Please, explain me.

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Please, explain me.

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I understand 3, but 2, how??
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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.

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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. \((k+1)(x\oplus y) = k (x\oplus y) \oplus (x\oplus y) \)
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 \(\oplus\)
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.
We have assumed that \(k(x\oplus y) = kx \oplus ky\). We have to show that the same holds for \(k+1\)
Ok, I temporarily "pretend" that I agree with you on that step, then??
\[kx \oplus ky \oplus x \oplus y\]Now collect the terms and you get \((k+1)x \oplus (k+1)y\) which was to be shown.
Anyway, I gotta go. So you see how this induction works, right?
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 \((n+1) (x\oplus y) = n(x\oplus y) + (x\oplus y)\\nx\oplus ny \color {red}{+} x\oplus y \)
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??
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".
I know induction.
me too, hes a jerk sometimes
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
@zzr0ck3r who is your "he" ??
fun
induction is my he
ok, :) good luck (for your final)
I am giving it, not taking it :)
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\[(k+1)\cdot (x\oplus y)\]\[=k\cdot (x \oplus y) \oplus (x\oplus y)\tag{definition of dot}\]\[= kx \oplus ky \oplus x \oplus y \tag{inductive assumption}\]\[= kx \oplus x \oplus ky \oplus y \tag{commutativity}\]\[= (k+1)\cdot x \oplus (k+1)\cdot y \tag{using definition of dot}\]
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.............
\[(k+1) \cdot x = k\cdot x \oplus x\]\[(k+1)\cdot \heartsuit = k \cdot \heartsuit \oplus \heartsuit\]\[(k+1)\cdot \spadesuit = k \cdot \spadesuit \oplus\spadesuit \]
Oh - never mind then...
Finally, you can crack my skull to put the logic in. Thanks for being patient to me.
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.
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...
Again, thank you so much.:)
Which is why it is useful to know that these operators are just addition and multiplication. lol :P No problem!

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