Author Topic: The Beauty of Math  (Read 4484 times)

Offline Serge

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The Beauty of Math
« Reply #20 on: October 16, 2008, 04:31:24 PM »
Quote from: Flying_Chao;17511
How did you get " a - 1 = a + 1 " ?

I divided both sides by (a - 1).

No, sh** dammit, I made a mistake. I'll fix the image in a sec.
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Offline Serge

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« Reply #21 on: October 16, 2008, 04:33:39 PM »
Here, fixed.
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Offline Flying_Chao

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« Reply #22 on: October 16, 2008, 04:36:07 PM »
That's better.

Offline Naryar

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« Reply #23 on: October 16, 2008, 04:49:20 PM »
I don't see how the hell do you jump from a-1=(a-1)*(a+1) to a-1=a+1...

And you forgot something important: the equation a^2 =1 admits TWO values of a: a=1and a =-1. So, a=1 => a^2=1 goes one way (the opposite isn't true) but all of the others goes both way (mathematical symbol <=>).

With a=-1 the thing works.

I bet that can be solved by imaginary numbers but I'm too lazy to use them now.

Yeah, I'm a math nerd.

Offline Serge

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« Reply #24 on: October 16, 2008, 04:55:12 PM »
Quote from: Naryar;17521
I don't see how the hell do you jump from a-1=(a-1)*(a+1) to a-1=a+1...

And you forgot something important: the equation a^2 =1 admits TWO values of a: a=1and a =-1. So, a=1 => a^2=1 goes one way (the opposite isn't true) but all of the others goes both way (mathematical symbol <=>).

With a=-1 the thing works.

I bet that can be solved by imaginary numbers but I'm too lazy to use them now.

Yeah, I'm a math nerd.

The first error was fixed, my bad, I am tired, refresh the image.

As for the confusion with a^2... I am nowhere telling that sqrt(a^2) = 1, so I do not need absolute values or deal with multiple cases.
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Offline Gigafrost

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« Reply #25 on: October 16, 2008, 07:51:09 PM »
An easy question:

Solve this problem without using the number 2 or anything that 2, 2.5, 1.5, or 7 is divisible by. Yet the numbers that create the product must be a whole number, equal to 2 and greater than 1.

1+1 = ?
« Last Edit: October 16, 2008, 08:08:31 PM by Gigafrost »

Offline Sage

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« Reply #26 on: October 16, 2008, 07:55:50 PM »
square root of 4.

you said '2 is divisible by' instead of 'divisble by 2'

otherwise just 5-3
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Offline Gigafrost

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« Reply #27 on: October 16, 2008, 07:57:22 PM »
Nu uh (smiley face)... Fixed =)

Offline andrewm0304

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« Reply #28 on: October 16, 2008, 08:37:26 PM »
Haha, thats a good one Serge ;)

Offline man manu

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« Reply #29 on: October 17, 2008, 01:09:06 PM »
-1+3?
Son of a fat bold guy!
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Offline Sage

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« Reply #30 on: October 17, 2008, 02:04:56 PM »
@man manu: 2. not even math really.

heres a question:

theres an undefined line and a line with the slope zero. they are perpendicular to each other. whats the slope of the undefined line?
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Offline Serge

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« Reply #31 on: October 17, 2008, 02:09:23 PM »
Quote from: Sage;17624
@man manu: 2. not even math really.

heres a question:

theres an undefined line and a line with the slope zero. they are perpendicular to each other. whats the slope of the undefined line?

The undefined line cannot be the graph of a function, since in a function no argument can result in more than one value (where in this case, there is an infinite numbers of values).
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Offline Sage

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« Reply #32 on: October 17, 2008, 02:25:11 PM »
incorrect
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Offline Naryar

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« Reply #33 on: October 17, 2008, 04:00:21 PM »
No, he's right.

This can't have any slope because if we consider it in a classic two-dimensional space (cartesian coordinates) with Ox and Oy as vectors, that (vertical) line would have an equation of x=k (k being a constant and y being free).

Problem, you can't have a slope (or can't get a derivative for being more precise) from such a function.

Common sense would say the value of the slope is fixed and infinite, but this is a math abomination.

So you can't have any slope on there, excepted if you're tricking us with 3D space.

Offline Sage

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« Reply #34 on: October 17, 2008, 04:26:41 PM »
well, from what i know, an undefined slope= a vertical line. if a slope = 0 then its horizontal, meaning they form a perpendicular. the answer was in the question.

the slope is undefined, because in the slope formula the answer would be something divided by zero, which is impossible, making it a vertical line.
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Offline R0B0SH4RK

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« Reply #35 on: October 17, 2008, 08:57:25 PM »
Basically, since slope is rise of the line over the run, having a completely vertical line is mathematically impossible from my understanding. The slope would be infinity divided by zero. In a distance over time graph (one measuring speed), such a line would mean that the object is everywhere in the universe at the same time.

Quote from: Serge;17509
(Image removed from quote.)


There is a flaw in the fourth line here. a-1 = (a-1)(a+1) should equal a-1 = a^2-1
which is 0=0, not 1=2 :P

Offline Gigafrost

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« Reply #36 on: October 18, 2008, 12:22:18 AM »
What is the square root of infinity?

Offline Sage

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« Reply #37 on: October 18, 2008, 12:23:35 AM »
Quote from: Gigafrost;17691
What is the square root of infinity?

pi.
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Offline Naryar

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« Reply #38 on: October 18, 2008, 02:44:45 AM »
Infinity too!

Offline Serge

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« Reply #39 on: October 18, 2008, 04:00:51 AM »
Quote from: R0B0SH4RK;17685
Basically, since slope is rise of the line over the run, having a completely vertical line is mathematically impossible from my understanding. The slope would be infinity divided by zero. In a distance over time graph (one measuring speed), such a line would mean that the object is everywhere in the universe at the same time.



There is a flaw in the fourth line here. a-1 = (a-1)(a+1) should equal a-1 = a^2-1
which is 0=0, not 1=2 :P


There is nothing wrong with this line. The error is somewhere else.
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