Phi day?

[Sangwa]

That's a funny one.

EDIT to explain nº2:

Hehe, my last try wasn't correct. I only cared about the end (a=2a) and didn't see it was (a=a^2)

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

This step here is wrong though. a(a-b) != a. Because a-b is 0 and a multiplied by 0 is 0.

[Taniquetil]

1 is equal to 2. Math clearly doesn't exist! Lie down and accept that nothing else can possibly exist! You are fake! 


a = b                              a = b
ab = b2                          ab = b^2
-(ab) = -(b2)                   -(ab) = -(b^2)
a2-ab = a2-b2                 a^2-ab = a^2-b^2
a(a-b) = (a+b)(a-b)         a(a-b) = (a+b)(a-b)
a = a+b                         a = a+b
a = 2a                           a = 2a
1 = 2                            1 = 2

i prolly may be definately wrong but i'm pretty sure a = a+b cannot transform to a = 2a....

[Velh Krome]

if a=b, a+b would be equal to a+a, which again indeed is the same as 2a.

but you mustnt divide by zero lol

[Sangwa]

If "a = b" then "a + b = 2a" is correct.

Oh, I just noticed something new. The step I noted was wrong, actually isn't. He simplified it correctly... Very strange! Intriguinging puzzle!

[Taniquetil]

if a=b, a+b would be equal to a+a, which again indeed is the same as 2a.

but you mustnt divide by zero lol

aha i see.....

wow thats crazeee stuuf 

[Velh Krome]

hm, i think you got it right, Sangwa, (a-b) is zero considering the included definitions, hence dividing by (a-b) is simply not allowed - or in different words, no valid mathematics lol

[Sangwa]

I believe the error here is in syntax (order of processes). The thought process seems to me to be erroneous because after you establish that a = b, you treat it like b != a and then you turn back to treating it like a = b. Like in a sentence, the math process here loses sense if there is not a logic succession of terms.

EDIT:
The same happens when you have (a-b) which is already established like 0 (because a=b), but is treated like something different.

[Velh Krome]

these operations are valid and correct to this certain step. apparently random transformations, except for the invalid division by zero, that simple. after all its about definitions (and hiding by variables).

[Sangwa]

Hey, so many people around this forum and we don't have math teachers? Can't you see we're having trouble here?

[Socius Rockus]

I do have one

[Izzabella]

I thought pi day was over...weeks ago.. why is this thread still so active! math sucks!

[SerqFeht]

  I had a version of that formula, but I didn't want to type it up, so I googled it. That was the wrong one! 

Here's mine:

x=0
a=x
b=x
a=x=b
a=b
(a-b)=(a-b)=0
(1)(a-b)=2(a-b)=0
Now take out those pesky (a-b)'s with some division
1=2

This is possible because there is some formula that lets the division by 0 work in a similar situation.

There are also other 1=2's that don't involve division, if you must...

[hitancrias]

Hey, so many people around this forum and we don't have math teachers? Can't you see we're having trouble here?

I do some tutoring in maths, I'll try to explain it.

The only problem is the division by 0, as has been said. It is not allowed to take the (a-b) part out of the equation because (a-b) has been defined as 0. Formally, when you divide both sides of an equation by an undefined variable, you should mention that the new form is only valid under the assumption that the variable does not equal 0. In fact this remark should be made every time you take out a variable by means of division.

For example, when you want to simplify xy=6x you get:
"y=6, under the assumption that x does not equal 0", and not just: "y=6x"

The reasoning of the puzzle can be summarized to:
x=0
1x=2x
1=2

The last line should include the "under the assumption that x does not equal 0"-remark. The assumption is not met, and hence the last form is not valid.

I believe the error here is in syntax (order of processes). The thought process seems to me to be erroneous because after you establish that a = b, you treat it like b != a and then you turn back to treating it like a = b. Like in a sentence, the math process here loses sense if there is not a logic succession of terms.

This is not the real problem. It's not illegal to maintain both the variables, even if the equation can be simplified by expressing one of them in terms of the other.
I hope it's more clear now.

[Taniquetil]

Hey, so many people around this forum and we don't have math teachers? Can't you see we're having trouble here?

I also have a math teacher

Many math teachers in fact

As you can see, none of them were any good

[Velh Krome]

SerqFeht,
your last construct comes to the same invalid result by dividing by 0, (a-b) that is. same thing. showing (a-b) to be 0 makes no difference: dividing 0 by zero ((a-b) that is) is invalid too.
why is dividing by 0 invalid? well, how often does 1 contain 0? yep, its not definable.