# Foo - The MATH thread

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View Full Version : The MATH thread

red house
07-26-07, 12:42 PM
Okay, you computer programming savantes and self-proclaimed ''mathemajicians'' of foo, this is your chance to prove your mettle. Now, these mathematical conundrums are not for the faint of heart, infact most of them probably have never been solved.. but, I invite you to try your hand at them anyway.. haha, best of luck (you're going to need it) :D :beer:

Math riddle #1.

If an integer n is greater than 2, provide the 'proof' that the equation a^n + b^n = c^n has no solutions in non-zero integers a, b, and c.

Math riddle #2.

Provide the proof that the circumference of a circle c divided by it's diameter d is infact always equal to a 'nonrational' real integer.

Math riddle #3.

Find a way to factor out (x+4) from the denominator of the trinomial;

(6x^3 + 23x^2 + 6x + 40) / (x+4)

good luck.. and pls show your work. :) :beer:

Falkon
07-26-07, 12:52 PM
I have a feeling for the 3rd, I'd probably have to factor out the top and multiply by the conjugate.

red house
07-26-07, 12:55 PM
I have a feeling for the 3rd, I'd probably have to factor out the top and multiply by the conjugate.

I have the feeling you're probably on to something.. -go with it (!).. :beer:

jschen
07-26-07, 12:56 PM
First two are interesting problems in their own right. Don't expect to come up with an answer yourself. They are classic problems in mathematics that took the mathematics community a REALLY long time to solve.

The third one, just do your long division. It's straightforward, with no tricks involved.

red house
07-26-07, 01:00 PM
I think Falkon was warmer.. call it a hunch.

Diggidy
07-26-07, 01:10 PM
3) 6x^2-x+10

red house
07-26-07, 01:26 PM
3) 6x^2-x+10

:eek: ding! ding! ding! ding! ding! ..

Diggidy, you have earned your merit in mathematical excellence. :D :beer:

Okay. . can anyone shed some light on how this feat was accomplished?

asherlighn
07-26-07, 01:40 PM

jschen
07-26-07, 01:41 PM
Okay. . can anyone shed some light on how this feat was accomplished?

We already told you. Long division. Do your homework already, before you run out of time to seek assistance like you did last time.

mwrobe1
07-26-07, 01:42 PM
I remember #1 from long ago...it is impossible to separate any power higher than the second into two like powers. Its a variation of the Pythagorean theorem (a squared + b squared = c squared) ie. (3, 4, 5) ( 5, 12, 13) ( 7, 24, 25) etc.

There are no solutions to this where n as stated above is greater than 2

If you plugged in (3,4,5) into a^n+b^n=c^n where n=3...it wouldn't work, in fact...nothing would work.

EDIT: I just realized I never "proofed it"...I'm too lazy.

squegeeboo
07-26-07, 01:42 PM
you take the original numerator 6x^3 + 23x^2 6x + 40 and work on reducing it down into this format:
(ax + b)(cx + d)(ex + f)
where a - f are positive or negative integers
You have a denominator thats already reduced in x + 4 so it's a resonable assumptino that ax + b is either going to be x + 4 itself, or a multiple of that y(x + 4) where y is a positive or negative integer, and you just crunch numbers until a sequence hits. I don't remeber if theres a quicker short cut to it, or if I just used to do enough of them that I could see the patterns better than I can now.

Ken B.
07-26-07, 01:54 PM
... Math riddle #2.

Provide the proof that the circumference of a circle c divided by it's diameter d is infact always equal to a 'nonrational' real integer.
:beer:

Wait a sec, a nonrational real integer? Da hell does that mean? An integer is rational by definition. Or does 'nonrational' have a different definition than the more common term 'irrational'? And by 'real' I assume you mean 'noncomplex' rather than the computer science meaning of 'noninteger'.

Now, it's not difficult to derive the value of pi, but is that what you are asking? Your terminology is corn-fusing.

red house
07-26-07, 01:57 PM
Wait a sec, a nonrational real integer? Da hell does that mean? An integer is rational by definition. Or does 'nonrational' have a different definition than the more common term 'irrational'? And by 'real' I assume you mean 'noncomplex' rather than the computer science meaning of 'noninteger'.

Now, it's not difficult to derive the value of pi, but is that what you are asking? Your terminology is corn-fusing.

I think perhaps I meant ''nonrational real number'' .. sorry, my bad. :)

Nicodemus
07-26-07, 01:58 PM
I'm so glad I've finished my schooling.

red house
07-26-07, 01:58 PM
you take the original numerator 6x^3 + 23x^2 6x + 40 and work on reducing it down into this format:
(ax + b)(cx + d)(ex + f)
where a - f are positive or negative integers
You have a denominator thats already reduced in x + 4 so it's a resonable assumptino that ax + b is either going to be x + 4 itself, or a multiple of that y(x + 4) where y is a positive or negative integer, and you just crunch numbers until a sequence hits. I don't remeber if theres a quicker short cut to it, or if I just used to do enough of them that I could see the patterns better than I can now.

Yeah, I tried that.. unfortunately I didn't have any luck at all. :P

red house
07-26-07, 02:03 PM
We already told you. Long division. Do your homework already, before you run out of time to seek assistance like you did last time.

Who is 'we' ? .. you are the only one who has mentioned long division ? .. haha, 'long division' of a trinomial.. -that's a good one jschen. :beer: . .now I've heard it all.

Hey, why don't you show us how to perform this 'magical' long division method of yours for trinomial functions. You are too funny sometimes.. ''long division'' :rolleyes: :lol: -?

squegeeboo
07-26-07, 02:04 PM
Yeah, I tried that.. unfortunately I didn't have any luck at all. :P

Yah it took me a while of trial and error, I started out figuring out with how many ways of getting 40(the last part of the numerator), with one of the numbers being a 4(due to the 4 in the denominator), so in this case 10x1x4 5x2x4 1x10x4 2x5x4 were the only ones I could come up with. tried a bunch of the 5x2 combo's none of those got even close for the 23x^2 part of the original, so I moved to the 10x1 combo, and stumbled into it.

jschen
07-26-07, 02:05 PM
You don't try to factor the whole thing from scratch. That's a MUCH tougher problem. Set up your division.

******___________________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40

This long division is easy since they already told you what to divide by. Factoring from scratch or solving by trial and error is hard work.

dragracer
07-26-07, 02:07 PM
Ok thanks..... I'm feelin' like a real Jethro Bodine over here.

red house
07-26-07, 02:07 PM
You don't try to factor the whole thing from scratch. That's a MUCH tougher problem. Set up your division.

******___________________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40

This long division is easy since they already told you what to divide by. Factoring from scratch or solving by trial and error is hard work.

Seriously, wtf is that? :eek: :rolleyes:

jschen
07-26-07, 02:09 PM
It's a division problem. Work left to right, just as you would a standard long division problem. I'm trying to in ASCII draw out what I would write on paper to set up the division.

Work with me... What do you have to multiply (x + 4) by to get a 6 x^3 term?

Diggidy
07-26-07, 02:09 PM
:eek: ding! ding! ding! ding! ding! ..

Diggidy, you have earned your merit in mathematical excellence. :D :beer:

Okay. . can anyone shed some light on how this feat was accomplished?

If you have the answer and the problem, it shouldn't be too hard to figure out the steps in between. If you are doing this for answers for homework, stop being lazy and do it by yourself. If it is indeed for school, then you will probably see the material again; and you will want to know it then. If it's just for funzies, then thanks; it's been a few months since I got out of high school, and I'm saddened at how hard that was for me.

red house
07-26-07, 02:10 PM
Yah it took me a while of trial and error, I started out figuring out with how many ways of getting 40(the last part of the numerator), with one of the numbers being a 4(due to the 4 in the denominator), so in this case 10x1x4 5x2x4 1x10x4 2x5x4 were the only ones I could come up with. tried a bunch of the 5x2 combo's none of those got even close for the 23x^2 part of the original, so I moved to the 10x1 combo, and stumbled into it.

Yeah, 23 being a prime was really discouraging and made me think I was going about it all wrong.

So, what does the ( . . )( . . )( . . ) look like?

red house
07-26-07, 02:12 PM
It's a division problem. Work left to right, just as you would a standard long division problem. I'm trying to in ASCII draw out what I would write on paper to set up the division.

Work with me... What do you have to multiply (x + 4) by to get a 6 x^3 term?

I dunno? 6x^2 ? .. but then how do you get the 4 ?

red house
07-26-07, 02:14 PM
If you have the answer and the problem, it shouldn't be too hard to figure out the steps in between. If you are doing this for answers for homework, stop being lazy and do it by yourself. If it is indeed for school, then you will probably see the material again; and you will want to know it then. If it's just for funzies, then thanks; it's been a few months since I got out of high school, and I'm saddened at how hard that was for me.

I am not 'lazy' - I am resourceful. And there is no rule saying one must do his or her homework by theirself w/o any help is there? No, I didn't think so, -(unless you happen to reside in North Korea). But thanku much for the help, it's much appreciated.

jschen
07-26-07, 02:16 PM
We'll get there. Yes, you multiply by 6 x^2 to get a 6 x^3 term.

******__6 x^2__________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40

Now 6 x^2 * (x + 4) = 6 x^3 + 24 x^2, so...

******__6 x^2__________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40
********6 x^3 + 24 x^2

You see what we do next?

red house
07-26-07, 02:19 PM
Wow.. thankyou Jschen! :beer: I really take back everything (bad) I said regarding anything concerning you. Thanks man for the help.. :)

Diggidy
07-26-07, 02:21 PM
We'll get there. Yes, you multiply by 6 x^2 to get a 6 x^3 term.

******__6 x^2__________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40

Now 6 x^2 * (x + 4) = 6 x^3 + 24 x^2, so...

******__6 x^2__________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40
********6 x^3 + 24 x^2

You see what we do next?

That +4 is bound to the X though, so you can't just do 6x^2 x X = 6x^3, because you have that four hanging in there. You'd be better off factoring the top, it's really not that hard.

I didn't look close enough, my apologies; correct so far. I still stick with my factoring is easier though.

jschen
07-26-07, 02:24 PM
Whatever... I really don't care what you say about me, good or bad. Just do your homework and stop asking for help under the pretense of trying to test us. That's getting really old, and I don't need to prove my abilities to you.

red house
07-26-07, 02:24 PM
We'll get there. Yes, you multiply by 6 x^2 to get a 6 x^3 term.

******__6 x^2__________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40

Now 6 x^2 * (x + 4) = 6 x^3 + 24 x^2, so...

******__6 x^2__________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40
********6 x^3 + 24 x^2

You see what we do next?

******__6 x^2__________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40
********6 x^3 + 24 x^2

but this would now equal = -x^2 . . -which would be divided back into (x + 4) ??

jschen
07-26-07, 02:26 PM
That +4 is bound to the X though, so you can't just do 6x^2 x X = 6x^3, because you have that four hanging in there. You'd be better off factoring the top, it's really not that hard.

I didn't look close enough, my apologies; correct so far. I still stick with my factoring is easier though.

This approach is systematic and guarantees a result no matter how big the initial polynomial or how complex the factor. Taking educated guesses at the answer and back testing them gets to be impractical on much bigger problems.

red house
07-26-07, 02:27 PM
Whatever... I really don't care what you say about me, good or bad. Just do your homework and stop asking for help under the pretense of trying to test us. That's getting really old, and I don't need to prove my abilities to you.

Then.. you need 'not' (?) . .But that does not mean that there aren't other's who may appreciate and accept the oppertunity to do so, jschen.

jschen
07-26-07, 02:29 PM
******__6 x^2__________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40
********6 x^3 + 24 x^2

but this would now equal = -x^2 . . -which would be divided back into (x + 4) ??
Bring down the next term in your polynomial, the + 6 x. Just like a normal long division problem. And yes, divide again. What do you have to multiply (x + 4) by to get a - x^2 term?

******__6 x^2__________________
x + 4 | 6 x^3 + 23 x^2 + 6 x + 40
********6 x^3 + 24 x^2
******************-x^2 + 6 x

squegeeboo
07-26-07, 02:30 PM
Yeah, 23 being a prime was really discouraging and made me think I was going about it all wrong.

So, what does the ( . . )( . . )( . . ) look like?

Wow, it looks like I had some sloppy math in mine. I'm at a loss for how to reduce the (6x^2 -x + 10) thankfully, I've already completed all the math courses I had to take.

But the division some one had suggested does work out. It really is just long division, it just feels weird to do it with variables instead of actual numbers.

jschen
07-26-07, 02:31 PM
I'm at a loss for how to reduce the (6x^2 -x + 10)

You don't. The problem doesn't ask you to. If you really care, stick it in the quadratic formula for some complex (ie with real and imaginary components) roots.

red house
07-26-07, 03:24 PM
Okay.. hey this is great. 9 HW problems down, two more to go. :beer:

Anybody up for one more?

If f (x) = 1 / (x+5) . . then the quotient f (1 + h) - f (1) / h can be simplified to; -1/ (ah + b) where a = ___ and b = ___ . This is a tricky one that has defied thus far defied my best efforts.

BananaTugger
07-26-07, 03:27 PM
1/0 = No.

Prove it.

red house
07-26-07, 03:31 PM
1/0 = No.

Prove it.

0/1 = 'yes' .. the opposite of 'yes' = 'no', therefore the reciprical of 0/1 must be equal the to opposite of 'yes' (which is 'no').. Now you do mine. :)

BananaTugger
07-26-07, 03:36 PM
0/1 = 'yes' .. the opposite of 'yes' = 'no', therefore the reciprical of 0/1 must be equal the to opposite of 'yes' (which is 'no').. Now you do mine. :)

I'll do you alright...

Wait...

what?

jschen
07-26-07, 03:40 PM
Anybody up for one more?

If f (x) = 1 / (x+5) . . then the quotient f (1 + h) - f (1) / h can be simplified to; -1/ (ah + b) where a = ___ and b = ___ . This is a tricky one that has defied thus far defied my best efforts.

Sure... show some work first. Even if it's a dead end.

Falkon
07-26-07, 03:44 PM
Wait a sec, a nonrational real integer? Da hell does that mean? An integer is rational by definition. Or does 'nonrational' have a different definition than the more common term 'irrational'? And by 'real' I assume you mean 'noncomplex' rather than the computer science meaning of 'noninteger'.

Now, it's not difficult to derive the value of pi, but is that what you are asking? Your terminology is corn-fusing.

it means pi

red house
07-26-07, 03:44 PM
suxz0rz... mai... n00b?

I don't speak English. :p

I heard it in P&R dood, I'm not sure what it means. :D :beer:

BananaTugger
07-26-07, 03:48 PM
I heard it in P&R dood, I'm not sure what it means. :D :beer:

Those pundits.

What will they think of next? :D

red house
07-26-07, 03:53 PM
Sure... show some work first. Even if it's a dead end.

Okay, I show you my work - you give me the answer, or the right way to find it?

deal. :beer:

f(x) = 1 / (x+5) . . so the quotient f (1 + h) - f (1) / h = (( 1 / ( 1 / (1 + h) + 5) - ( 1 / 6 )) / h

= (6 - (1 + h) + 5) / (6 (1 + h) +5 ) / h = (6 - (1 + h) + 5) / h (6 (1 + h) +5 ) ..okay, now what? :) ?

can be simplified to; -1/ (ah + b)

red house
07-26-07, 04:06 PM
Those pundits.

What will they think of next? :D

You've heard the expression before? I just recieved a warning because apparently one of the moderators has also heard of it as well.. I thought that chipcom and kris were just making that **** up. :D -? But sorry, didn't mean to offend if I did. :beer:

jschen
07-26-07, 04:53 PM
f(x) = 1 / (x+5) . . so the quotient f (1 + h) - f (1) / h = (( 1 / ( 1 / (1 + h) + 5) - ( 1 / 6 )) / h

= (6 - (1 + h) + 5) / (6 (1 + h) +5 ) / h = (6 - (1 + h) + 5) / h (6 (1 + h) +5 ) ..okay, now what? :) ?

Let's try that again. You correctly simplified f(1) = 1/6
Let's redo f(1 + h). Get that term as simplified as you can.

alainp
07-26-07, 05:39 PM
f(1) = 1/6, which is simple enough. f(1+h), when simplified, comes out to 1/(h+6).

Plug in those results into the numerator [f(1+h) - f(1)] to get (1/(h+6) - 1/6) which gives you [6-(h+6)]/[6(h+6)]. Remember this from jschen's hint in the other thread with the limits? That is, (1/a + 1/b) = (b+a)/ab.
Okay, simplify this even further to get -h/6(h+6).

Now that you know the numerator, plug it all into the overall equation to get {-h/[(6(h+6)]**/h. The h's cancel out leaving -1/[6(h+6). Expand the numerator to get -1/(6h+36). Now, compare this to -1/(ah + b). Notice how they are in the same "form".

Assuming I haven't screwed anything up with my simplification, you should be able to figure out a and b. In fact, they're staring right at you shouting "Here I am!!!!":D