1 = 2 ...?

[freQ]

While all of that makes for an amusing equation I fail to see how it is relevant to the discussion. BTW, Welcome to the forum. Have you read the FAQ yet?

 

 

Please accept my apologies for the "hit and run". =)

[Qfwfq]

Nothing, it was just a nitpick on wording.

Quite right, no sense in word picking... let's see if I can put the whole thing more sensibly; previously I strove to be brief and concise, as I am accustomed to doing so.

 

The very paradox to be resolved means that, to be brief, I used a bit of Orwellian doublethink. As we know that x can't be 1 and also 2, when we regard 1 as a value of x, we must regard 'x = 2' hypothetically as an equation, without considering its solution x = 2. The same goes the other way around: when we regard 2 as a value of x, we must regard 'x = 1' hypothetically as an equation, without considering its solution x = 1. This is why I don't say:

the statements x=2 and x(x-1) = 2(x-1) are only equivalent for x = 2.

while I do say that they are equivalent only if x does not equal 1, and I don't consider them

not equivalent for other values of x either.

in the sense that, if the value of 'x - 1' isn't zero, I can call it 'a' so the above step is like:

 

x = 2 ==> ax = 2a

 

Let's try a better and less confusing treatment of the problem: we know that the two equations:

 

x = 2

 

and

 

x = 1

 

are not equivalent, as they trivially have incompatible solutions. The answer is to show that, although the intermediate steps aren't actually wrong, they aren't fully right either. At least one of the steps isn't between two equivalent equations, this happens for the steps that Zadojla and I pointed out. Indeed, the following two equations:

 

x = 2

 

and

 

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

 

are not equivalent because the second one has two solutions, the first only has one of them. Much the same goes for the step Zadojla pointed out.

[MortenS]

x = 2

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

x2-x = 2x-2

x2-2x = x-2

x(x-2) = x-2

x = 1

 

it looks right but there is a mistake in there -> can you see it?

 

Let me first write this out so I understand it :)

 

You start with x = 2

You then multiply with (x-1) on both sides

 

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

 

which leads to:

x^2 - x = 2x - 2

 

then you add (x-2x) on both sides

 

x^2 - x + (x-2x) = 2x -2 + (x-2x)

which leads to

x^2 - 2x = x - 2

 

You factorize the left side

 

x*x - 2*x = x - 2

x(x-2) = x -2

 

 

 

You then divide by (x-2) and this is the fallacy...You have defined x= 2, so dividing by x-2 equals division by zero.

 

it is important to remember that x=2 during the whole procedure, no matter how much the expressions looks like equations.

 

Substiting with 2 for every x we get:

 

2 = 2

2(2-1) = 2(2-1)

2*2 -2*1 = 2*2 -2*1

2^2 - 2 = 2*2 - 2

 

2^2 - 2 + (2-2*2) = 2*2 -2 +(2-2*2)

2^2 - 2*2 = 2 - 2

2 (2 - 2) = 2 - 2

 

2(2-2)/(2-2) = (2-2)/(2-2)

2*(0/0) = 0/0

 

The error lies in claiming that 0/0 equals 1...

[Qfwfq]

it is important to remember that x=2 during the whole procedure, no matter how much the expressions looks like equations.QUOTE]Who says they aren't equations? :o

 

To me, the simplest way is to say that:

 

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

x2-x = 2x-

x2-2x = x-2

x(x-2) = x-2

 

are equivalent equations, of second degree, having the following two solutions:

 

x = 1

x = 2

 

:)

[MortenS]

it is important to remember that x=2 during the whole procedure, no matter how much the expressions looks like equations.QUOTE]Who says they aren't equations? :o

 

To me, the simplest way is to say that:

 

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

x2-x = 2x-

x2-2x = x-2

x(x-2) = x-2

 

are equivalent equations, of second degree, having the following two solutions:

 

x = 1

x = 2

 

:o

 

I spoke to quick...of course it is equations....