ok can anyone explain to me how this proof is not true
set -1/1=1/-1
take the square root of both sides
i/1=1/i
multiply both sides by 1/2
i/2i=1/2i
add 3/2i to both sides
i/2+3/2i=1/2i+3/2i
multiply both sides by i
i2/2+3i/2i=i/2i=3i/2i
simplify
and you should get 2=1 (or 1=2 depending on what sides you used)
i posted this in the general forum and people recomended i take it here
At 3/24/04 09:39 PM, bulletproofdisaster wrote: ok can anyone explain to me how this proof is not true
set -1/1=1/-1
take the square root of both sides
i/1=1/i
multiply both sides by 1/2
i/1 * 1/2 = i/2 not i/2i
i/2i=1/2i
Funny story, I have to deal with amplification for my job, and the instructors use the rule of thumb, for every 3 dB, the signal doubles.
So I had to figure out how much amplification occurs if there is a 20dB gain. I immediately computed 100 times higher. As for every 10 dB there is a tenfold increase. The instructor disagreed with me and began furiously working on the paper. 20/3 =.... 6 ish... 2*2*2*2*2*2=... 64.
It took me fricking forever to explain my math.
My friend told me there was a math club, but I didn't really believe him. Can I join? Haha, I LOVE math. I plan to major in it, and I'm taking 3 math classes next year in school, so this kind of stuff is totally fun to me. =) Please???
At 3/24/04 09:39 PM, bulletproofdisaster wrote: ok can anyone explain to me how this proof is not true
set -1/1=1/-1
take the square root of both sides
i/1=1/i
multiply both sides by 1/2
i/2i=1/2i
add 3/2i to both sides
i/2+3/2i=1/2i+3/2i
multiply both sides by i
i2/2+3i/2i=i/2i=3i/2i
simplify
and you should get 2=1 (or 1=2 depending on what sides you used)
-1/1 = 1/-1
sqrt both sides ---- i/1 = 1/i
multiply by 1/2 ---- i/2 = 1/2i
add 3/2i to both sides ------ i/2 + 3/2i = 1/2i + 3/2i
multiply by i ---- i^2/2 + 3i/2i = i/2i + 3i/2i
simplify (step 1) ----- 1/2 + 3/2 = 1/2 + 3/2
simplify (step 2) ----- 4/2 = 4/2
Conclusion: well it seems you have proved something equals itself
error made ---- in the step of multiplying by i you did that step for eveything except the first component of the problem i/2; it seems there you multiplied by 2i
irony in the error ------ if you performed this in that way then you should conclude the answer is 2i^2/2 (1) +3i/2i (3/2) = i/2i (1/2) + 3i/2i (3/2)
which would equal ----- 5/2 = 4/2 or 2.5 = 2, not 2=1 or 1=2
At 3/24/04 11:04 PM, Dr_Arbitrary wrote:At 3/24/04 09:39 PM, bulletproofdisaster wrote: ok can anyone explain to me how this proof is not truei/1 * 1/2 = i/2 not i/2i
set -1/1=1/-1
take the square root of both sides
i/1=1/i
multiply both sides by 1/2
i/2i=1/2i
that is an irrelevant statement given that error is corrected in the next step by accident
At 3/25/04 01:08 AM, miket311 wrote:
that is an irrelevant statement given that error is corrected in the next step by accident
Didn't catch that, I stopped looking after the first error was made.
Okay, a few people have asked to join the Math club. Personally, I am completely unimpressed when I look at other clubs and see that they start off with a list of titles that will be given out to people based on their order of membership, I feel the same way about numerical titles too "Yay, I am teh 43rd member of the Club." If anyone is interested in joining the club either state that you're a member, or put something in your signature linking to this club.
The reason I ask this is because of the nature of the club, often, people who do not have much interest in mathematics will post in here because they are having difficulty with math. By putting something in your signature, you make yourself differentiable from other people who are asking questions. It's sort of a mark of authority.
If anyone has any problems with this, please let me know through E-mail or AIM, I have made mistakes in the past, and there is likely a better way to handle this, however, I don't want to spend the next two pages arguing over this.
Thanks to all of you who have expressed interest in the Mathematics Club. Mathematics is powerful, and although not all of us have the skill or interest necessary for it, it can benefit us all. Furthermore, the best way to solidify the concepts in mathematics is to force ones self to understand it so well that it can be explained to others.
Here's a bit of a puzzle I read about a long time ago.
There are x number of chess players
none is at exactly the same skill level as another. Furthermore, each can be put in order of skill, a person who has higher skill will always beat a person of lower skill (basically, it's not paper rock scissors, it's more like slingshot, rifle, tank... the tank always wins)
The question is: is it possible to, before any matches are played, create a system that will in as few matches as possible, rate each player relative to the others.
Here's an example:
Three players.
The winner of a game, marked by a box takes the high route out, the loser takes the lower one.
(see picture at end of post)
How do you do this with higher values for X, I know how to do it with 4, and I think I have an answer for 5.
To clarify the picture, the lines on the left side represent players, the order from top to bottom is arbitrary. The Right side represents an ordered list of players, best ones at the top.
Screw that. Have any geometry type puzzles that involve finding the area of something. Those are kinda fun.
Okay, two cylinders of radius 1 intersect eachother at 90 degree angles. What is the volume of the solid that is within both cylinders?
And now to make it difficult, don't use calculus to figure it out.
At 3/29/04 12:18 AM, Dr_Arbitrary wrote: Okay, two cylinders of radius 1 intersect eachother at 90 degree angles. What is the volume of the solid that is within both cylinders?
And now to make it difficult, don't use calculus to figure it out.
Okay. Im not sure what you mean, so my calclations are based on my picture.
The distance between the the two radius' (I forget how to spell it) is one. The distance from the radius to intersection is also one. So, the triangle has sides of 1, 1/2, and root3/2. The angle between the intersections is 120 degrees, or 2pi/3. So, The area inclosed is given by 2{(2pi/2)/(2)-sin[(2pi/3)/2]cos[(2pi/3)/2]}. This is going to become [(16pi-3*root3)/24].
Right? I went by the trig law that the slice will be given by r^[((angle/2)-sin(angle/2)cos(angle/2)]
Here is my diagram:
At 3/24/04 01:03 PM, BWWWWWWWWWWWWWWWWWWS wrote:At 3/24/04 12:41 PM, mr_trivia wrote: Is there a such thing as the rule of 72? I think I heard about it in my senior year of high school, but I can't remember exactly how it worked.It has to do with interest rates, and the amount of time needed to double the investment. It is a crude/shitty approximation...youll be better off forgetting it to tell you the truth.
You have the longest Alias I have ever seen...plus it's kinda annoying cause it stretches the page.
No, two cylinders intersect eachother at 90 degrees. Like two pipes hitting connecting at an X junction.
At 3/29/04 03:01 AM, Dr_Arbitrary wrote: No, two cylinders intersect eachother at 90 degrees. Like two pipes hitting connecting at an X junction.
Fuck that. Im not doing anything in three dimensions without my dear calc. Actually, im just really busy this week. For all I know, I got the last one wrong; I havent checked it, and I did it in a bit of a hurry. Im talking about the little pictures that people make. You know, like a square with a bunch of diagonal lines, circles, and semi-circles. Whatever. And yeah, my name is pimp dizzle.
is there ne chance that i can join this club, i aint fantastic at maths, well ot yet neways,
At the moment, i am studying Maths, Further Maths, Physics and ICT at school (UK A-Levels)
so i can proboly help out with some problems, and i will progressivelyt get a lot better over the next few months...
sooooo..can i?
At 3/29/04 05:24 PM, isthatlegal wrote: is there ne chance that i can join this club, i aint fantastic at maths, well ot yet neways,
There really isnt any need to "join." You are much more than welcome to post here, offer help, and/or ask questions. So, have at it and welcome.
At 3/29/04 10:22 PM, BWWWWWWWWWWWWWWWWWWS wrote: There really isnt any need to "join." You are much more than welcome to post here, offer help, and/or ask questions. So, have at it and welcome.
ok, cheers...is thread dead most of the time??? cos it has only had one post since yesterday..
Yeah, it's not a very active club, mostly because it's a very specific target audience.
If you find a difficult puzzle, or want help with something, feel free to post it.
I've got mathematica, so if you need any difficult problems evaluated, I'd be happy to do so.
Here is one. It may or may not give you trouble. Not too hard, but not too obvious either.
Tthe rectangle at the corner measures 7 cm x 14 cm. What is the radius of the circle in cm?
At 3/31/04 02:31 AM, Aasha wrote: Er...35cm
I think that's the right answer, tricky problem!
Here's a very difficult one:
The sum of a certain number of consecutive positve integers is 1000. What are these integers?
I'm not sure if there's a way to do this other than trial and error.
for real tho i got 1....
you have a triangle with angles a,b, and c and angle b is obteuse...
x is a point between a and c and the measurment of the altitude
b>x is 12 cm and x-c is 16 degrees... angle a is 12 deg.
find da perimeter!
At 3/31/04 07:52 PM, lexus3 wrote: e=mc
hehehe...
solve dat shit
First of all, its: E=MC^2
Second, even if someone did, would you even know what the fuck theyre talking about? Th answer is no, because youre obviously an immature little twit. You know, nobody is going to think that your last post was funny, and if that was the intention you did a pretty shitty job. If you need some help or something go ahead and post, otherwise spare us from reading anymore ignorant bullshit.
At 3/31/04 07:55 PM, lexus3 wrote: you have a triangle with angles a,b, and c and angle b is obteuse...
x is a point between a and c and the measurment of the altitude
b>x is 12 cm and x-c is 16 degrees... angle a is 12 deg.
find da perimeter!
That wouldnt be too hard, but in order to do it I would need to know what the hell youre talking about. The triangle ABC has an obtuse angle B and x is across from it...the rest is like?(huh)
