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/sci/ - Science & Math

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>> No.8523881 [View]
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8523881

>>8523878
>loser
>fool
>liar
>FAGGOT
you must be a joy at parties

>> No.8453638 [View]
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8453638

>>8453630
good job! now set that equal to 1

>> No.8419864 [View]
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8419864

>>8419707
>>8419713
'yes' is the reasonable answer but one should remember the zero vector itself is never considered an eigenvector

>> No.8367746 [View]
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8367746

>>8367726
you need to be careful while doing modular arithmetic, all those calculations are mod 5, and so the symbol 0 really represents the equivalence class of 0 mod 5, which consists of every number congruent to 0 mod 5, i.e. all the usual integers {...-25, -20, -15, -10, -5, 0, 5, 10 ,15, 20, 25...} that are divisible by 5 with remainder of 0

by the definition of congruence given in the question, a = b mod 5 means that a-b is divisible by 5 (i.e. it's some number in the set i listed above), and 0 is also divisible by 5, so a-b = 0 mod 5

but this property i used to claim a-b=0 mod 5 is actually what you're trying to prove, so you wouldn't want to use that in your proof (to be more clear, if you want to say that a=b mod 5 implies a-b = 0 mod 5 then you need to know what you're trying to prove since that's essentially using a=b mod 5 and -b=-b mod 5 to imply a-b = b-b = 0 mod 5)

and so a more appropriate proof would be this one from my other post where you do in fact transition to division properties of integers:
a=b mod 5 implies 5 divides (a-b) and x=y mod 5 implies 5 divides (x-y).
therefore 5 divides [(a-b)+(x-y)] and so a-b+x-y=0 mod 5 or in other words a+x=b+y mod 5

>> No.8335119 [View]
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8335119

>>8335060
it's not quite true, you might want to use a different textbook then whereever you're getting this info from. you can find it on page 7 here (https://eprint.iacr.org/2005/374.pdf)) but basically you just:
1) write a=#E(F_5)-(5+1) and x^2-ax+5=(x-b)(x-c)
2) by theorem 4.12 in reference 15 in the above link we have #E(F_(5^n))=(5^n+1)-(b^n+c^n)
3) manually check that there's 6 points over F_5
4) so 6=#E(F_5)=5+1-a implies a=0
5) so x^2+5=(x-b)(x-c) implies b = -c = sqrt(-5)
6) so #E(F_5^n)=5^n+1 if n is odd, and #E(F_5^n)=5^n +- 2*5^(n/2)+1 if n is even

>> No.8253162 [View]
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8253162

>great at writing scripts
wat
name one theorem ever discovered by a computer

>> No.8252088 [View]
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8252088

>>8252047
no, just a low-grade master's student, and that description of the canonical deformation/theta-link is probably frighteningly inaccurate

>> No.8240418 [View]
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8240418

in case you're using griffiths shitty book you might want to pick up a different one btw

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