Comments on: Math Optical Illusion

By: dido1983

dido1983 — Thu, 22 Oct 2009 09:55:30 +0000

Besides, 2a=a equals to 2a-a=0 a=0, not to 2=1.

If two dollars have the same value as one dollar, then those dollars are FAKE!

By: Balli

Balli — Sat, 03 Oct 2009 02:50:21 +0000

It is not given that a=b=1. That is the only thing that is wrong. And then there is the problem that the only solution is that a=b=0 so you are dividing with 0 which is a problem as the answer infinite.

By: anonymous

anonymous — Sun, 27 Sep 2009 20:28:10 +0000

It is because they divided by a-b in the 4th step and because a=b that means they divided by zero, which gets UNDEFINED not any real answer. Duh!

By: Ms.Observant

Ms.Observant — Wed, 23 Sep 2009 12:58:12 +0000

them dudes try devide by zero. Tards. I IS IN 8TH GRADE AND ME FIND ANTSA! NO ZERO DIVISION PPL!

By: ben

ben — Sun, 09 Aug 2009 15:27:05 +0000

(a+b)(a-b)=b(a-b) ….. isnt tht bit rong?

By: A Math Major

A Math Major — Tue, 14 Jul 2009 21:46:09 +0000

Sigh… the answer, which very few people have correct, is that you cannot divide by 0. ALL of the math here is sound, EXCEPT that one of the steps is not reversible (divide by 0). That is in going from line 4 to 5. If you notice, lines 1-4 are all true as long as a = b, which was the premise. Starting at line 5, after the division by 0, everything is only correct if a = b = 0, which was not a stated condition.

By: Samantha

Samantha — Fri, 03 Jul 2009 07:14:07 +0000

Don’t blame me for not getting this i’m still in primary school…(age 11) tho some of my classmates have that a – b = ? some sort of things…

By: qqqq

qqqq — Sun, 21 Jun 2009 02:16:35 +0000

if a=b , then a and b are equal, meaning that THEY ARE THE SAME NUMBER!
therefore, the answer cant be 2=1

By: Eluem

Eluem — Sat, 20 Jun 2009 06:41:19 +0000

“this is not an optical illusion,they just try to confuse you and check if you at lest remmember 6grade maths.everything that’s below line 3 is not right:
(a+b)(a-b) will never equale the statement b(a-b)
always (a+b)(a-b)=a2+b2″

First of all this must be corrected:
“always (a+b)(a-b)=a2+b2″
should read:
“always (a+b)(a-b)=a^2-b^2″

In this situation, this isn’t the error.
b(a-b) is equal to ab-b^2 and since a=b this could also read a^2-b^2.

The real error is as it has been said before, when they divide by zero by trying to cancel by dividing the (a-b) over since a=b, a-b = 0. You cannot divide by zero.

By: Callum Teh Robertson

Callum Teh Robertson — Fri, 24 Apr 2009 22:51:00 +0000

They missed out the parameters – "a=b=1" (or if you prefer "a=1", "b=12)

Need to state what a & b are ;) otherwise it is technically correct and then the last line is also wrong… you cant just assume 'a' is equal to one – needs to be stated… yes, im a uni student with nothing better to do ;)