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a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1


id like to see you figgure that one out

Bochek

LOL man thats old!

Custommx3 wrote:LOL man thats old!

just dont give it away for thoes who cant get it.

Bochek

Good to see this old thing still making rounds - we used to do things like this in the 8th grade i think.

Another really simple one
(x-a)(x-b).......(x-z) = 0

Wouldn't it be proof that 2=1 (I'm more than one way )

I don't know if there's some hidden message that your Implying (that I'm not seeing), but for the math, it's been many years since I've worked with it, it basically all makes sence, except for this part:

Bochek wrote: a^2-b^2 = a*b-b^2

Nd4SpdSe wrote:Wouldn't it be proof that 2=1 (I'm more than one way )

I don't know if there's some hidden message that your Implying (that I'm not seeing), but for the math, it's been many years since I've worked with it, it basically all makes sence, except for this part:

Bochek wrote: a^2-b^2 = a*b-b^2

no that still makes sence, remember A=B so if you assign a number to A, lets say 2 then A and B = 2

so then its just

2^2-2^2=2*2-2^2
witch is
4-4=4-4
wich is
0=0
wich is true.

Bochek

it all works with subsitution of real numbers until you get to

a+b=b

because, if a=2 like you said, and a=b, then the equation is

2+2=2
4=2
2=0

same thing with a+a=a...then there is

2a=a

2(2)=2
4=2
2=0

so your actually proving that 2=zero, if infact a and b equal 2.

Therefore this formula proves that any given value that is subsituted into these equation will end up equal to zero. If 4=a and b, then 4=0. same with 5, 6,7,8,9, whatever.

Now, if your going by simply manipulating the variables, then lets turn b into a

then,

a = a
a^2 = a*a < - also known as a^2, so a^2=a^2
a^2-a^2 = a*a-a^2 <- also known as a^2-a^2=a^2-a^2 and finaly0a=0a
(a+a)(a-a) = a(a-a) <- (2a)(0)=a(0) which is again, 0+0
a+a = a <-2a = a or if you minute a from each side, then a=0

a is the variable, therefore anygiven variable=0

an you can't devide by a variable, that would be improper because you are remouving a variable.

FOol

Vanished wrote: so your actually proving that 2=zero, if infact a and b equal 2.

Therefore this formula proves that any given value that is subsituted into these equation will end up equal to zero. If 4=a and b, then 4=0. same with 5, 6,7,8,9, whatever.

Now, if your going by simply manipulating the variables, then lets turn b into a

then,

a = a
a^2 = a*a < - also known as a^2, so a^2=a^2
a^2-a^2 = a*a-a^2 <- also known as a^2-a^2=a^2-a^2 and finaly0a=0a
(a+a)(a-a) = a(a-a) <- (2a)(0)=a(0) which is again, 0+0
a+a = a <-2a = a or if you minute a from each side, then a=0

a is the variable, therefore anygiven variable=0

an you can't devide by a variable, that would be improper because you are remouving a variable.

FOol

ummm.. your on the right track for all the wrong reasons.

if 4=2, then you know you have a false statement, and any result is invalid. You could argue (like you said):
4-2=2-2 leads to 2=0
but at the same time
4/2=2/2 which ends with 2=1, both of which are false.

Also, A is not neccicarily a variable. It is assumed to be an unknown constant, but even if it was a variable, there is no reason you can't divide by it. Division of variables is used all the time to reduce multivariable equations down to a single variable that can then be solved for.

mr1in6billion wrote:

Vanished wrote: so your actually proving that 2=zero, if infact a and b equal 2.

Therefore this formula proves that any given value that is subsituted into these equation will end up equal to zero. If 4=a and b, then 4=0. same with 5, 6,7,8,9, whatever.

Now, if your going by simply manipulating the variables, then lets turn b into a

then,

a = a
a^2 = a*a < - also known as a^2, so a^2=a^2
a^2-a^2 = a*a-a^2 <- also known as a^2-a^2=a^2-a^2 and finaly0a=0a
(a+a)(a-a) = a(a-a) <- (2a)(0)=a(0) which is again, 0+0
a+a = a <-2a = a or if you minute a from each side, then a=0

a is the variable, therefore anygiven variable=0

an you can't devide by a variable, that would be improper because you are remouving a variable.

FOol

ummm.. your on the right track for all the wrong reasons.

if 4=2, then you know you have a false statement, and any result is invalid. You could argue (like you said):
4-2=2-2 leads to 2=0
but at the same time
4/2=2/2 which ends with 2=1, both of which are false.

Also, A is not neccicarily a variable. It is assumed to be an unknown constant, but even if it was a variable, there is no reason you can't divide by it. Division of variables is used all the time to reduce multivariable equations down to a single variable that can then be solved for.

Your on the right track with this idea.

Bochek

This reminds me of the Women = Evil proof:

women = time + money
>since time is money
women = money^2
>since money is the root of all evil
women = sqrt(evil)^2
>sqrt() to the power of 2 cancles
Women = Evil

i love that equation...its so amazing, and so true.

so no one can figgure out how 1=2?

Bochek

happyclown wrote:This reminds me of the Women = Evil proof:

women = time + money
>since time is money
women = money^2
>since money is the root of all evil
women = sqrt(evil)^2
>sqrt() to the power of 2 cancles
Women = Evil

For this to work out - it should be
women = time * money
not
women = time + money

Bochek wrote:so no one can figgure out how 1=2?

Bochek

It only took my a min to get. I stared at it for a min and couldn't get it at first, but if you write out all the steps it's obvious why it's false.

(( how come there are no spoiler tags here? ))

Bochek wrote:a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1


id like to see you figgure that one out

Bochek

The problem comes when you devide both sides by (a-b); because, if a=b, then a-b=0. Thus, you are dividing both sides of the equation by zero which automatically makes the equation undefined.

So, here is a better way of looking at it.

a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b)(0) = b(0)
0 = 0