EPT![[image loading]](http://kr.img.blog.yahoo.com/ybi/1/56/ca/jangshpc05/folder/3/img_3_50_3?1109469444.jpg)
![[image loading]](http://kr.img.blog.yahoo.com/ybi/1/56/ca/jangshpc05/folder/3/img_3_52_0?1109470291.jpg)
![[image loading]](http://kr.img.blog.yahoo.com/ybi/1/56/ca/jangshpc05/folder/3/img_3_52_1?1109470291.jpg)
HAHA^^
Post some funny pics - Page 6
| Forum Index > Closed |
|
jangsh
Korea (South)172 Posts
HAHA^^ | ||
|
Malmis
Sweden1569 Posts
On February 26 2005 18:14 jangsh wrote: HAHA^^ EWWWWWWWWWWWWWWW | ||
|
ieatkids5
United States4628 Posts
| ||
|
pooper-scooper
United States3108 Posts
And I also realized that I spelled rejoicing wrong =( maybe I still did | ||
|
froZen_wYnd
Canada270 Posts
On October 13 2004 17:06 cav wrote: sorry, that is not the reason. the equation you are given is a=b. then it is a?=ab. what happened was the entire equation was multiplied by a variable, thus adding another answer (0). When you get down to the line 2a=a, 'a' must equal 0; that is the only solution that works. this comes from multiplying the entire equation by 'a' in the first place. saying there is something wrong this this is like saying 5x=2x 5=2 OMGOMG hope that englightened someone. ur not very bright... jtan was right, lol | ||
|
Cambium
United States16368 Posts
On February 26 2005 20:33 froZen_wYnd wrote: ur not very bright... jtan was right, lol It's not that hard. and jtan is absolutely right. You start out with an equation that is linear, then you turn it into a quadratic (ignoring the negative effect), you end up with something bizzare. Simiarly, I can 'prove' x = -x, x belongs to real numbers. You start off with x = x, root(x^2) = root((-x)^2) x = -x | ||
|
inkblot
United States1250 Posts
(a^2)-(b^2)=0 ab-(b^2)=0 Nothing more you can do with the equation. | ||
|
HnR)hT
United States3468 Posts
On February 26 2005 20:48 Cambium wrote: You start off with x = x, root(x^2) = root((-x)^2) x = -x I don't get it. The third line doesn't follow from the second line  edit: If that actually was your whole point than ignore this post. | ||
|
PuertoRican
United States5709 Posts
![[image loading]](http://www.challenge-tv.com/hosted/pr/Pimp%20Daddy1.jpg) | ||
|
koit
United States450 Posts
 | ||
|
Cambium
United States16368 Posts
On February 26 2005 21:07 HnR)hT wrote: I don't get it. The third line doesn't follow from the second line edit: If that actually was your whole point than ignore this post. From second to third line: The root sign and the square sign cancel. | ||
Liquid`Ret
Netherlands4529 Posts
| ||
|
HnR)hT
United States3468 Posts
On February 26 2005 21:14 Cambium wrote: From second to third line: The root sign and the square sign cancel. Since the process of squaring a real number and then taking the square root of the obtained quantity is not generally commutative, that's not a legal operation. Also, the real function f = x^2 is not injective, and the real function g = x^.5 is not surjective, so unless I'm mistaken the composite function f(g) must be neither, which means you can never cancel things (and neither f nor g can have inverses). I'm mostly just writing this for my own benefit though. I'm pretty annoyed, lol. Why is it said that the rule (a^b)^c = a^(b*c) holds for all real a, b, c? | ||
|
1337
United States171 Posts
| ||
|
theognis1002
United States38 Posts
On February 26 2005 21:08 PuertoRican wrote: ROFL | ||
|
1337
United States171 Posts
On February 26 2005 18:14 jangsh wrote: HAHA^^ that is fucking hilarious! | ||
|
Cambium
United States16368 Posts
On February 26 2005 21:40 HnR)hT wrote: Since the process of squaring a real number and then taking the square root of the obtained quantity is not generally commutative, that's not a legal operation. Also, the real function f = x^2 is not injective, and the real function g = x^.5 is not surjective, so unless I'm mistaken the composite function f(g) must be neither, which means you can never cancel things (and neither f nor g can have inverses). I'm mostly just writing this for my own benefit though. I'm pretty annoyed, lol. Why is it said that the rule (a^b)^c = a^(b*c) holds for all real a, b, c? That's the whole point -_- Ok, I guess I misinterpretted your thread in the first place -_- It looks 'right', but it's not. | ||
|
HnR)hT
United States3468 Posts
 . That's why I was a little worked up t.t | ||
|
tiffany
3664 Posts
On February 26 2005 22:15 1337 wrote: you have g2b kidding me. how is there a discusion about math happening right now? how can there actualy be an intelligent math thread? hell has frozen over. :O actually math and other academically focused topics are talked about thoroughly and intelligently quite often at this site. the number of intellectual forum members here eclipses that of the typical forum, believe it or not | ||
|
yeehaw
San Marino888 Posts
On February 26 2005 22:35 HnR)hT wrote: Oh, it took me like 10 minutes of googling "rules of exponents" to find *one* place that says the (a^b)^c = a^(b*c) rule holds only for positive bases. Many sites just say it holds whenever all numbers involved are real, which is just plain wrong . That's why I was a little worked up t.tI would like a counter example please. | ||
| ||
![[image loading]](http://www.veryfunnypics.com/cartoons/misc/images/bizarre%20math.jpg)
