# Two Cans & String

ForumTechnical Corner ► Math Problem Thread
woah, I didn't realize you could set custom colors like that
I feel like that shouldn't be my main takeaway, it's just pretty cool

Yeah I only discovered the colors recently

i fixed the issue. i accidently had the initial direction multiplied by the velocity

Ok so i have 2 pretty abstract questions

if a number is chosen completely randomly between 0 and 1, there is a probability of 0 of the number being rational (this is my first assumption that might be wrong). For the number to be rational, it has to either terminate, and therefore have an infinite number of 0s after the point of termination, or it repeats, meaning the same repeating sequence has to be randomly chosen infinitely. i think the most likely possible fallacy in my logic is that I am assuming infinitely generating digits is the same as choosing a number completely randomly. Is that enough to prove that the number will always be irrational?

My second math rambling: For any finite range of numbers, there is the same number of numbers as all finite ranges of numbers. this can be proved by correlating the numbers. For any number between upper/lower limit, U_1 and L_1 of range 1, it can be correlated to a number between upper/lower limit U_2 and L_2 of range 2. (x-L_1)(U_2-L_2)/(U_1-L_1)+L_2. Is it also the same number of number for an infinite range on numbers?

I'm pretty sure those are both correct.

The first one is actually a statistical law, for any probability curve P, P(x)=0 for any x. That's why you have cumulative functions instead of instantaneous functions in statistics. The probability of any specific number is 0.

If you need proof, take the integral of any probability curve where the limits of integration are equal to each other. It's always 0, by the laws of Integral calculus

It's not just impossible to get rational numbers, it's impossible to get any number.

(You can trust me, I'm a statistics major)

Edit, Redux: On reviewing your theory, what you have proposed isn't a number at all, since it fails to be algebraic or transcendent due to a lack of a defined, consistent value or entanglement between digits. It's just a concatenated sequence, so no algebraic or mathematical rules apply. TL;DR, It can't, by nature, be either rational or irrational, because it's not a number

Can you give a simpler explanation of why it is not a number?

A sequence of digits is not inherently a number.

Basically, yes. It's defined randomly and not by any formula. It's an infinite sequence of random digits; a "number" must have a predictable, set value.

In other words, randomly generating digits ≠ selecting random numbers because math, probability, and statistics.

so does that mean it is impossible to generate a completely random number?

Yes, because there are infinite numbers within any range where a ≠ b. Thus, it is impossible to generate any number fairly without discrete intervals.

Logically, it makes sense. With your example, you could never generate even one number because you would be generating digits for eternity

I guess that makes sense.

anyways have a desmos related, math-adjacent question.

if I have a function f(x), how would I go about making a function g(x,y) = f(f((...x))) ?
the ... representing nesting f(x) a total of y times.

i could make specific definitions of g(x,y) based on the definition of f(x), but I want to make a generalized g(x,y)

Solve the recursive function for a closed form. Otherwise, you are probably SOOL.

Does it let you define a recursive function at all? I haven't tried.

Nope.

Okay this one is a bit of a doozy and I'm having difficulty manipulating it

Let * be an associative function with identity "e". Prove that if both a and b are inversible, a * b is inversible.

I've gotten decently far
Let a's inverse be a-1 and b's inverse be b-1

e = a * a-1 = a-1*a (and similarly for b)

It's pretty easy to prove that:
e = a-1*(a * b) * b-1
given e * e = e
but I don't know where to go from here. Commutativity isn't here here so I'm kind of stuck unless I can find a way to move the a * b to one side.

Edit:
I think I got it, I can perform a * to begin with on both sides, then since e is the identity a*e=e*a then I can apply * a-1 to both sides, then by association I can prove that (a*b)-1=b-1*a-1 and then by finding it I have proven it exists

Thought this one was fun. Thought I'd share it here, for the fellow math nerds.
From The Q&A:
How many numbers have at least one '1' among their digits? (I.E. '213', '12', '1', '1573', and '12141' are all valid)

Solution: (If you do not wish to solve it on your own)
Overly Complex Solution
Spoiler text below. Highlight to read.
Generalize for all numbers between 1 and 10n:
• Define it as a recursive function, where cn is the count of numbers with '1' in them
• Base case is 1
• For n > 1, 10% of the numbers are guaranteed to have at least one '1', since 10% of the numbers will start with 1 (out of the ten possible digits, 0-9). Thus, 10n-1.
• The remaining numbers are just 9 * cn-1. For 1000, for example, it already contains all of the numbers from 100. However, 200, 300, ..., 900 also contain the same amount of valid numbers as 100. Thus, 9 * cn-1
• This means c1 = 1, cn = (10n-1) + (9 * cn-1)

Solve for the closed form of this and simplify
• Spoiler alert: It's 10n - 9n

Divide by 10n (the total number of all candidates) to get the percentage between 1 and 10n
• We get 1 - (9/10)n
Take the limit
• Value is 1.

For all numbers between 1 and inf, ~100% of the numbers have at least one '1' in them.
• Positive and negative spaces are symmetric, so you can generalize from -inf to inf.
• So for all natural numbers, ~100% have at least one '1' in them.
QED.

Simple Solution
Spoiler text below. Highlight to read.

Of the available numbers, 9/10 will not have a '1' in the first place.
• Of those remaining numbers, 9/10 will not have a '1' in the second place.
• Of those remaining, 9/10 will not have a '1' in the third place
• As 9/10 < 1, limit is clearly 0.

To find the fitting numbers, remove the non-fitting numbers from all numbers
• 1 - 0 = 1

For all numbers between 1 and inf, ~100% of the numbers have at least one '1' in them.
• Positive and negative spaces are symmetric, so you can generalize from -inf to inf.
• So for all natural numbers, ~100% have at least one '1' in them.
QED.

Kylljoy said:
In other words, randomly generating digits ≠ selecting random numbers because math, probability, and statistics.
I'm late, but I disagree. If you accept that one can select a countably infinite sequence of random digits, then I think you have to accept that that process selects a random real between 0 and 1. If we've chosen each digit di, then x = d1/101 + d2/102 does define a real number (uniquely, I think).

I don't think that this set is countable, though.

If we've chosen each digit di, then x = d1/101 + d2/102 does define a real number (uniquely, I think).

Only exceptions to uniqueness are with trailing 9s and 0s representing all the same numbers as each other (which itself has two exceptions: 0 is only .000…, and 1 is only .999…).
Kylljoy said:
I don't think that this set is countable, though.

Which set specifically do you mean, and why does its countability matter?

Ah, great point, justabitjaded. I was thinking about 1 = .99..., which doesn't apply, but the same ambiguity exists for any terminating digit sequence. So the mapping from random digit sequences to [0, 1] is onto but not 1-1.

I guess this is a bump.

Find the zeroes.
x4+1

Here's what I got so far.
x4+1
(x2+i)(x2-i)
(x+i)(x-i)(?)(?)

ok so basically it is x4=-1 so x is equal to the 4 4th roots of -1

I am pretty sure you made a mistake on your last step,

anyways with roots of complex numbers, you just divide the angle by the root

in this case, -1 has an angle of pi. your first solution is ei(pi/4)

an angle of pi is also an angle of 3pi, which means your 2nd solution is ei(3pi/4)

its also 5pi, so your third solution is ei(5pi/4), and your final solution is ei(7pi/4)

i prob explained that really bad so let me know if i need to explain it again. Or it is quite likely i made a mistake somewhere too

I am pretty sure you made a mistake on your last step,

So you factored x2-i as (x-i)(x+i)

you forgot to take the sqrt of i, you have to take the sqrt of both terms

in reality it would factor to (x-sqrt(i))(x+sqrt(i))

those 2 solutions are the first and the last solutions that I stated in my explanation above

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