I proved 2=1 , logicaly it cant be - but mathematicaly I made no mistake - can you assure me on that !!?

Question

Is this derivation really correct? Or is there somewhere a mistake? If so, where?

2 = 1
******

Suppose: a = b

It then follows that

a * a = a * b

a^2 = a * b

a^2 - b^2 = a * b - b^2

(a - b) * (a + b) = (a - b) * b

a + b = b

a + a = a

2 * a = 1 * a

2 = 1

Answer 1

Hey there!

You divided both sides by a-b. Since a=b, then literally you are dividing both sides by a-a or 0. This is where the error occurred.

Hope it helps!

Answer 2

Hie Green day - Great Question !!!!!!!

No matter how you look at it, the problem in the "proof" is that zero behaves differently, and no account is taken of that.

You are suggesting that it is in the step I would simply call
"factoring" that something "goes wrong." But let's look at the whole proof with an eye on the fact that a=b, so that we are aware of what is zero:::::::::

Written
********
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

Reality
********
a = a
a^2 = a * a
a^2 - a^2 = a*a - a^2
( a + a )( a - a ) = a( a - a )
a + a = a
a + a = a
2a = a
2 = 1

Judgment
*************
True
True
True { = 0 }
True { = 0 }
false { unless a = 0 } !!!!!
false
false
false

Go through that carefully; you see that the factored step is the last one that is true. It has not introduced the error, but it sets things up for an error, by putting the equation into a form in which we are multiplying by zero. It also introduces the 2, in a hidden form; so you are right in saying that this is not a coincidence. It's sort of like a burglar entering your house. When he walks up your street, he has not yet broken in, but it is no coincidence that he is there. At this point in the "proof," the burglar is on the scene, but isn't in yet.

Now, you are right that at this step we have, formally,

2a * 0 = a * 0

and it already shows signs of the problem to come; but it is not wrong yet, just as

2 * 0 = 1 * 0

is not wrong. It is only if you forget that 0 times anything is 0, and that zero is special at this point, that there is a problem, and that is exactly what is forgotten in the next step.



The problem, of course, is that what you call the "division property of equality" has a condition::::::::::::::

if ab = ac, AND a is not zero, then b = c
If we omit that condition, then it is not true, because
0b = 0c = 0 for ALL b and c, regardless of whether b = c

The error in the proof is that this condition has not been checked, so that a false "property" was applied.

Your reasoning depends entirely on your condition:::

If you momentarily assume that 0a may not equal zero for all a, ...

That is just what I am saying, except that you seem to be considering it possible that this assumption could be true. Since it is not, all the rest of what you say really doesn't mean much. I think it's a lot cleaner to say, as I have, that factoring just sets us up to make the mistake, by unwittingly making this assumption, rather than that there is something in the factored equation itself that implies that
2=1.

Answer 3

I don't quite agree / satisfied with what other answerers answer !!

I was browsing the YAHOO ANSWER and saw the "proof" that 2=1. I read how, since two does not equal one, somewhere along the line an error was made. It was then concluded that the source of error was division by zero. Here is the proof I am discussing:

a=b Given

a^2=ab Multiplication Property of Equality

a^2-b^2=ab-b^2 Subtraction Property of Equality

(a+b)(a-b)=b(a-b) Sum & Difference Pattern/Distributive
Property

a+b=b Division Property of Equality

b+b=b Substitution Property

2b=b Combination of like terms

2=1 Division Property of Equality

I suppose division by zero makes sense as a valid reason for fallacy, but how do we know for sure that it is the cause? The Sum and Difference Pattern also makes sense, but when a=b I found that the solution actually changes:

a+b)(a-b)=a^2-b^2 Sum and Difference Pattern

(a+a)(a-a)=a^2-a^2 Substitution Property

a(1+1)a(1-1)=a^2(1-1) Distributive Property

2a^2*0=a^2*0 Combination of Like Terms, Inverse -
Axiom of Addition

Notice that here the LHS is exactly twice the RHS if you disregard the Property of Zero in Multiplication, which states that 0a=0. If you momentarily assume that 0a may not equal zero for all a, then the fact that the LHS is twice the RHS accounts for the fact that the solution to the false proof has an LHS twice the RHS. Therefore, if you assume that 0x=0y is only true when x=y, the Sum and Difference Pattern is revised to (a+b)(a-b)=2(a^2-b^2) when a=b. If you substitute this into the equation in step four, the solution is a true identity.

Is it possible that this is the reason that the proof is false?
Otherwise it would seem hard to believe that the fact that the Sum and Difference Pattern's "identity" when a=b and the "identity" of the false proof are proportional is just a coincidence. Perhaps the difference between 0x and 0y where x does not equal y is so minute that it cannot be perceived from our perspective and is impossible to
tell apart, but can be viewed from our perspective and made apparent by canceling out, or dividing, by zero? Is it possible to assume that the error in this false proof may have resulted from something other than division by zero?

Thank you.

Answer 4

What the others have said is almost correct. Division by 0 isn't defined, but tends towards infinity. The limit of 5/x as x goes to 0 is infinity. So this isn't why the proof fails.

The problem is that you have (a-b)/(a-b).

This is 0/0. The subtle difference is that this doesn't tend towards infinity. This is what is known as an indeterminate form. Other examples are infinity/infinity, 0^0, 0^infinity and a few more.

As you can probally tell from the good old English usage of indeterminate, 0/0 doesn't tend to any specific quantity, rather it can go to anything depending on the context.

Answer 5

Hi,
I've seen several claims that supposedly prove that 2=1, but all of them have some violation of math procedures in them. Yours has one also:

Step 4) (a - b) * (a + b) = (a - b) * b

Step 5) a + b = b (Divide by (a-b) )
To get to step 4 from step 5 you divided by (a-b) as I have noted. Since a=b; then a-b = 0, and we all know that division by zero is not allowed.

Nice try though.

FE

Answer 6

When you went from this:
(a - b) * (a + b) = (a - b) * b

To this:
a + b = b

you divided by (a - b). But a- b is zero, and division by 0 is not defined. It is not defined because if it was, it would create problems like this.

If you go through the steps with actual numbers, it's easy to spot the mistake.

Answer 7

i know it has been proven somehow that 2=1, my algebra 2 teacher has demonstrated it before. so yes, it is mathematically possible. *however* i don't think you have enough evidence to say that a+b=b or a+a=a. i could be wrong. good luck.