/showt ... genumber=2
Check it out, I am not what you might call a math wizard so am hoping some of the math wizards here can share their opinion of that statement.
Check it out, I am not what you might call a math wizard so am hoping some of the math wizards here can share their opinion of that statement.
Last edited by nigma on Sun Mar 21, 2004 1:12 pm, edited 1 time in total.
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Paddy
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Nigma: That was a funny question coming from a programmer. No harm intended. b and a are just variables like $temp would be. In this particular case it is essential to declare a to equal b so that the proof works.
I have been trying to get my head around why this works actually. the 2=0 one at least. I think when they say (a-b) and cancels them they are essentially dividing by zero, a-b = 0 if a and b are the same, which is not allowed.
I can't find a hole in the second one. But it has been like 5 years since I have done imaginary numbers.
I have been trying to get my head around why this works actually. the 2=0 one at least. I think when they say (a-b) and cancels them they are essentially dividing by zero, a-b = 0 if a and b are the same, which is not allowed.
I can't find a hole in the second one. But it has been like 5 years since I have done imaginary numbers.
I ask these questions, fully willing to expose how much I really know (or for that matter, don't know) in hopes that I might benefit by asking questions.
How can one number be equal to another number without being the same number?
How can 2 be 0? They are different numbers?
If they were the same wouldn't
2 + 2 be the same as 0 + 0? Which would mean 4 = 0?
How can one number be equal to another number without being the same number?
How can 2 be 0? They are different numbers?
If they were the same wouldn't
2 + 2 be the same as 0 + 0? Which would mean 4 = 0?
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Stoneguard
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I believe the fallacy for the first proof lies in the order of substitutions.
I think one of the first rules for solving equations would cause you to reaplace either a with b or b with a BEFORE proceeding on to any other steps in the process, therefore elinimating the misrepresentation of unequal parts.
I think one of the first rules for solving equations would cause you to reaplace either a with b or b with a BEFORE proceeding on to any other steps in the process, therefore elinimating the misrepresentation of unequal parts.
ok, now resubstitute b = aBoth 2=1 and 1=0 can be proven.
Given: b = a
ab = a^2 :::::::: multiply both sides by a
ab - b^2 = a^2 - b^2 ::::::::subtract b^2 from both sides
b(a-b) = (a+b) (a-b) ::::::::::factor both sides
b = (a+b) ::::::::::::::cancel common term
b = 2b :::::::::: a = b is given
2 = 1 :::::::::divide by b
- aa = a^2 :::::::: multiply both sides by a
- aa - a^2 = a^2 - a^2 ::::::::subtract b^2 from both sides
- a(a-a) = (a+a) (a-a) ::::::::::factor both sides
- a = (a+a) ::::::::::::::cancel common term, i.e. dividing both sides by (a-a)=0
- ouch
Both lines are supposed to cause each other (if line 1 is true then line 2 is true as well and the other way around)sqrrt(1/-1) = sqrrt(-1/1) :::::::square root both sides
sqrrt(1)/sqrrt(-1) = sqrrt(-1)/sqrrt(1) ::::::split the roots - yes you can do that
But if you take the square root you have to test both possibilities +sqrt(term) and -sqrt(term)
sqrt(1) is either +1 or -1, (-1*-1)=1 as well. This test has been forgotten because it spoils the proof