[wolfgang]

I can prove

1 = 2 !! or 0= 1,, or 3 = 7 and so on....i. e. x = y( where x and y are normally not equall ).

mathmatically and without breaking any mathmatical roles !

can you ?

[Guest]

Let x=y.

Then x^2=xy.

Then x^2+x^2=x^2+xy.

Then 2x^2=x^2+xy.

Then 2x^2-2xy=x^2+xy-2xy.

And then 2x^2-2xy=x^2-xy.

You can re-write that last step as 2(x^2-xy)=1(x^2-xy). After you cancel the (x^2-xy)'s you get 2=1.

Easy!

[wolfgang]

Let x=y.

Then x^2=xy.

Then x^2+x^2=x^2+xy.

Then 2x^2=x^2+xy.

Then 2x^2-2xy=x^2+xy-2xy.

And then 2x^2-2xy=x^2-xy.

You can re-write that last step as 2(x^2-xy)=1(x^2-xy). After you cancel the (x^2-xy)'s you get 2=1.

Easy!

but x^2-xy = 0

and u can`t cancel it (devide on 0).

[Guest]

but x^2-xy = 0

and u can`t cancel it (devide on 0).

But I didnt subtract the xy from both sides, i added a x^2.

[Guest]

But I didnt subtract the xy from both sides, i added a x^2.

Your last step you said cancel (x^2-xy). How do you do that? You divide both sides by (x^2-xy). But since x^2-xy = 0 you can't divide by x^2-xy. So you can't cancel. Otherwise I could say simply well 1*0=2*0. Therefore 1=2. I wouldn't need to go through the manipulations you used. But 1*0=2*0 does not imply that 1=2.

[Guest]

No I can't...

[Guest]

No I can't...

Haha then did i do all that math for no reason?

[wolfgang]

Your last step you said cancel (x^2-xy). How do you do that? You divide both sides by (x^2-xy). But since x^2-xy = 0 you can't divide by x^2-xy. So you can't cancel. Otherwise I could say simply well 1*0=2*0. Therefore 1=2. I wouldn't need to go through the manipulations you used. But 1*0=2*0 does not imply that 1=2.

thank you Maurice ..!

[wolfgang]

No I can't...

you are very intelligent !!...ty again

[araver]

x=y (mod g.c.d.(x,y))

or simpler x=y (mod 1)

[Guest]

ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10 + ...

2 * ln(2) = 2 - 2/2 + 2/3 - 2/4 + 2/5 - 2/6 + 2/7 - 2/8 + 2/9 - 2/10 + ...

2 * ln(2) = 2 - 1 + 2/3 - 1/2 + 2/5 - 1/3 + 2/7 - 1/4 + 2/9 - 1/5 + ...

2 * ln(2) = 2 - 1 - 1/2 + 2/3 - 1/3 - 1/4 + 2/5 - 1/5 - 1/6 + 2/7 -1/7 + ...

2 * ln(2) = (2 - 1) - 1/2 + (2/3 - 1/3) - 1/4 + (2/5 - 1/5) -1/6 + (2/7 -1/7) - ...

2 * ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - ...

2 * ln(2) = ln(2)

2 = 1

q.e.d.

[wolfgang]

x=y (mod g.c.d.(x,y))

or simpler x=y (mod 1)

The sentence x = y (mod n) means that n is a divisor of x - y. This

sentence is read, "x is congruent to y modulo n." It is something

like a remainder, because if you subtract a remainder from the

dividend, the divisor will go into the result evenly.

try again

[wolfgang]

Hi friends....

do you know that (1)exponent(0) =1

and (1)exponent(1) =1

so....

1 exponent 0=1exponent 1

There is a rule that says:if the bases are the same, then the exponents must be the same also,

thus..

0=1

or 1=2

Edited October 31, 2010 by wolfgang

[Guest]

Hi friends....

do you know that (1)exponent(0) =1

and (1)exponent(1) =1

so....

1 exponent 0=1exponent 1

There is a rule that says:if the bases are the same, then the exponents must be the same also,

thus..

0=1

or 1=2

1^m=1^n does not imply that m = n since the rule you are using does not apply when 1 is the base because ln(1)=0. So what you are really saying is

1^m = 1^n => ln(1^m) = ln(1^n) => m*ln(1) = n*ln(1) => m = n. Of course the last implication is false because as stated earlier in this thread you cannot divide by 0.