Maths
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- Attaboy! You finished my labyrinth and I'm proud of you!
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Re: Maths
If you're going to be 2 boxing, please change prophylactics before switchin
Re: Maths
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
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
no the last digit of their product
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- The original good boy.
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Re: Maths
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.
Edit: I both can't read and can't think, yeah that's exactly what this is.
Re: Maths
Time Cube except nothing but positivity
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Re: Maths
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
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
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
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
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
ok this is more complicated than i thoughtDoug wrote: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
Re: Maths
sqrt game
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Re: Maths
-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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