A=e is it true?

[hammertime]

O.K. - I'm doing the same to both sides of the equation. Why am I getting this answer?

 

a=x

a+a=a+x

2a=a+x

2a-2x=a+x-2x

2(a-x)=a-x

2(a-x)/a-x=a-x/a-x

2=1

We all know that two does not equal one.

[hammertime]

Does it have something to do with dividing by zero?

[CraigD]

Does it have something to do with dividing by zero?

Yes, you have answered your own question. :) Dividing by zero isn't allowed for ordinary numbers, so your 2=1 result isn't correct.

[hammertime]

Thanks..

 

Although it does look weird and like it should work. I felt like I was missing something easy.

[Tormod]

Yeah, you were missing something easy. :)

[ughaibu]

Hammertime:

There's a short thread on this kind of thing: http://hypography.com/forums/watercooler/7099-false-proofs-mathematical-philosophical.html

[hammertime]

Thanks Ughaibu. In that thread, you say that you know how this falsehood works. Ron says he knows too, but neither one of you elaborated. Tormod just said that I am missing something easy. ???

[ughaibu]

Think about what you did in line 6.

[hammertime]

Dividing by (a-x) which in essence is zero?

[ughaibu]

Yes.

[hammertime]

Thanks. I got that. I thought there was something else.

[CraigD]

Thanks Ughaibu. In that thread, you say that you know how this falsehood works. Ron says he knows too, but neither one of you elaborated. Tormod just said that I am missing something easy. ???

The “how it works” of the contradiction generating algebra in this thread and the older 7099 share explanations involving division by zero. There are several common-sense and more formal explanations of why division by zero isn’t allowed in algebra involving complex or real numbers:

• Algebra isn’t guaranteed to work when you permit division by zero. If you think of using the rules of algebra as something that guarantees correct results, allowing a division by zero step “voids the guarantee”, and you can no longer rely on its correctness
  
• An equation involving division by zero results has one or more indeterminate variables, for which we can’t determine a value. Note that an indeterminate variable is not necessarily a bad thing, mathematically, just a statement that a particular equation involving the variable doesn’t tell us exactly it’s value, that is, for an indeterminate variable [math]a[/math], the equation can be manipulated into the tautology [math]a = a[/math].
  
• Dividing by zero requires a infinite, or transfinite number, for which the usual rules of arithmetic and algebra are different than with complex and real numbers. For example, for a transfinite number [math]b[/math], [math]\frac{2b}{b} \not= 2[/math] and [math]b=b+1[/math] can be considered true.
  
• Having zero terms (eg: [math](a-b)[/math] when [math]a=b[/math]) in an equation result in lost of information. This is sometimes also described as performing a non-reversible transformation.
  

How a given person explains it to their satisfaction is rather an individual choice, so long as the result is that they are conscious of the limitations it imposes on algebra using particular kinds of numbers.

[Little Bang]

If a=x then a-x=0 therefore a-x/a-x=0, 0=0

[CraigD]

If a=x then a-x=0 therefore a-x/a-x=0, 0=0

LB, from where do you get the idea that [math]\frac{0}{0} = 0[/math]? It’s not conventionally accepted math.

 

There are several possible lines of reasoning concerning the value of a variable, [math]y = \frac{0}{0}[/math], including:

• [math]y = 1[/math]. Reason: [math]\frac{\mbox{anything}}{\mbox{anything}} = 1[/math]. This is the reasoning that lead hammertime, in post #1, to the contradictory conclusion [math]1=2[/math]
  
• [math]y = 0[/math]. Reason: [math]\frac{0}{\mbox{anything}} = 0[/math]. This is what Little Bang is asserting
  
• [math]y[/math] is indeterminate – that is, it’s value can’t be determined from the equation [math]y = \frac{0}{0}[/math]. This is equivalent to saying “you can’t divide by zero”, and is the usual, accepted conclusion for arithmetic of real numbers.
  

If we accept that [math]\frac{0}{0} = 0[/math], the sequence of steps in post #1 do avoid a contradictory result, however:

[math]a=x[/math], [1] given

[math](a) +a = (x) +a[/math], [2] you can add the same number to both sides of an equation

[math]2a = x+a[/math], [3] you can group common terms

[math](2a) -2x = (x+a) -2x[/math], [4] add the same number to both sides

[math]2(a-x) = a - x[/math], [5] extract common constant, group common terms

[math]\frac{(2(a-x))}{a - x} = \frac{(a-x)}{a - x}[/math], [6] you can divide both sides by the same number

[math]0 = 0[/math], [7] assuming [math]\frac{0}{\mbox{anything}} = 0[/math].

 

You can avoid the division by zero controversy by just noting that in [5], [math]2(a-x) = 0 = a-x[/math], and stopping there by recognizing that [math]0 = 0[/math] is a tautology (always true) and doesn’t contain any information about the variables. You might call this the “stop when the equation doesn’t contain any variables” rule. :lightbulb

[Rade]

Is the OP question an example of deriving the Law of Identity ?

 

So, back to step # 1, before you can claim a = x, first you must start with as axioms, a = a and x = x. So:

 

START

(1) a = a ; axiom (Law of Identity)

(2) x = x ; axiom (Law of Identity)

let,

(3) a = x

(4...6)

and as stated above by CraigD you reach step:

(7) 0 = 0 ; axiom (Law of Identity)

STOP

 

which is where you started in steps 1 & 2--the Law of Identity is derived from the Law of Identity, a tautology (which is not a bad thing, much better than deriving a contradiction).

 

So I think the OP is a good example of how we can get into trouble in mathematics (such as moving to step #8 trying to divide by 0) when we ignore the importance of stopping all logical derivations when a step yields the Law of Identity (a more general rule than stopping when the equation does not contain any variables as suggested by CraigD).