Math Brain Teaser

[Mears]

Taking the derivative of x^3 = x^2 + x^2 + x^2+...+x^2 (x terms) yields: 3x^2 = 2x + 2x + 2x + ... + 2x = x(2x) = 2x^2
Hence: 2x^2 = 3x^2 so 3 = 2

So 3 = 2.

 

[Legendary]

What?

 

[her209]

How does x^3 = x^2 + ... x^2 (x terms) ?

 

[MacBaine]

Where did you prove that x^3 = x^2 + x^2 + ... + x^2?

 

[notfred]

2x+2x+2x+2x is not x(2x). If you have an arbitrary number of "2x"s, you can call it n(2x) if you want, but definitely not x(2x).

 

[mEarles]

that is so untrue it makes me spit....
now that i've cleaned off my monitor, GET A FRICKIN ALGEBRA BOOK FROM THE LIBRARY!!!

 

[Mears]

How does x^3 = x^2 + ... x^2 (x terms) ?

For example: 5x = x + x + x + x + x (x added 5 times)

 

[her209]

Originally posted by: Mears

How does x^3 = x^2 + ... x^2 (x terms) ?

For example: 5x = x + x + x + x + x (x added 5 times)

So how does x^3 = x^2 + ... x^2 (x terms) ?

 

[MacBaine]

Originally posted by: Mears

How does x^3 = x^2 + ... x^2 (x terms) ?

For example: 5x = x + x + x + x + x (x added 5 times)

Hmm, now I see... I didn't understand the X terms for a minute.... but I think he's right.

so with x^3 = x^2 + ... x^2 (x terms)

It's

x*x*x = (x*x) + (x*x) + (x*x) + ... + (x*x)

For example,

5^3 = (5*5) + (5*5) + (5*5) + (5*5) +(5*5)

125 =125

As for the rest of it, I donno yet.

 

[ArmenK]

Originally posted by: her209

Originally posted by: Mears

How does x^3 = x^2 + ... x^2 (x terms) ?

For example: 5x = x + x + x + x + x (x added 5 times)

So how does x^3 = x^2 + ... x^2 (x terms) ?

he means x^2 + x^2 ... + x^2 (x times) as in you have x terms total so it would be the same as x * ( x^2)

 

[Mears]

he means x^2 + x^2 ... + x^2 (x times) as in you have x terms total so it would be the same as x * ( x^2)

Right

 

[her209]

so when x = 1

x^3 = x^2

The inductive step as failed!!!!

 

[MacBaine]

Originally posted by: notfred
2x+2x+2x+2x is not x(2x). If you have an arbitrary number of "2x"s, you can call it n(2x) if you want, but definitely not x(2x).

I understand it now, and he's right

If you have a number X, and X many terms...

say X=3

X^3 = X^2 + X^2 + X^2

(dy/dx)==> 3x^2 = 2x +2x +2x

And because you are always going to have X many terms on the right side, 2x +2x +2x = x(2x)

3x^2 = x(2x)

3x^2 = 2x^2

3=2

At least that's what I get out of it... there's obviously something that is being done wrong... I'm just too tired to care

EDIT: I just saw that after you take the derivative, the two sides are no longer equal if you plug in the value for X. There's where it goes wrong.

 

[MacBaine]

Originally posted by: her209
so when x = 1

x^3 = x^2

The inductive step as failed!!!!

1=1

 

[gopunk]

Taking the derivative of x^3 = x^2 + x^2 + x^2+...+x^2 (x terms) yields: 3x^2 = 2x + 2x + 2x + ... + 2x = x(2x) = 2x^2

i'll prove you wrong by contradiction... let x = 2

 

[notfred]

Originally posted by: MacBaine

Originally posted by: notfred
2x+2x+2x+2x is not x(2x). If you have an arbitrary number of "2x"s, you can call it n(2x) if you want, but definitely not x(2x).

I understand it now, and he's right

If you have a number X, and X many terms...

say X=3

X^3 = X^2 + X^2 + X^2

(dy/dx)==> 3x^2 = 2x +2x +2x

And because you are always going to have X many terms on the right side, 2x +2x +2x = x(2x)

3x^2 = x(2x)

3x^2 = 2x^2

3=2

At least that's what I get out of it... there's obviously something that is being done wrong... I'm just too tired to care

EDIT: I just saw that after you take the derivative, the two sides are no longer equal if you plug in the value for X. There's where it goes wrong.

x^3 doesn't equal x^2+x^2+x^2 except when x=3. For the general case, 3x^2 = x(x^2)

x * x^2 = x^3, so that makes perfect sense. Differentiate them both, and you get 3x^2 for each one.

If you want to differentiate the specific case where x=3, you do this
derivative of 3^3 = 0
derivative of 3^2+3^2+3^2 = 0+0+0
0=0, not 3=2

 

[GoingUp]

yea, you just keep taking the derivative, if you have just 2 on the right side the derivative of 2 would be 0 or non existant. I dont see how this thing can work....

 

[bleeb]

Guys don't even get my started on even trying to reach 1000 posts with this thread!

 

[GroundZero]

so in essence you are saying x to the third power= x to the second power plus x to the second power infinite times?
and you say that x=2? you are wrong! the answer to that q? is
if i got the question correct then x=1

 

[Kev]

You guys are hurting my brain with this wacked out math. I think notfred is the only one who has said anything correct in this thread.

 

[GroundZero]

Originally posted by: GroundZero
so in essence you are saying x to the third power= x to the second power plus x to the second power infinite times?
and you say that x=2? you are wrong! the answer to that q? is
if i got the question correct then x=1

am i correct or not or is the question missunderstood on my end?

 

[gopunk]

Originally posted by: notfred

Originally posted by: MacBaine

Originally posted by: notfred
2x+2x+2x+2x is not x(2x). If you have an arbitrary number of "2x"s, you can call it n(2x) if you want, but definitely not x(2x).

I understand it now, and he's right

If you have a number X, and X many terms...

say X=3

X^3 = X^2 + X^2 + X^2

(dy/dx)==> 3x^2 = 2x +2x +2x

And because you are always going to have X many terms on the right side, 2x +2x +2x = x(2x)

3x^2 = x(2x)

3x^2 = 2x^2

3=2

At least that's what I get out of it... there's obviously something that is being done wrong... I'm just too tired to care

EDIT: I just saw that after you take the derivative, the two sides are no longer equal if you plug in the value for X. There's where it goes wrong.

x^3 doesn't equal x^2+x^2+x^2 except when x=3. For the general case, 3x^2 = x(x^2)

x * x^2 = x^3, so that makes perfect sense. Differentiate them both, and you get 3x^2 for each one.

If you want to differentiate the specific case where x=3, you do this
derivative of 3^3 = 0
derivative of 3^2+3^2+3^2 = 0+0+0
0=0, not 3=2

it's not supposed to, x^3 for 2 for example, is supposed to equal x^2 + x^2

 

[AgentDib]

Originally posted by: Mears
Taking the derivative of x^3 = x^2 + x^2 + x^2+...+x^2 (x terms) yields: 3x^2 = 2x + 2x + 2x + ... + 2x = x(2x) = 2x^2
Hence: 2x^2 = 3x^2 so 3 = 2

The problem in this is that it is taking the derivative of an equation that is only defined at a single point. Every X you choose is a new equation and obviously there is an undefined change of rate of a function identified with a single point for any X that you pick.

But yes, algebraically it works out fine if you can avoid thinking about what a derivative is.

 

[MacBaine]

While it doesn't matter really... I think some of you are still misunderstanding it. Whatever number X represents, is how many X^2 terms you have on the right side of the equation. if X=5, then X^3 = 5(x^5)

 

[GroundZero]

ok i may be old, but this is new, new math.
someone spell out the equation for me in words and not symbols...
thanks