Silent Witch Chinmoku no Majo no Kakushigoto - Vol. 1 Ch. 3.2
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So apparently the poor girl was beat as a child and would go to reciting math as a coping mechanism. That is bad and probably explains a lot about her social anxiety...
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Her pouch of lunch was actually full of nuts because she doesn't eat a lot (in an unhealthy way)
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Rather than saying there's no way that's her lunch, there's a rather easy solution for that.
Her backstory is a bit common, in a way, but with a different way to cope.
Can't even stand someone she likes and trusts.
Her backstory is a bit common, in a way, but with a different way to cope.
Can't even stand someone she likes and trusts.
Well, he's a cat, so therefore a villain by default, though I'm not sure I'd call her a "perfectly decent human". Morally, maybe, but her sense...
Yeah, that's about it. There's more to it regarding why she got beat up and why she uses math to cope, but suffice to say, she's met some bad people in her younger days.
Nah, it's someone who hasn't appeared yet. Cat's black haired.
The Fibonacci sequence's adjacent elements are coprime?
That sounds fun to prove.
Let's see... Aha! Take any segment (a b c) in the sequence. We can see that a=c-b. Thus if b and c are both divisible by x, then so is a. We can see that the preceding segment (z a b) will have a and b divisible by x, so z is also divisible by x. We can keep repeating this until we reach the start of the sequence. We can form a statement: if an adjacent pair is both divisible by x, then all preceding elements are also divisible by x. Since the sequence contains 1, then such x other than 1 cannot be found, since 1 would have to be divisible by it.
Or, if b is divisible by x and a isn't (iff this applies for all b's divisors, then b and a are coprime), then c=a+b also isn't divisible by x. So if a and b are coprime, then b and c must also be coprime.
2 and 3 are coprime, so by induction 3 and 5 are, so 5 and 8 are, etc. To infinity. (hmm... I wonder if the statements "0 and 1 are coprime" and "1 and 1 are coprime" are valid.)
That was pretty simple, but fun.
That sounds fun to prove.
Let's see... Aha! Take any segment (a b c) in the sequence. We can see that a=c-b. Thus if b and c are both divisible by x, then so is a. We can see that the preceding segment (z a b) will have a and b divisible by x, so z is also divisible by x. We can keep repeating this until we reach the start of the sequence. We can form a statement: if an adjacent pair is both divisible by x, then all preceding elements are also divisible by x. Since the sequence contains 1, then such x other than 1 cannot be found, since 1 would have to be divisible by it.
Or, if b is divisible by x and a isn't (iff this applies for all b's divisors, then b and a are coprime), then c=a+b also isn't divisible by x. So if a and b are coprime, then b and c must also be coprime.
2 and 3 are coprime, so by induction 3 and 5 are, so 5 and 8 are, etc. To infinity. (hmm... I wonder if the statements "0 and 1 are coprime" and "1 and 1 are coprime" are valid.)
That was pretty simple, but fun.
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