# Two Cans & String

ForumTechnical Corner ► Math Problem Thread
Edit: Nvm u changed it.

oh ok so i was on the right track. I had done a little something using square roots on the side.

If you wanted to factor the x2+i you would rewrite it as x2- (-i), which would factor to (x+sqrt(-1))(x-sqrt(-i)) and those are the 2nd and 3rd solutions

What?

rewrite the plus as subtracting a negative, so you can do the difference of squares factoring

W_Licky said:

x4+1 = 0

x4+1 = 0
x4-(-1) = 0
(x2 - i)(x2 + i) = 0
(x - √i)(x + √i)(x2 - (-i)) = 0
(x - √i)(x + √i)(x - i√i)(x + i√i) = 0

Check:
(√i)4 = i2 = -1 √
(-√i)4 = i2 = -1 √
(i√i)4 = (-i)2 = -1 √
(-i√i)4 = (-i)2 = -1 √

KylIjoy said:
i4 = -1
what???
i^4 = 1

Ummm i4 is 1, not -1.

Hm. Why didn't the square root symbols for those two copy over?

Edit: fixed. The root symbols for the two i's just didn't show up for some reason.

I’m also not 100% sure where i * sqrt(i) came from

Oh also, it's y=x4+1

edit: demented i think its bc √-i needs to be simplified to i√i. Like how √-9 is i√9.

I’m also not 100% sure where i * sqrt(i) came from

Difference of squares with (x2 - (- i)) = (x2 - (i2)(√i2))

Ohh that makes sense I didn’t see it that way

It's kinda annoying, because it's a multiplicity-one unique root problem.

Here are the roots plotted on the complex plane Ok I am trying to solve this problem:

How many rational numbers on the open interval (0,1) have a numerator and denominator that add to 1000 when in simplest form?

My first thought would be however many primes there are between 500 and 999 since that’s how many denominators there would be that would have a numerator less than them that would not simplify, but there are most likely nonprimes between 500 and 999 that are coprime with their compliment.

I honestly have no clue how to approach this now

For now I'll just leave a hint, but if I've been too cryptic please do let me know and I can write up a full explanation:

For which integers x with 0<x<10 are x and 10-x coprime?
For which integers x with 0<x<12 are x and 12-x coprime?
For which integers x with 0<x<7 are x and 7-x coprime?
For which integers x with 0<x<n are x and n-x coprime?

10: 1,3,7,9
12: 1,5,7,11
7: 1,2,3,4,5,6
n: all numbers that are not multiples of factors of n

the prime factorization of 1000 is 2^3 * 5^3, so all factors of 1000 are either even or multiples of 5

500 even numbers + 200 multiples of 5 - 100 overlap = 600 non-viable x values

and since we are looking for pairs of numbers, we need to divide the 400/2, giving an answer of 200

Is that correct?

200 is the answer I came to as well.

Problems like this, while interesting to think about, are ultimately why I prefer my degree in Statistics over pure mathematics. Multi-Dimensional Probability spaces are much more comforting that guessing and hoping for pattern-recognition.

I'm curious what kind of Multidimensional Probability Space questions you could pose that I wouldn't approach by leaning on similar principles of pattern recognition.

Extremely arbitrary ones, lol

There are numbers 0-9 on a keypad that can be made into 6-digit codes. How many different possible codes are there?

10^6 = 1,000,000. Am I missing something?

Actually, I guess you did say "numbers," not "integers" or something of the like, so the other answer I can come up with is infinity. Although that's an obviously wrong answer, even if it is technically correct in a stupid sort of way.

Or you also don't specify that you're looking for 6-digit codes when you actually ask the question, which might again imply that the answer is infinity, because codes of any length are acceptable. This feels slightly less wrong than my previous infinity, but is still obviously dumb.

Nope, it's just 1,000,000. Remember that he's asking a high school math problem, so it's going to be relatively clean by default.

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