Motivational Posters Thread

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Ross

Staff member
Allen said:


It's a pity the math is terrible.
Only, the "Given a=b" part is a false, since it's impossible for a to equal b. Hence, while the algebra is correct, the initial assumption is incorrect.
 
Rasputin said:
Allen said:


It's a pity the math is terrible.
Only, the "Given a=b" part is a false, since it's impossible for a to equal b. Hence, while the algebra is correct, the initial assumption is incorrect.
2a=a is perfectly valid if a=0.
therefore, it's obvious that dividing both sides by a (0) would produce a nonsensical result.
 
Tinwhistler said:
Rasputin said:
Allen said:
[bad math]

It's a pity the math is terrible.
Only, the "Given a=b" part is a false, since it's impossible for a to equal b. Hence, while the algebra is correct, the initial assumption is incorrect.
2a=a is perfectly valid if a=0.
therefore, it's obvious that dividing both sides by a (0) would produce a nonsensical result.
They also divide by zero when they divide to remove (a-b) since a-b=0
 
Allen said:
Tinwhistler said:
Rasputin said:
[quote="Allen, who is Quiet":bxsoxely][bad math]

It's a pity the math is terrible.
Only, the "Given a=b" part is a false, since it's impossible for a to equal b. Hence, while the algebra is correct, the initial assumption is incorrect.
2a=a is perfectly valid if a=0.
therefore, it's obvious that dividing both sides by a (0) would produce a nonsensical result.
They also divide by zero when they divide to remove (a-b) since a-b=0[/quote:bxsoxely]

Which is the main flaw in the deduction.
 

Ross

Staff member
I still stand by my previous post... That the assumption that a=b is invalid. As previously said, although it is possible at the end, a has to not be 0 in order to get to that point, therefore a=0 is not a viable answer.

The mathematicians on this forum SHOULD agree with me on this one... :|
 
Rasputin said:
I still stand by my previous post... That the assumption that a=b is invalid. As previously said, although it is possible at the end, a has to not be 0 in order to get to that point, therefore a=0 is not a viable answer.

The mathematicians on this forum SHOULD agree with me on this one... :|
a=b is a perfectly valid assumption. Variables can be equal to each other.
 
Mav said:
less math more posters?
a^2+b^2=c^2 in a right triangle
cos(-x)=cos(x)
sin(2x)=2sinxcosx
c^2=a^2+b^2-2abcosC

Prove: cos2y x secy=2cosy-secy
Given: cos2y x secy
=(2cos^2y-1) x secy
=(2cos^2y x secy)-secy
=(2cos^2y x 1/cosy)-secy
=2cosy-secy

 
W

wana10

Rasputin said:
I still stand by my previous post... That the assumption that a=b is invalid. As previously said, although it is possible at the end, a has to not be 0 in order to get to that point, therefore a=0 is not a viable answer.

The mathematicians on this forum SHOULD agree with me on this one... :|
as allen who is quiet said the problem isn't that a=b, which is perfectly valid. the problem is that for any value of a and b you will have to divide by zero and then...


 
Allen said:
Tinwhistler said:
Rasputin said:
[quote="Allen, who is Quiet":1mkfgykj][bad math]

It's a pity the math is terrible.
Only, the "Given a=b" part is a false, since it's impossible for a to equal b. Hence, while the algebra is correct, the initial assumption is incorrect.
2a=a is perfectly valid if a=0.
therefore, it's obvious that dividing both sides by a (0) would produce a nonsensical result.
They also divide by zero when they divide to remove (a-b) since a-b=0[/quote:1mkfgykj]

This should have been posted first.
 
figmentPez said:
New ones!



is this guy shirtless on a plane or bus? where is he leaving? Arkansas? Hey guy just make yourself comfortable, there's probably room to put your trans am up on cinderblocks in the back
 

Ross

Staff member
Allen said:
Rasputin said:
I still stand by my previous post... That the assumption that a=b is invalid. As previously said, although it is possible at the end, a has to not be 0 in order to get to that point, therefore a=0 is not a viable answer.

The mathematicians on this forum SHOULD agree with me on this one... :|
a=b is a perfectly valid assumption. Variables can be equal to each other.
The problem isn't that they can't be equal. I'm saying that they ASSUME IN THE BEGINNING BEFORE THEY EVEN START that a=b AND IS TRUE. Through the rest of the problem, we know that a=b IS NOT true, and the assumption is therefore incorrect from the very start.

It is an invalid assumption, since the assumption is that it is TRUE, when it is, in fact, FALSE.

EDIT:
While a can equal b in OTHER examples, it cannot in THIS example, which is what the whole argument is about.
 

GasBandit

Staff member
Rasputin said:
Allen said:
Rasputin said:
I still stand by my previous post... That the assumption that a=b is invalid. As previously said, although it is possible at the end, a has to not be 0 in order to get to that point, therefore a=0 is not a viable answer.

The mathematicians on this forum SHOULD agree with me on this one... :|
a=b is a perfectly valid assumption. Variables can be equal to each other.
The problem isn't that they can't be equal. I'm saying that they ASSUME IN THE BEGINNING BEFORE THEY EVEN START that a=b AND IS TRUE. Through the rest of the problem, we know that a=b IS NOT true, and the assumption is therefore incorrect from the very start.

It is an invalid assumption, since the assumption is that it is TRUE, when it is, in fact, FALSE.
Math is all kinds of wack.

x = 0.99999...
10x = 9.99999....
=> 9x = 9 (subtracting 1x from both sides of the equation)
=> x = 1

Oh, I'm sure you'll have nits to pick there, so let's do it this way...

1/3=0.333...
(multiply both sides by 3)
3/3=0.999...
3/3=1
1=0.999...

MAAAAJIIIIC!
 
GasBandit said:
Rasputin said:
Allen said:
Rasputin said:
I still stand by my previous post... That the assumption that a=b is invalid. As previously said, although it is possible at the end, a has to not be 0 in order to get to that point, therefore a=0 is not a viable answer.

The mathematicians on this forum SHOULD agree with me on this one... :|
a=b is a perfectly valid assumption. Variables can be equal to each other.
The problem isn't that they can't be equal. I'm saying that they ASSUME IN THE BEGINNING BEFORE THEY EVEN START that a=b AND IS TRUE. Through the rest of the problem, we know that a=b IS NOT true, and the assumption is therefore incorrect from the very start.

It is an invalid assumption, since the assumption is that it is TRUE, when it is, in fact, FALSE.
Math is all kinds of wack.

x = 0.99999...
10x = 9.99999....
=> 9x = 9 (subtracting 1x from both sides of the equation)
=> x = 1

Oh, I'm sure you'll have nits to pick there, so let's do it this way...

1/3=0.333...
(multiply both sides by 3)
3/3=0.999...
3/3=1
1=0.999...

MAAAAJIIIIC!

These only work because the decimal is repeating. The more repeats you use, the more true it is that .99999999=1


We assume in your example, that the repeats go out infinitely, which means that we really do approach 1 if you have .9 with an infinite number of nines repeating..

If you don't use repeating decimals, we end up with something like this:
X = 0.99999
10x = 9.9999
10X - X = 9.9999 - 0.99999 (note that 10X without repeats is 9.9999 not 9.99999)
9X = 8.99991
X = 0.99999
 
Rasputin said:
The problem isn't that they can't be equal. I'm saying that they ASSUME IN THE BEGINNING BEFORE THEY EVEN START that a=b AND IS TRUE. Through the rest of the problem, we know that a=b IS NOT true, and the assumption is therefore incorrect from the very start.

It is an invalid assumption, since the assumption is that it is TRUE, when it is, in fact, FALSE.

EDIT:
While a can equal b in OTHER examples, it cannot in THIS example, which is what the whole argument is about.
a can still be equal to b throughout the problem. The only thing wrong is they divide by 0 and get 1=2. a=b is still valid. If there's a problem in one step (in this case, from 3 to 4), that doesn't mean earlier steps are incorrect.

Example:
1+1=2
1=3

That's wrong. 1 doesn't equal 3. That means the math is wrong, not the assumption that 1+1=2 is wrong.
 
It's not an assumption that a=b, it's given. To do the proof, you have to work within the framework that a=b from the very start. At any point where a cannot equal b, you're done. You cannot go any further.
 

ElJuski

Staff member
Okay, I'm banning anyone who makes posts another math poster that gets all the number dorks in a tizzy -_-. Where's the FUNNY, DAMNIT
 
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