Find the flaw in this.. (solving a=b)

[nahya^^]

I came across this problem and am VERY perplexed. So I guess 2 does equal 1.

Let a=b

(a^2) = ab
(a^2) + a^2 = (a^2) + ab
2(a^2) = (a^2) + ab
2(a^2) - 2ab = (a^2) - ab
{2[(a^2) - ab]} / [(a^2) - ab] = [(a^2) - ab] / [(a^2) - ab]
2 = 1

My notion is that in the process of dividing the equation by (a^2) - ab is the flaw.

On the left side of the equation of the 4th step (1st step being (a^2) = ab), it would seem that there are no flaws if there wasn't the coefficient 2 for ab. so (a^2) - ab will cancel out on both sides and 2 and 1 are left.

I also think that there are no flaws, it's just that you can't derive numerical answers from an equation composed of variables and nothing else. Therefore the actual answer to a=b is no solution.

What do you guys think?

P.S.: How do you guys make the number 2 in A^2 a superscript?
P.P.S.: Whoops. Sorry, guys, I forgot to type the dividing part.

[techgnome]

You are correct. Mathmatically, there is no flaw in it, as long as you leave it a & b. I first came across this in a college class 10yrs ago. We stepped through it one line at a time. It breaks no rules. However, we found that once you try to set a=some#, it begins to fall apart since your first instinct is to then start actually calculating things out, rather than letting it flow.

[opus]

You are breaking no rules?
I think you do, by dividing tru 0!

[krtxmrtz]

Originally posted by opus
You are breaking no rules?
I think you do, by dividing tru 0!

Opus is right, that's precisely the point. If this helps you folks to better grasp the idea, consider this way to see it:

0 = (A)(0) for any value of A -> 0/0 = A

i.e., if you divide 0 by 0 you can get anything or, put it another way, 0/0 is said to be "undefined". This is precisely what you have:

[(a^2) - ab] / [(a^2) - ab] = 0/0

This fallacy has been widely used to prove things like 1=2 and many others.

[Spooner]

One of the common ones takes this form:

25 - 25 = 5 - 5

The lhs is the difference of 2 squares, factorise it:

(5+ 5) ( 5- 5)= (5 - 5)

Cancel:

10 = 1

Once again, a division by 0 from the 5-5 term

[bugzpodder]

no one has yet found the error in my "proofs":

http://www.vbforums.com/showthread.p...hreadid=224922

[sql_lall]

How do you guys make the number 2 in A^2 a superscript?

just use the HTML tags [ sup ] and [ /sup ] before & after
e.g:
x [ s u p ] 2 [ / s u p ] --(without the spaces)
=
x²

you can also use [ s u b ]

[Shaggy Hiker]

While the point about dividing by 0 is valid, I don't really think that that is the crux of the problem. I think that the real issue is that (A^2)-AB=0 when A=B. Hence 2(A^2)-2AB=0 because 2*0 = 0. This comes back to the old saw that all equations are easy to solve since you can validly multiply both sides of an equation by a constant: Multiply both sides by 0...problem solved.

If equation1=equation2 then x(equation1)=x(equation2). This is true, but if x=0, then all equations evaluate to 0=0, which is true, but not very useful. In the example that started this thread, that is exactly what is happening, since the A^2 - AB term is 0 when A=B. I contend that it is this fact, not the divisor that makes the problem fail.

After all, this could be stated for any number of unreasonable situations that don't use division at all. Consider:

If
xA=xB (and all more complicated variants).

then A=B.

However, if x=0, then the original equation is true for all A and all B, but the second statement is not true if A<>B.


Well, I think I have beaten this subject long beyond death.

[Osiris]

Hey, i dont know if you guys solved this but
i think i saw an error in the beginning, i saw this kinda late huh?

Let a=b

(a^2) = ab
(a^2) + a^2 = (a^2) + ab
2(a^2) = (a^2) + ab
2(a^2) - 2ab = (a^2) - ab
{2[(a^2) - ab]} / [(a^2) - ab] = [(a^2) - ab] / [(a^2) - ab]
2 = 1

i got until:

a² = ab
a² + a² = a² + ab
(2a²) = (a² + ab)

but then, shouldnt the next part be:

(2a²) - 2ab = (a² + ab) - 2ab

inside the parenthesis are the numbers that equal each other...
because on the this line you are subtracting 2a² - 2ab = (should equal 0)
and (a²) - ab is not what you did on the other side of the
equation....
so the flaw is that the confusion was because when 2ab was
subtracted on 1 side, it wasnt done properly on the other...

[Osiris]

ok, just in case you are trying to play with ratios/proportions

2a² - 2ab = a² - ab //ok, i can see why you wrote it like this

2(a² - ab) = a² - ab //works for me!

[2(a² - ab)] / (a² - ab) = (a² - ab) / (a² - ab) //now this doesnt work for me

since you guys wanted to make the equation on the left twice as
much, you should also divide by TWICE as much so it should look
like this...

[2(a² - ab)] / [2(a² - ab)] = (a² - ab) / (a² - ab)
1 = 1

[MXK]

I'm not quiet sure that I see your point Osiris. However, if we go back to the idea that division by 0 results in infinity, the most basic explanation to this problem is that for any value a, given that a=b, 2*infinity does = infinity.