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Solve this math equation!

Posted: Fri Nov 06, 2009 11:55 pm
by Boredness
I found this on the net...

Given: a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b) (a-b) = b(a-b)
(a+b) = b
(a+a) = a
2a = a
2 = 1


WTF?

Re: Solve this math equation!

Posted: Sat Nov 07, 2009 11:59 am
by iSean
I never knew Singapore Math so hard O_O....

Re: Solve this math equation!

Posted: Sun Nov 08, 2009 6:07 pm
by flee0308
division by 0 gives an error. If you do not understand what I mean, let (a-b)=0, thus you cannot divide by (a-b)

Re: Solve this math equation!

Posted: Mon Nov 16, 2009 11:36 am
by iZenna
(a+b) = b---(1)
(a+a) = a---(2)

from
(1) you cannot get to (2) like that

should be like this

(a+b)/b = 1

[a(a+b)]/b = a

thus,

[(a^2)+(ab)]/b = a


LOL boredness the one you found on the net FAIL la,
you cannot get an equation of (2) from (1)

a^2+ab = ab
a^2 = 0
a = 0

if a=0 b=0

Re: Solve this math equation!

Posted: Mon Nov 16, 2009 11:38 am
by Boredness
iZenna wrote:(a+b) = b---(1)
(a+a) = a---(2)

from
(1) you cannot get to (2) like that

should be like this

(a+b)/b = 1

[a(a+b)]/b = a

thus,

[(a^2)+(ab)]/b = a


LOL boredness the one you found on the net FAIL la,
you cannot get an equation of (2) from (1)

a^2+ab = ab
a^2 = 0
a = 0

if a=0 b=0
fail, a=1.

Re: Solve this math equation!

Posted: Mon Nov 16, 2009 11:41 am
by iZenna
Boredness wrote:
iZenna wrote:(a+b) = b---(1)
(a+a) = a---(2)

from
(1) you cannot get to (2) like that

should be like this

(a+b)/b = 1

[a(a+b)]/b = a

thus,

[(a^2)+(ab)]/b = a


LOL boredness the one you found on the net FAIL la,
you cannot get an equation of (2) from (1)

a^2+ab = ab
a^2 = 0
a = 0

if a=0 b=0
fail, a=1.
walao, you tell mine fail, yours more fail lor, 2=1, how can it be?

Re: Solve this math equation!

Posted: Mon Nov 16, 2009 11:47 am
by Boredness
fuck.. i said what's wrong with the eqation not SOLVE.

Re: Solve this math equation!

Posted: Wed Nov 18, 2009 6:39 am
by Susashiden
Boredness wrote:fuck.. i said what's wrong with the eqation not SOLVE.
by right (a-b)(a+b) = ab square ... not some random things .. he is trying to confuse us by doing one step wrong and without us knowing it :D

Re: Solve this math equation!

Posted: Wed Nov 18, 2009 7:27 pm
by guyz92
Given: a = b

a^2 = ab
a^2 - b^2 = ab - b^2
(a+b) (a-b) = b(a-b)
(a+b) = b

(a+a) = a <------- From this part, it does no have any logic.
2a = a <------- This part is the find it's ratio.
2 = 1

Adding both side will make (-b)^2 =0.
When both side have like terms (a-b), you can cross out(Law of Indices).
This question should be a question asking about ratio. It cannot be a proving question.

--------------------------------------------------------------------------------------
But i feel this way it should work out if it is a proving question:

a=b, (a^2= a x a)

a^2=ab
axa=axb
a=(axb)/a
Therefore
a=b

Re: Solve this math equation!

Posted: Thu Nov 19, 2009 8:27 pm
by flee0308
;) No, no, thats not the point. This is not to show a=b. The point is to find the mistake in the equaion, and this is it.
Boredness wrote:I found this on the net...

Given: a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b) (a-b) = b(a-b)
(a+b) = b
(a+a) = a
2a = a
2 = 1


WTF?

If you look at step 3, you would know that both sides of the equation is divided by (a-b). That cannot be done, as a=b, which means (a-b)=(a-a)=0. Division by Zero is not allowed in any case. Thus, the equation should be continued like this:

Given: a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b) (a-b) = b(a-b)
(a+b)(a-a)=b(a-a)
(a+b)*0=b*0
0=0

guyz92: There is NO such thing as "crossing out" in maths. "crossing out" is a just a more convenient mean of division. Further more, there is no Law of Indices about crossing out. Oh and there is a shorter step of solving the equation if it was a proving sum.

Proof that a=b from a^2=ab.

a^2=ab
Dividing both sides by a, a=b