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can some one please elaborate on this?

YOu can do alot of crazy shit with math.

Like 1+1=3 etc.

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

although the ba is also the same as a squared, the ba shouldn't count for both the ba and a squared at the same time

At 5/1/09 02:27 AM, X0 wrote: can some one please elaborate on this?

You suck at math.

At 5/1/09 02:32 AM, Hornby wrote: (a+b)(a-b)=b(a-b)

although the ba is also the same as a squared, the ba shouldn't count for both the ba and a squared at the same time

umm what?

At 5/1/09 02:36 AM, X0 wrote:

At 5/1/09 02:32 AM, Hornby wrote: (a+b)(a-b)=b(a-b)

although the ba is also the same as a squared, the ba shouldn't count for both the ba and a squared at the same time

umm what?

Yeah, and it REALLY shouldn't count for a+b at the same time. Check what you did, you said ab=a^2, then you turned a^2 into a+b by removing the parentheses.

At 5/1/09 02:27 AM, X0 wrote: can some one please elaborate on this?

Mah GAWD!!! I'm surprised i never thought of that! of couse! so genius!

At 5/1/09 02:41 AM, GodsBitch wrote: B(A-B)

The 4th line is your problem. b(a - b) on the right hand side of the equation means that you are multiplying the b on the left by 0 since a = b that means that (a - b) = 0 so from there on your theory is false at the left side of the equation b(a-b) = 0.

It's hard to explain, but I hope you understood.

yeah i get it now

thanks!

Fixed.

At 5/1/09 02:49 AM, GodsBitch wrote:

At 5/1/09 02:41 AM, GodsBitch wrote: Stuff...

The point I was trying to make is that the 4th line shows nothing more than 0 = 0. I noticed after re-reading my previous post that I got a few words mixed up.

So if the 4th line is invalid, then so is the rest of the proof.

So zero does not equal zero?

a=b
a-b=0

and you divide by a-b in the 3rd to 4th line which is like dividing by 0

You can actually make 2+2=5 be correct you know.

You're dividing by zero in the fourth line, which is why your equation is fucked up.

At 5/1/09 03:46 AM, The-Great-One wrote: You can actually make 2+2=5 be correct you know.

Example please?

A can never be B
theyre diffirent letters dumbass

lmao, that is the most retarded math problem I have ever seen. If a=b then when you subtract them it's 0... the problem should end at like line 3. terrrrible attempt at trying to blow our minds, whoever made this.

At 5/1/09 07:03 AM, ghostface619 wrote: A can never be B
theyre diffirent letters dumbass

You're either sarcastic, or an idiot.

The error is in the step where you divide by (a-b). Since a=b, a-b=a-a=0. Thus, you divide by zero and the universe implodes and is replaced with an even stranger version of itself.

At 5/1/09 02:27 AM, X0 wrote: can some one please elaborate on this?

The transition from the 4th to the 5th line is incorrect.
(a+b)(a-b) b(a-b)
= (a^2-b^2)/b = a-b
= (a^2)/b - b = a-b
= (a^2)/b = a
So if B = A and B = 1,
= (1^1)/1 = 1
= True :)

At 5/1/09 07:21 AM, Montycarlo wrote:

At 5/1/09 02:27 AM, X0 wrote: can some one please elaborate on this?

The transition from the 4th to the 5th line is incorrect.
(a+b)(a-b) b(a-b)
= (a^2-b^2)/b = a-b
= (a^2)/b - b = a-b
= (a^2)/b = a
So if B = A and B = 1,
= (1^1)/1 = 1
= True :)

^^^^^this

At 5/1/09 07:09 AM, 4urentertainment wrote: You're either sarcastic, or an idiot.

I think he's an idiot.

The fourth line is completely flawed. And even then, none of these are possible unless using flawed math. At some point there's always a value equaling to another value, which is never possible.

At 5/1/09 08:08 AM, Yiffy wrote: The fourth line is completely flawed.

Absolutely not.

(a+b)(a-b) = b(a-b)
(1+1)(1-1) = 1(1-1)
2 * 0 = 1 * 0
0 = 0

At 5/1/09 02:27 AM, X0 wrote: can some one please elaborate on this?

Yes, they probably could and they already have. In the future, do your own homework. Newgrounds isn't the best place to get help on this subject if you need it. People will intentionally give you wrong answers and the like.

I think u failed

keep trying though

Its basic algebra ie X=5, Y=56, O=83 stuff like that.

At 5/1/09 08:24 AM, Montycarlo wrote:

At 5/1/09 08:08 AM, Yiffy wrote: The fourth line is completely flawed.

Absolutely not.

(a+b)(a-b) = b(a-b)
(1+1)(1-1) = 1(1-1)
2 * 0 = 1 * 0
0 = 0

But a and b cannot be the same. The way you put it is that a = 1 and b = 1, Which would render using either a or b useless. You are better off using just a or just b then.

Fuck maths.

At 5/1/09 08:57 AM, SuperFlonic wrote: But a and b cannot be the same. The way you put it is that a = 1 and b = 1, Which would render using either a or b useless. You are better off using just a or just b then.

Fuck maths.

First line of the problem, a = b.
Last line depicts b = 1.