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I understand 3, but 2, how??

Hey, you can probably use induction for proving II.

The given information doesn't include \(n* (x \oplus y)\), hence we can't define the product of it.

We don't have \(k(x\oplus y) = kx\oplus ky\)

But that's the inductive step!

But first off, the operation must be defined, right?

It has been defined recursively...

\[(k+1)x = kx \oplus x\]

Hey, you could observe that the operators are + and * though, right?

ok, you can say if \(k (x\oplus y)\), then \((k+1) (x\oplus y)= k(x\oplus y)+(x\oplus y)\)

That \(+\) should be \(\oplus\). Now \(k(x\oplus y)=kx \oplus ky\).

hihihi... I don't see anything allows me to distribute like that.

That is the inductive step, man.

Ok, I temporarily "pretend" that I agree with you on that step, then??

Anyway, I gotta go. So you see how this induction works, right?

I don't know what given condition allows me pick like terms to add up. :(

Hey, that is the definition of \(\cdot\).\[(k+1)x = kx\oplus x\]given in your question

That \(\color{red}+\) sign should be \(\oplus\).

OK, let's start from scratch. Sorry for being a little casual the last time.

Do steps please.

Question: is it a Set theory problem??

I know induction.

me too, hes a jerk sometimes

fun

induction is my he

ok, :) good luck (for your final)

I am giving it, not taking it :)

We've thus proven \(P(k+1)\).

Try proving all those properties for rational numbers as an exercise.

the line " definition of dot" !!!

oh, got it. hihihih

Thanks a ton.............

Oh - never mind then...

Finally, you can crack my skull to put the logic in. Thanks for being patient to me.

Again, thank you so much.:)