So... Who likes math?
- Thread starter OneJoeZee
- Start date
(a+b)(a-b) = b(a-b) does not mean (a+b) = b. What you are doing is dividing the first equation by zero (a-b) to give yourself the second equation, which is mathematically incorrect.
And THAT's why this proof is false.
It didn't take abstract algebra in university for nothing
And THAT's why this proof is false.
It didn't take abstract algebra in university for nothing
OneJoeZee said:
It being equal to zero just means that a and b are equal.
At every step they're equal...
a^2 = a*b ... a^2 - a*b = 0 ... they're equal
a^2-b^2 = a*b-b^2 ... (a^2-b^2) - (a*b - b^2) = 0... they're equal
(a+b)(a-b) = b(a-b) ... (a+b)(0) = b(0) = 0 ...expanded they're equal
(a+b) = b ... wrong because third step is wrong
a+a = a ... wrong
2a = a ... wrong
2 = 1 ... wrong
I don't see where the problem in step 3 is.
(a+b)(a-b) = b(a-b)
=
(2a)(0)=b(0)
0* anything is 0, so that step has no errors.
On a side note, I love figuring out stuff like this, but I get so freaking frustrated with proofs cause I'm never right! DAMN YOU MATH!
edit, and I still suck at math. Good going Rob for getting it right, except for I'm still convinced that its bullshit.
(a+b)(a-b) = b(a-b)
=
(2a)(0)=b(0)
0* anything is 0, so that step has no errors.
On a side note, I love figuring out stuff like this, but I get so freaking frustrated with proofs cause I'm never right! DAMN YOU MATH!
edit, and I still suck at math. Good going Rob for getting it right, except for I'm still convinced that its bullshit.
annoyingrob said:(a+b)(a-b) = b(a-b) does not mean (a+b) = b. What you are doing is dividing the first equation by zero (a-b) to give yourself the second equation, which is mathematically incorrect.
And THAT's why this proof is false.
It didn't take abstract algebra in university for nothing
^Winner.
It really was that simple. You cannot divide by zero. That would cause an undefined solution. Therefore, you could not go any further than step 3. but if you COULD like the fautly proof shows, you would get the remaining steps that follow.
Derk, you were so close when you were talking about zeros a few posts ago.
At least you identified that step 3 is where the problem lies but Rob gave the correct reason why it's not right.It's really hard for most to see. The proof just looks like basic cancellation at step 3 to get rid of (a-b).
Last edited:
The problem isn't in step 3. The problem is going from step 3 to step 4. You cannot simply take away a factor of (a-b), as you would be dividing by zero.speed dodici said:
OneJoeZee said:^Winner.
It really was that simple. You cannot divide by zero. That would cause an undefined solution. Therefore, you could not go any further than step 3. but if you COULD like the fautly proof shows, you would get the remaining steps that follow.
Derk, you were so close. when you were talking about zeros a few posts ago.
Let's break it down...
a^2 = a*b ... a*b (b=a) = a^2... ok
a^2-b^2 = a*b-b^2 ... a*b is equivalent to a^2 from above... that's ok too
(a+b)(a-b) = b(a-b) ... ahhhh snap... here's the problem
(a+b) = b ... all this junk is wrong
a+a = a ... wrong since the above is wrong
2a = a ... still wrong...
2 = 1 ... WRONG!
a-b = a - a = 0... (a+b)(a-b) = 2a*0 = b*0... so the whole thing is zero
I came to that conclusion 2 pages ago!!!
That wasn't a fight for money, only Joes friendship was involved, and that isn't a big thing, so don't get irritated
The solution is not zero. Read what Rob wrote. Step 3 to 4 shows cancellation by division but it's dividing by zero. That results in an undefined solution, not zero.
Whether step 4 and below are right or wrong is incosequential.
Whether step 4 and below are right or wrong is incosequential.
valgekotkas said:
If in the middle of the proof you end up with both sides equaling zero... but in the rest of the proof afterwards it shows something else... I think that would show that there's something very wrong...
Yeah, ALL of the steps are CORRECT. ALL of them. The incorrect part is reducing step 3 to step 4.
(a+b)(a-b) = b(a-b) != (a+b) = b
(a+b)(a-b) = b(a-b) != (a+b) = b
Sry, I deleted the post, cause I couldn't rephrase it. lets try... You showed, that something is wrong, but you didn't show why exactly that was wrong. The reason why 0 isnot 2=1.SupraDerk said:
SupraDerk said:
You have to look at the proof in relation to the steps. Telling me that the expanded equation is equal to zero, which is ok, doesn't explain why step 4(in particular) and the following steps that result in 2 = 1 are false.