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# Why 1 = 2

30 replies to this topic

### #1 rookie1ja

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Posted 30 March 2007 - 03:37 PM

Why 1 = 2 - Back to the Cool Math Games
Find the mistake in these mathematical equations.
x = 2
x(x-1) = 2(x-1)
x2-x = 2x-2
x2-2x = x-2
x(x-2) = x-2
x = 1

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.
Pls visit New Puzzles section to see always fresh brain teasers.

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### #2 kola

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Posted 20 April 2007 - 07:17 PM

x(x-2) = x-2

implies that either x = 1 or x = 2. Since x = 2, x = 1 does not have to be true. Therefore, concluding x = 1 from above equation is wrong.
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### #3 rookie1ja

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Posted 20 April 2007 - 07:43 PM

x(x-2) = x-2

implies that either x = 1 or x = 2. Since x = 2, x = 1 does not have to be true. Therefore, concluding x = 1 from above equation is wrong.

x(x-2) = x-2 ... then dividing the equation by (x-2) ... this is what can't be done ... we can't divide by zero
x(x-2)/(x-2) = (x-2)/(x-2) ... which can be simplified to
x = 1
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### #4 Raven

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Posted 28 April 2007 - 04:07 AM

Technically there is no error.

If anything is wrong, it was introducing x into the multiple in the second line. By including x, a quadratic equation was created, which will always have 2 solutions for x.

If you solve the quadratic equation, the 2 solutions are 1 and 2.

As we already knew the stated value for x from line 1, applying this to line 2 removes x as an unknown variable and hence removes the possibility of multiple potential values for x.
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### #5 fosley

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Posted 07 May 2007 - 09:55 PM

There are two solutions. Both can be gotten with the quadratic equation (I don't remember how anymore), or they can be gotten the way you did it. There's your solution which gets x = 1, then there's this solution which gets x = 2:

x(x - 1) = 2(x - 1)
x(x - 1) / (x - 1) = 2(x - 1) / (x - 1)
x = 2

Notice that I did the exact same thing you did, but I didn't re-arrange it to get the other solution. Now, we can verify that both x = 1 and x = 2 are correct:

Substitute for x = 2:
x(x - 1) = 2(x - 1)
2(x - 1) = 2(x - 1) which is true just because both sides are the same
2(2 - 1) = 2(2 - 1)
2(1) = 2(1)
2 = 2
Since the result is true, X = 2 is true.

Substitute 1 for x = 1:
x(x - 1) = 2(x - 1)
1(1 - 1) = 2(1 - 1)
1(0) = 2(0)
0 = 0
Since the result is true, X = 1 is true.

Of course, since we were already given that x = 2, there was no need to do anything but substitute to verify that x = 2 is correct. So the obvious mistake is that you were trying to solve for x when you already knew its value.
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### #6 TheOnlyOne

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Posted 19 May 2007 - 09:37 PM

This has nothing to do with why 1=2 it just shows how you can input a one or a two into and equation and it comes out equal on both sides, and since plugging in one gave you zero=zero and plugging in two gave you two=two they are not equal.
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### #7 Garrek99

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Posted 22 May 2007 - 12:29 AM

Fellows, that's not what the puzzle is about at all.

Roockie is introducing a proof that shows that 1 = 2.
There is obviously something wrong with the proof.
AND that what is wrong is that you can not divide both sides of an equation with (x-2) when x=2 because you'll get ZERO (0) and you are not allowed to divide by "0". { x/0 = undefined }

Tangents: the equation x(x-1)=2(x-1) has 2 solutions but that is unrelated to anything. Because from the get go the host announced that x=2. And this equation was a manipulation of x=2. And you can not get a new value for x because x only = to 2 and nothing else.
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### #8 Incognitum

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Posted 23 June 2007 - 12:45 PM

Alright, try this one on for size:

X=.9999999...

10x=9.99999...
(10x)-x=9
(9x)/9=1
x=1

.999999...=1
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### #9 Garrek99

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Posted 24 June 2007 - 02:21 AM

Alright, try this one on for size:

X=.9999999...

10x=9.99999...
(10x)-x=9
(9x)/9=1
x=1

.999999...=1

I see where the problem could lie. This one sure is an interesting one.
I was starting to think that some of the arithmetical principals we accept are dogmatic.

^^^^^^^^^^^^^^^^^^^^^^^^^^__
The difference between 1 and 0.9 we can not represent in a number format. At least the level of math that I have reached I have never encountered a way of putting a number like that down on paper. (unless you accept 1 * 10^-infinity)
^^^^__
1 - 0.9 = is a theoretical number.
I could use some feedback here. Am i correct here or no?

Therefore:
10 * that theoretical number can not be the same as the "theoretical number" unless that number is 0.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^__
So, 10x - x /= (does not equal) 9x if and only if x = 0.9.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^__
10x = 10 - 10 * "theoretical number (1 -0.9)" and
^__^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^__
9.9 = 10 - 1 * "theoretical number (1 -0.9)" and since those 2 things are not equal that's where I think the problem lies.
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### #10 Incognitum

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Posted 25 June 2007 - 07:59 AM

9.9 = 10 - 1 * "theoretical number (1 -0.9)" and since those 2 things are not equal that's where I think the problem lies.

Sorry dude,

In mathematics, the recurring decimal 0.999… denotes a real number equal to 1. In other words, "0.999…" represents the same number as the symbol "1". The equality has long been accepted by professional mathematicians and taught in textbooks. Various proofs of this identity have been formulated with varying rigour, preferred development of the real numbers, background assumptions, historical context, and target audience.

I do not understand your objection, and I think you're attempting to be contrarywise. However, here it is in proof form to overcome your objections.

Proof: 0.9999... = Sum 9/10^n
(n=1 -> Infinity)

= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)

= lim .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)

= lim .9(1-10^-(m+1))/(9/10)
(m -> Infinity)

= .9/(9/10)

= 1

Perhapse you'd like to add two 0.99999 and get 2?

0.999... + 0.999... = Sum[n = 1 to infinity](9*10^-n)
+ Sum[n = 1 to infinity](9*10^-n)

= Sum[n = 1 to infinity](18*10^-n)

= Sum[n = 1 to infinity](10*10^-n + 8*10^-n)

= Sum[n = 1 to infinity](10^(-n+1))
+ Sum[n = 1 to infinity](8*10^-n)

= Sum[n = 0 to infinity](10^-n)
+ Sum[n = 1 to infinity](8*10^-n)

= 1 + Sum[n = 1 to infinity](10^-n)
+ Sum[n = 1 to infinity](8*10^-n)

= 1 + Sum[n = 1 to infinity]((8+1)*10^-n)

= 1 + Sum[n = 1 to infinity](9*10^-n)

= 1.999...

But you can't, you get 1.9999... interesting eh?

Perhapse if that does not satisfy, you could see it this way.

1.000000000...
-0.999999999...
=0.000000000...

You may ask about the "1" in the last decimal place, and I'll write that when I'm done writing an infanate string of "0"s. So the 4th decimal place has a "0", and the 4 billionth has a "0", and the decimal place corrisponding to any number you could or could not name or imagine has a "0." It can be seen that a number with a "0" in every decimal place must be equal to 0.

Or there is a quite technical and compleate work up of many difrent proofs and systems to arrive at such conclusions at Wikipedia

http://en.wikipedia.org/wiki/0.999...
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