inverse modulo

[umbral canyon]

Why such property works for any k in mod p?
I want to find some proof different from the fact that 1/k + -1/k = 0

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[hard gyro]

I don't understand your concern, for 1/(-k) = 1/(-1) x 1/k =- 1/k.

you could also multiply everything by -k and check this is 0

[umbral canyon]

indeed xD thx

+close

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[wet haven]

Because it's modular arithmetic.

[wet haven]

Is this actually true?

[hard gyro]

is there a difference ?

[wet haven]

Yes. Modular arithmetic is only defined over integers.

[hard gyro]

if k is any element invertible in a ring, then 1/(-k) = -1/k

it doesn't matter which ring

[wet haven]

Basically what you want to prove is that (-k)^(-1) == -(k)^(-1) (mod p).

[wet haven]

When you said 1/k, I thought you were talking about a rational number.

[hard gyro]

no, it is just another notation for k^(-1)

(faster to type, and arguable in a field)