Maths

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Probably what *this* should be called.
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Re: Maths

Post by Khaos » Sat May 08, 2021 11:01 pm

lol

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Re: Maths

Post by the_zoomies » Wed May 12, 2021 3:35 am

If you're going to be 2 boxing, please change prophylactics before switchin

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Re: Maths

Post by Crunchums » Wed May 12, 2021 3:36 am

pfffffff
u gotta skate

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Re: Maths

Post by Doug » Thu Jan 06, 2022 3:22 pm

I thought this was a cool problem but maybe it's dumb. I'm paraphrasing

Fred tells Lucas, I'm thinking of two different one-digit whole numbers. Can you tell me their sum

Lucas says, the sum of the numbers you happen to be thinking of? Of course not

Fred says, ah but what if I told you that when we multiply the two numbers, the last digit of the product is the same as the number you played on the roulette wheel a moment ago, now can you tell me their sum

Lucas says, well actually yes

What is their sum
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Re: Maths

Post by Crunchums » Thu Jan 06, 2022 4:04 pm

Spoiler!
0 -> lots of ways to get here obv
1 -> 3 and 7 only
everything greater than 1 will have X and 1 so any other pairing with a different sum makes it impossible for it to be that one
2 -> 2 and 6
3 -> 9 and 7
4 -> 7 and 2
5 -> 5 and 3
6 -> 3 and 2
7 -> 3 and 9
8 -> 2 and 4
9 -> 9 and 1 only

so 10
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Re: Maths

Post by Khaos » Thu Jan 06, 2022 4:43 pm

Spoiler!
if the number is 36, the factors are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

we can eliminate 12, 18, and 36 since they are two digit numbers. and thus eliminate 1, 2, and 3. we can eliminate 6, because the number can't be duplicated. so the answer is 4 + 9 = 13. but that's one case. i don't wanna do all the cases. but 36 is the most highly composite number within the provided range so it's probably the worst case

BUT! i looked at a table of factors and it looks like it fails for the following roulette numbers:

24
18
12
8
6

weird

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Re: Maths

Post by Khaos » Thu Jan 06, 2022 4:44 pm

oh the last digit of their sum. woops

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Re: Maths

Post by Doug » Thu Jan 06, 2022 4:48 pm

Khaos wrote: Thu Jan 06, 2022 4:44 pm oh the last digit of their sum. woops
no the last digit of their product
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Re: Maths

Post by Khaos » Thu Jan 06, 2022 5:11 pm

Doug wrote:
Khaos wrote: Thu Jan 06, 2022 4:44 pm oh the last digit of their sum. woops
no the last digit of their product
right. that's what i meant. anyway the important part is that i did it wrong

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Re: Maths

Post by groff_enthusiast » Fri Jan 07, 2022 3:12 am

Totally a barenstein effect here, but I could have sworn there were two different "pairs" with an identical sum but different product. I got the same answer as crunchums

Edit: I both can't read and can't think, yeah that's exactly what this is.
Ashenai wrote: Sat Jul 16, 2022 5:08 pmoh no

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Re: Maths

Post by Luna » Thu Aug 04, 2022 5:43 am

Image

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Re: Maths

Post by Ashenai » Thu Aug 04, 2022 5:47 am

that is a good looking riemann surface

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Re: Maths

Post by Jeb Bush 2012 » Thu Aug 04, 2022 2:22 pm

thank you arti

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Re: Maths

Post by Khaos » Thu Aug 04, 2022 2:43 pm

you've done a wonderful job

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Re: Maths

Post by Doug » Thu Aug 04, 2022 3:01 pm

Time Cube except nothing but positivity
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Re: Maths

Post by Doug » Thu Apr 20, 2023 8:40 pm

You guys are math people so you probably find this easy and dumb. I find it fascinating, though

i = i
=i^1
=i^(4/4)
= (i^(4))^(1/4)
=1^(1/4)
=1

which step is illegal
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Re: Maths

Post by Crunchums » Thu Apr 20, 2023 8:46 pm

i would guess this one
Doug wrote: =i^(4/4)
= (i^(4))^(1/4)
but idk
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Re: Maths

Post by Rylinks » Thu Apr 20, 2023 9:26 pm

1^1/4 equals +/- 1 and +/- i

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Re: Maths

Post by Doug » Thu Apr 20, 2023 9:28 pm

Rylinks wrote: Thu Apr 20, 2023 9:26 pm 1^1/4 equals +/- 1
That's very true, but neither of those is actually equal to i even though the erroneous steps would seem to prove they are
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Re: Maths

Post by Rylinks » Thu Apr 20, 2023 9:28 pm

Doug wrote:
Rylinks wrote: Thu Apr 20, 2023 9:26 pm 1^1/4 equals +/- 1
That's very true, but neither of those is actually equal to i even though the erroneous steps would seem to prove they are
it has four roots, see the edit

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Re: Maths

Post by Doug » Thu Apr 20, 2023 9:32 pm

Crunchums wrote: Thu Apr 20, 2023 8:46 pm i would guess this one
Doug wrote: =i^(4/4)
= (i^(4))^(1/4)
but idk
That is it!

It turns out that while x^(m/n) is trivially (x^m)^(1/n) for real numbers x, you can't do this if x is imaginary or complex

In fact, (i^4)^(1/4) equals four different numbers while (i^(1/4))^4 only ends up equaling i

This is another way of saying that (x^m)^(1/n) does not necessarily equal (x^(1/n))^m for imaginary numbers
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Re: Maths

Post by Doug » Thu Apr 20, 2023 9:33 pm

Rylinks wrote: Thu Apr 20, 2023 9:28 pm
Doug wrote:
Rylinks wrote: Thu Apr 20, 2023 9:26 pm 1^1/4 equals +/- 1
That's very true, but neither of those is actually equal to i even though the erroneous steps would seem to prove they are
it has four roots, see the edit
Right, and one of them is i

Of course the drama comes from the other three not being i
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Re: Maths

Post by Khaos » Thu Apr 20, 2023 10:02 pm

Doug wrote: =i^(4/4)
= (i^(4))^(1/4)
you cant just move numbers into the exponent willy nilly

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Re: Maths

Post by Khaos » Thu Apr 20, 2023 10:04 pm

Doug wrote:
Crunchums wrote: Thu Apr 20, 2023 8:46 pm i would guess this one
Doug wrote: =i^(4/4)
= (i^(4))^(1/4)
but idk
That is it!

It turns out that while x^(m/n) is trivially (x^m)^(1/n) for real numbers x, you can't do this if x is imaginary or complex

In fact, (i^4)^(1/4) equals four different numbers while (i^(1/4))^4 only ends up equaling i

This is another way of saying that (x^m)^(1/n) does not necessarily equal (x^(1/n))^m for imaginary numbers
ok this is more complicated than i thought

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Re: Maths

Post by Rylinks » Thu Apr 20, 2023 10:16 pm

-1
sqrt(-1^2)
1

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Re: Maths

Post by s. goblin » Thu Apr 20, 2023 10:37 pm

i'm dying sqrt

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Re: Maths

Post by Doug » Thu Apr 20, 2023 10:38 pm

sqrt game
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Re: Maths

Post by Magical » Thu Apr 20, 2023 10:54 pm

Rylinks wrote: Thu Apr 20, 2023 10:16 pm -1
sqrt(-1^2)
1
Yeah Rylinks is right, it's not about the base being imaginary or complex because this shows it breaks for reals too

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Re: Maths

Post by Rylinks » Thu Apr 20, 2023 11:02 pm

i think another relevant concept is the distinction between solving for a variable, which is what most people are used to, and proofs

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Re: Maths

Post by Doug » Thu Apr 20, 2023 11:41 pm

Magical wrote: Thu Apr 20, 2023 10:54 pm
Rylinks wrote: Thu Apr 20, 2023 10:16 pm -1
sqrt(-1^2)
1
Yeah Rylinks is right, it's not about the base being imaginary or complex because this shows it breaks for reals too
-1 isn't a variable, so I think order of operations requires you to simplify the sqrt and the ^2 before you do anything else

Wait no that's not right

X^AB should equal (X^A)^B and (X^B)^A both among reals even if A or B are fractions, shouldn't it
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