Motivational Posters Thread
- Thread starter Vagabond
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Some that I had saved from the old boards.
figmentPez
Staff member
New ones!
R
Rubicon
someones been reading 4chan todayDenbrought said:
2a=a is perfectly valid if a=0.Rasputin said: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.Allen said:
It's a pity the math is terrible.
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=0Tinwhistler said:
I don't make them, I just post them.Dieb said:
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... :|
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.Rasputin said:
R
Rubicon
less math more posters?
a^2+b^2=c^2 in a right triangleMav said:
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
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...Rasputin said:
figmentPez
Staff member
No, I'd rather not get too cozy or sexy with math. I want more posters!Allen said:
A
Anubinomicon
Ross
Staff member
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.Allen said:
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
Math is all kinds of wack.Rasputin said:
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:
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
Z
Zarvox
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.Rasputin said:
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.
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