Hi @LeakyNun

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@LeakyNun Isn't every element of $\mathbb{N}$ the identity of element of the binary operation $\mathbb{gcd(a,b)}$?

that isn't what identity means

okay, got it.

lol

@LeakyNun Why does symmetry across the main diagonal of the binary operation table imply commutativity?

let * denote the operation

let f(i,j) be the (i,j)-th entry

symmetry across the main diagonal means f(i,j) = f(j,i) for all i and j

commutativity means i*j=j*i for all i and j

but the (i,j)-th entry of the cayley table is just i*j

@Abcd aka "by thinking about it for more than a second"

yes

@LeakyNun Why is $-5 \mod 3= 1$ and not $2$

division algorithm: -5 = 3*(-2) + 1

@LeakyNun isn't $1 \mod 5 = 0$?

1 = 5*(0) + 1

@LeakyNun $1=0.2\times 5 +0 $

bye

you clearly aren't serious

I realised its wrong.

mod is only for integers.

@LeakyNun mod hasn't been taught to us...but there's a question in relations and functions chapter that says $ab \mod 5$ means the remainder obtained after dividing by 5.

@LeakyNun While proving associativity in binary operations why don't we chose the "extra element" from $A\times A$? We rather choose it from $A$.

Isn't that wrong?

For example to prove addition of natural numbers is associative.

We take c from N and not $(c,d)$ from $N\times N$

but the function is supposed to take values from it's domain = $N\times N $ which consists of ordered pairs

are you there?

Never mind, I got it.