Thread: Math Fun

Well im pretty bored so how about making a quiz only on math? Like someone ask questions then when those are answered someone else may come with new questions so i start
How does this work
1:
2+2=irrational number (10,......)

2:
2+3=5
5+12=13
6+8=10

What base is this based on:
a: 1+1=10
b: 11=100

i´ll come with more later, lets see if you can figure this out, i bet someone will

Originally Posted by Zelos

2:
2+3=5
5+12=13
6+8=10

What base is this based on:
a: 1+1=10
b: 11=100

2. Base 14 works for all of these numbers.

a. binary
b. base 99

14 is wrong, becuase then i would have said base
a is right, b is wrong

Originally Posted by Zelos

14 is wrong, becuase then i would have said base
a is right, b is wrong

Then what are you asking for on 2 then? And for b, how can 2 numbers be the same without both sides being different bases?

I'm really confused. :?

that is the point, youa re supposed to figure it out, but i ensure you its the same base

I don't even understand your 2nd question

11=100

Do you mean in what base does 11 = 100 in decimal? If that's the question then yeah, base 99 is right.

no, i mean that 11 in one base is 100 in the same base

11=100 in Fibonacci base, right?

hmm, not familiar with that name, any other names its known under?

I've heard of a Fibonacci sequence, but I don't think it has anything to do with this. (been a few years though). As far as I know there's no base where 2 different numbers are equal to each other...

there is thoe

Well, golden ratio base then. (I thought I'd heard of fibonacci base before, but I think this is what I was thinking of.) This is related to fibonacci coding though.

Also, the first and second questions look like pythagorean triplets to me. Then '+' would be defined as a '+' b = sqrt(a*a + b*b). (Two different +'s there).

Ah, it figures that it'd be some wacky-ass irrational business like that :P

magimaster is correct, its the golden ration base and its pythagoras where you dont write the square. anyone able to figure out nr1?

Nice pic vampyr, very friendly.

Hmm... questions.

Prove e^2*pi*i-1 = 0

Originally Posted by MagiMaster

Prove e^2*pi*i-1 = 0

r u mad or a bot?????? i wld need to no the oiler identitties and the taylors series to prve it. my iq is 14.5 O RLY? YA RLY. ya, i wld also nede to no abt complexy type numbr type things.

hoo wants any pi anyways, but gimeee e any da. il ly down now

O RLY?

MikeNiv, you have an IQ of 145. You should be able to do it. It's not hard.

Well, in any case, if I can think of something better, I'll post it later.

Off-topic posts deleted. Please do not insult each other. Play nice in the Math section guys.

Originally Posted by MagiMaster

Prove e^2*pi*i-1 = 0

Not for the first time, I don't have a hot date tonight, so I thought this through. It is actually not trivial to prove from first principles. But I figured it. I wonder if you have. It is certainly not something I would expect a 15 yr old to do, however clever (and irritating)

[quote="Guitarist"]

Originally Posted by MagiMaster

. It is certainly not something I would expect a 15 yr old to do, however clever (and irritating)

Fifty seven year old geologists do not have much luck with it either. :?

Originally Posted by Ophiolite

Fifty seven year old geologists do not have much luck with it either. :?

Hmm. Drink has been taken, as they say in Scotland. Maybe I'll show you tomorrow, if I can work out how to make symbols here.

Anyway O. My cleverest (and richest) friend is a geologist. I trust you are both also?

Originally Posted by MagiMaster

Prove e^2*pi*i-1 = 0

e^x*i = cos(x)+i*sin(x)
e^2*pi*i = cos(2*pi)+i*sin(2*pi)
e^2*pi*i = 1+0*i
e^2*pi*i = 1
e^2*pi*i-1 = 0

good and easy evidence

Ok. I agree that it'd be hard to do without either e^x*i = cos(x) + i*sin(x) or the taylor series expansion of e^x. I was assuming either of those could be used. Hmm... I can't ever think of good math puzzles... :?

Originally Posted by MagiMaster

Ok. I agree that it'd be hard to do without either e^x*i = cos(x) + i*sin(x) or the taylor series expansion of e^x. I was assuming either of those could be used. Hmm... I can't ever think of good math puzzles... :?

And I was thinking we may not assume the Euler identity, which isn't that hard to prove, as you say, using the Taylor expansion. But then, once you switch to the complex plane you have to be really cautious.

Do you think of it as a "math puzzle"? It is one the most beautiful bits of math that I know!

IQ of 145 doesnt mean ive studied further maths - only GCSE

Ok... I think I've got one this time. (And yeah, I just like that identity.)

Some people are trying to build a step pyramid. Each level will be 1 meter high. The top level will be 2 meters by 2 meters. Each level below that will be 2 meters wider and 2 meters longer than the one above it.

Assume that the builders can get any number of 1x1x1, 2x2x2 and 3x3x3 blocks. What is the fewest number of blocks that would be needed to build a pyramid 4 meters tall? 7 meters? k meters?

145? didnt your parents tell you its bad to lie?

sounds great

please digg

http://digg.com/users/lizziekean/submitted

Originally Posted by Zelos

What base is this based on:
a: 1+1=10
b: 11=100

THis is a trick methinks!
the left had side of (a) could be any base, the right hand side binary
IF
the left hand side of (B) is base 99 then the right is decimal.
IF
if the left hand side of (b) is base 3 then the right hand side is binary

THus the only one that matches is LHS BASE 3, RHS Base 2

If that's not right then it must be 'borg thing'

Nice one Zelos! :wink:

Edit:

Just noticed there's 3 pages so if anyone else answered first........

no its not base 3 nor 99

Ok smarty

A borg who's up on quantum physics ought to get this in a few 'picoseconds'.


one squared = 16556761


How come?