So... Who likes math?

[OneJoeZee]

If

a = b

Then,

a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1



:biglaugh: :icon_mad: :icon_surp :icon_surp :icon_mad: :biglaugh:

 

[Big Wang Bandit]

I don't, thanks for reminding me of school.

 

[OneJoeZee]

Obviously, our knowledge of mathematics does not allow for two different numbers to be equal to each other so the proof is false.

To whoever finds where the proof fails, I'll be your friend forever.








Or at least say good job...

 

[GeneStarWindGSW]

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

there's something wrong here lol

 

[GotToyota?]

Fuck math! :icon_conf

If I was in school and actually had to think, I could figure it out.

-Matt

 

[OneJoeZee]

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

there's something wrong here lol

a+a = a just means that 2 = 1 and that is the solution, which is wrong, but that's not where the problem lies.



going from (a+b) = b to (a+a) = a is mathematically sound since the given information states a = b.

You're on the right track though. Keep going.

 

[speed]

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

there's something wrong here lol

exactly, everything is right except for those two. a+a=2a ; a+b=2b.

thus, 2a=2a and a=a. Not that hard, and I got kicked out of my engineering program cause I can't do tripple integrals or differentiate!

 

[speed]

a+a = a just means that 2 = 1 and that is the solution, which is wrong, but that's not where the problem lies.



going from (a+b) = b to (a+a) = a is mathematically sound since the given information states a = b.

You're on the right track though. Keep going.

That cannot be true. If a + b = b, then A would have to equal zero. Which would mean that b equals zero. Well, in that case yes the whole thing works out. But it can only work if a=b=0

 

[OneJoeZee]

The responses, so far, are not the right answer to the proof's falsehood.

 

[OneJoeZee]

But it can only work if a=b=0

Close... but no cigar

 

[SupraDerk]

I want to answer...

 

[OneJoeZee]

Feel free.

 

[SupraDerk]

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

 

[OneJoeZee]

You are 99% there but you haven't told me what rule used in the proof does not exist in math.


Everything you stated that is wrong is technically right given that the step you pointed out to be faulty, which is the correct step so good job, is a real mathematic rule.

Tell me what is wrong with the 3rd to 4th step and you will win the prize.

 

[GeneStarWindGSW]

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

if all that math is write you wanna do my math h.w for the rest of my life? lol

 

[SupraDerk]

You are 99% there but you haven't told me what rule used in the proof does not exist in math.


Everything you stated that is wrong is technically right given that the step you pointed out to be faulty, which is the correct step so good job, is a real mathematic rule.

Tell me what is wrong with the 3rd to 4th step and you will win the prize.

going from the first step to the second step... I have no idea what law/rule it is trying to be. (and if it is, then I've NEVER heard of it... unless completing a square... but even that's still different)

a^2 = a*b
a^2-b^2 = a*b-b^2

 

[OneJoeZee]

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

(a+b) = b <----this step is correct given that the above step is true. but it is not. why? (you're so close)

 

[OneJoeZee]

going from the first step to the second step... I have no idea what law/rule it is trying to be. (and if it is, then I've NEVER heard of it... unless completing a square... but even that's still different)

a^2 = a*b
a^2-b^2 = a*b-b^2

That's simple.

a = b

a^2 (a squared) = a*b <---- If a = b, then 'a' times 'a' or otherwords A^2 is equal to 'a' times 'b' (a*b).

See? That step is sound. The next step is where the proof is fubar, as you pointed out, but I still haven't heard the correct reason.

All steps after step 3 are theoretically correct... GIVEN that step 3 is right.

In reality, the proof stops at step 3 and cannot proceed any further.

 

[SupraDerk]

That's simple.

a = b

a^2 (a squared) = a*b <---- If a = b, then 'a' times 'a' or otherwords A^2 is equal to 'a' times 'b' (a*b).

See? That step is sound. The next step is where the proof is fubar, as you pointed out, but I still haven't heard the correct reason.

I don't doubt that hence the

a^2-b^2 = a*b-b^2 ... a*b is equivalent to a^2 from above... that's ok too

Hah, that's why I said going from the first to the second step (I guess I should have said, first line to second line)... adding in the -b^2 on both sides.

If you're going to do that it's fine I guess, but it's a constant and can't effect the values (the change will be the same on both sides... so if you try to move the added value to the other side it will negate itself)

 

[OneJoeZee]

Colder than you were a few posts ago. In hindsight, your post that the whole thing is zero is incorrect but you were on the right track. The real solution does not lead to zero.