Please, explain me.

- Loser66

Please, explain me.

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- schrodinger

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- Loser66

I understand 3, but 2, how??

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- ParthKohli

Hey, you can probably use induction for proving II.

- Loser66

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

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## More answers

- ParthKohli

OK, one method to do this problem is to recognize the operators as \(+\) and \(\cdot\).
Here is how you use induction.
It holds true for \(n = 1\).
Let \(k(x\oplus y) =kx \oplus ky\).
We have to prove that \((k+1)(x \oplus y) = (k+1) x \oplus (k+1)y\).
Now we can use the recursive definition. \((k+1)(x\oplus y) = k (x\oplus y) \oplus (x\oplus y) \)

- Loser66

Look at 1) we have if \(x^k\) defined, then .....
but to \((x\oplus y)\), we don't have any thing to define \((x\odot y)^k\), hence 1) is not the correct one.
And if it is so, the same logic for 2) which indicates to \(\oplus\)

- Loser66

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

- ParthKohli

But that's the inductive step!

- Loser66

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

- ParthKohli

It has been defined recursively...

- ParthKohli

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

- ParthKohli

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

- Loser66

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

- ParthKohli

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

- Loser66

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

- ParthKohli

That is the inductive step, man.

- ParthKohli

We have assumed that \(k(x\oplus y) = kx \oplus ky\). We have to show that the same holds for \(k+1\)

- Loser66

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

- ParthKohli

\[kx \oplus ky \oplus x \oplus y\]Now collect the terms and you get \((k+1)x \oplus (k+1)y\) which was to be shown.

- ParthKohli

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

- Loser66

I am sorry for being dummy. I don't get it still.
I rearrange what you said
\(Assume\) \(n(x\oplus y ) = nx \oplus ny\)
now prove \((n+1) (x\oplus y) = (n+1) x \oplus (n+1) y\)
Given condition gives us \((n+1) (x\oplus y) = n(x\oplus y) + (x\oplus y)\\nx\oplus ny \color {red}{+} x\oplus y \)

- Loser66

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

- ParthKohli

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

- ParthKohli

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

- ParthKohli

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

- Loser66

Do steps please.

- Loser66

Question: is it a Set theory problem??

- ParthKohli

OK, so we have to prove \(n\cdot (x\oplus y)\) for all positive integers \(n\). As a matter of fact, this holds for all real numbers (which I'm not going to cover), so proving it for positive integers is going to be simper than that.
What tells us to use induction is that the definition of \(\cdot\) is given recursively in terms of \(k\) and \(k+1\), so we're going to have to prove in such a way that we only have to work with \(k\) and \(k+1\). That method of proof is called induction.
Let \(P(n)\) be a given statement that you have to prove for all positive integers \(n\). We first prove that \(P(1)\) (known as the base-case) is true. Now comes the "inductive step". We first assume that \(P(k)\), where \(k\) is an arbitrary positive integer is true. We have to show that \(P(k+1)\) is true assuming \(P(k)\). This has a "domino-effect".

- Loser66

I know induction.

- zzr0ck3r

me too, hes a jerk sometimes

- zzr0ck3r

I am proctoring a test, so I will look at this problem and come back, not sure if you solved it yet because it looks like fin

- Loser66

@zzr0ck3r who is your "he" ??

- zzr0ck3r

fun

- zzr0ck3r

induction is my he

- Loser66

ok, :) good luck (for your final)

- zzr0ck3r

I am giving it, not taking it :)

- ParthKohli

Here is the given statement: \(P(n): n\cdot(x\oplus y) = n x \oplus ny\).
Vital assumptions:
1. commutativity, associativity and closure under \(\oplus\).
2. \((k+1)x = kx \oplus x\).
Let us first show \(P(1)\), which is trivial.
Now we assume that \(P(k): k\cdot(x\oplus y)= kx \oplus ky\) is true for \(k\). Let us use this to prove \(P(k+1)\).\[P(k+1): (k+1)\cdot (x\oplus y) = (k+1)\cdot x \oplus (k+1) \cdot y\]Now\[(k+1)\cdot (x\oplus y)\]\[=k\cdot (x \oplus y) \oplus (x\oplus y)\tag{definition of dot}\]\[= kx \oplus ky \oplus x \oplus y \tag{inductive assumption}\]\[= kx \oplus x \oplus ky \oplus y \tag{commutativity}\]\[= (k+1)\cdot x \oplus (k+1)\cdot y \tag{using definition of dot}\]

- ParthKohli

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

- ParthKohli

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

- Loser66

the line " definition of dot" !!!

- Loser66

oh, got it. hihihih

- Loser66

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

- ParthKohli

\[(k+1) \cdot x = k\cdot x \oplus x\]\[(k+1)\cdot \heartsuit = k \cdot \heartsuit \oplus \heartsuit\]\[(k+1)\cdot \spadesuit = k \cdot \spadesuit \oplus\spadesuit \]

- ParthKohli

Oh - never mind then...

- Loser66

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

- ParthKohli

Anyway, if you ever receive these questions on a multiple-choice test, spend a minute thinking if these operators are related to any that we already know. That makes things easier.

- Loser66

The problem is: I have to have the answer in about 2 minutes. I need tricks to figure out right after seeing the problem. To this problem, option 3 is trivial, but option 2 made no sense to me. Whatever, I got it now.
oh, yeah... life is beautiful now. lalalalala...

- Loser66

Again, thank you so much.:)

- ParthKohli

Which is why it is useful to know that these operators are just addition and multiplication. lol :P
No problem!

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