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Mathmeticians: Do you believe 1 = 0
Posted: Wed Oct 08, 2003 10:59 pm
by nigma
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.
Posted: Wed Oct 08, 2003 11:22 pm
by volka
- 2 = 1:
undo the substiution Given: b = a (replace any b by a) and you will find that there is a division by 0 - ouch - 1=0:
::::::split the roots - yes you can do that
but -a*-a == a*a - ouch
Posted: Thu Oct 09, 2003 1:36 am
by nigma
Volka: Why would b = a in the first place? I do not mean this sarcastically or in a smart ass way, I am really curious and would like to truely understand this.
Posted: Thu Oct 09, 2003 1:44 am
by Paddy
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.
Posted: Thu Oct 09, 2003 2:26 am
by nigma
I realize b and a are variables, I do know a a wee bit of algebra

I guess I am just a bit confused, and I cannot put my finger on what is confusing me, maybe the whole damned thing !?!
Posted: Thu Oct 09, 2003 2:33 am
by nigma
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?
Posted: Fri Oct 10, 2003 7:23 am
by devork
ok let's put it this way
if
a=b AND b=c
THEN a=c
if
[my hand] touching [table] AND [table] touching [floor]
THEN
[my hand ] is touching [floor]
everything can be expressed in math form [dumps up for Mathematicians]
.....
Posted: Fri Oct 10, 2003 10:00 am
by Stoneguard
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.
Posted: Fri Oct 10, 2003 11:05 am
by volka
Both 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
ok, now resubstitute b = a
- 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
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
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)
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 
Posted: Fri Oct 10, 2003 11:25 am
by nigma
Thanks for "enlightening" me

Apreciate the timely and easily understandable explanation!
Posted: Fri Oct 10, 2003 11:33 am
by volka
everything can be expressed in math form [dumps up for Mathematicians]
The more cheese the more holes. The more holes the less cheese. Ergo: The more cheese the less cheese
hope the translation is somewhat understandable...
Posted: Fri Oct 10, 2003 1:39 pm
by devork

yep it was ...