math gurus needed again............

[wetback]

I told my friend about how you guys helped me out and she needs some help also.

I am mathematically retarded! Someone please help me with this question!!


The following mathematical solution contains one critical error, since we all
know that the number 1 is not equal to the number 2. Use your knowledge
of solving equations, multiplication, division, general mathematical
knowledge, and a little research, to explain the error in the solution.
Hint: You may have to research factoring to determine what happened
between lines 3 and 4 on both sides of the equation.x = x
x2 = x2
x2 - x2 = x2 - x2
x(x - x) = (x + x)(x - x)

x(x - x) = (x + x)(x - x)
(x-x) (x-x)


(those are fractions, but they won't line up!)

x = (x + x)
x = 2x
If we let x =1, then upon substitution we have
(1) = 2(1)
1 = 2

 

[wootoom]

go to the genius forum fuck

 

[wetback]

go to the genius forum fuck

That was insiteful thanks.

 

[wootoom]

That was insiteful thanks.

no prob bro just tryin to help

 

[wetback]

Bump

 

[mrplunkey]

In the fourth line, you divided by a guaranteed zero for all values of "x". You can multiply by zero all you want, as you did in lines 1, 2 and 3 -- but you cannot divide by zero.

 

[drrman]

In the fourth line, you divided by a guaranteed zero for all values of "x". You can multiply by zero all you want, as you did in lines 1, 2 and 3 -- but you cannot divide by zero.

bingo

 

[mrplunkey]

edit... need to work on that a sec

Working on a good way to mess with the prof

 

[pdaddy]

In the fourth line, you divided by a guaranteed zero for all values of "x". You can multiply by zero all you want, as you did in lines 1, 2 and 3 -- but you cannot divide by zero.

That's the complicated version, this is a bit easier to follow:

 

[mrplunkey]

Nerd alert... ok, I confess... i'm a former engineer who has turned to the dark side.

It's time to mess with the teacher...

Technically, writing [x*(x-x)]/(x-x) is what's known as an "Indeterminate Form". It's better studied as a mathematical limit tho... so let's write:

Here's the left hand side of line 4:

Limit {[x*(x-y)]/(x-y)} = Limit {(x2-xy)/(x-y)}
x -> y.......................... x -> y

Since this in the form of f(x)/g(x) and yields an indeterminate form in the limit, you can apply L'Hopital's rule which says:

Limit f(x)/g(x) = Limit f'(x)/g'(x)

Therefore:

f'(x)/g'(x) = 2x / 1 = 2x

Here's the right-hand side of line 4:

Limit {[(x-y)*(x+y)]/(x-y)} = Limit {(x2-y2)/(x-y)}
x -> y ............................... x -> y

so f'(x)/g'(x) = 2x / 1

So alas... 2x = 2x

So see how a little calculus sees right past a divide by zero error?

 

[Lumberg]

actually you're putting two mutually exclusive solutions to x^2-x^2 on either side of the equals sign. It's an equation with two solutions, but you can't use both. It's like having your cake and eating it, too. You can't do it.

 

[mrplunkey]

actually you're putting two mutually exclusive solutions to x^2-x^2 on either side of the equals sign. It's an equation with two solutions, but you can't use both. It's like having your cake and eating it, too. You can't do it.

But x*(x-x) = (x+x)(x-x) anyway. That part is legitimate.

It would be like saying:

y*(x-x) = (y+x)*(x-x)

That's true as well, regardless of the value of y.