faux proof?

[littlelaurabee]

WHATS WRONG WITH THIS PROOF???

a = b (neither a or b are equal zero)

multiplying by a on both sides:                 a^2 = a*b

subtacting b^2 on both sides:      (a^2 - b^2) = (a*b - b^2)

factoring

Comments

Re: Ummmm....

[littlelaurabee]

*buzz* sorry your wrong...try again though kay.

First line says: "a = b ( a not equal zero)"-->thus b does not equal to zero either! thus your answer stating "What this all means is the above proof is only valid if a=b=0" is not the answer we are looking for here.

Come on danny, your an engineer...you can figure this out *wink*

Re: Ummmm....

[littlelaurabee]

changed...better? or still invalid?

[asphalteyedew]

and to further elaborate on that... (after some discussion with margaret)

i see where (b^2 - b^2) is basically 0 ... so then you have:

1(0) = 1(0)

well that could just be 0=0 or... you could divide each side by 1(0)... giving 1(0)/1(0) on each side which is 1... thus, 1=1...

but I still don't know quite where the problem is lol.

[asphalteyedew]

and anyway... i don't know if you could do this... but to check formulas, you would plug the numbers back in to make sure they work and these most certainly would not. in that case, you could also prove that a could never equal b, because we damn sure know that 2 does not equal 1. eh, but still not what you're looking for.. I'm sure. hah.

[chodemnky]

what we all need to realize is that:

2 + 2 = 5

for extremely large values of 2

[littlelaurabee]

yeah of course every computer, number-maching and calculator takes .999999 equal to 1 since we can mathematically prove that it is. We can use series but here is my dumbed down proof:

X = X

9*X/9 = X

(10X-X)/9 = X

Plugging in X = 0.99999999 we get:

(10 * 0.99999999 - 99999999)/9 = 99999999

1 = 99999999

Of course we have to assume there are an infinate amount of 9's....thats why i guess the proof with series is better fitting and more acurate.