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Loser66
 one year ago
Please, explain me.
Loser66
 one year ago
Please, explain me.

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I understand 3, but 2, how??

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Hey, you can probably use induction for proving II.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0The given information doesn't include \(n* (x \oplus y)\), hence we can't define the product of it.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4OK, 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
 one year ago
Best ResponseYou've already chosen the best response.0Look 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
 one year ago
Best ResponseYou've already chosen the best response.0We don't have \(k(x\oplus y) = kx\oplus ky\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4But that's the inductive step!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0But first off, the operation must be defined, right?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4It has been defined recursively...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4\[(k+1)x = kx \oplus x\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Hey, you could observe that the operators are + and * though, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0ok, you can say if \(k (x\oplus y)\), then \((k+1) (x\oplus y)= k(x\oplus y)+(x\oplus y)\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4That \(+\) should be \(\oplus\). Now \(k(x\oplus y)=kx \oplus ky\).

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0hihihi... I don't see anything allows me to distribute like that.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4That is the inductive step, man.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4We have assumed that \(k(x\oplus y) = kx \oplus ky\). We have to show that the same holds for \(k+1\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Ok, I temporarily "pretend" that I agree with you on that step, then??

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4\[kx \oplus ky \oplus x \oplus y\]Now collect the terms and you get \((k+1)x \oplus (k+1)y\) which was to be shown.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Anyway, I gotta go. So you see how this induction works, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0I don't know what given condition allows me pick like terms to add up. :(

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Hey, that is the definition of \(\cdot\).\[(k+1)x = kx\oplus x\]given in your question

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4That \(\color{red}+\) sign should be \(\oplus\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4OK, let's start from scratch. Sorry for being a little casual the last time.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Question: is it a Set theory problem??

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4OK, 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 basecase) 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 "dominoeffect".

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0me too, hes a jerk sometimes

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0@zzr0ck3r who is your "he" ??

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0ok, :) good luck (for your final)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I am giving it, not taking it :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Here 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
 one year ago
Best ResponseYou've already chosen the best response.4We've thus proven \(P(k+1)\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Try proving all those properties for rational numbers as an exercise.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0the line " definition of dot" !!!

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thanks a ton.............

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4\[(k+1) \cdot x = k\cdot x \oplus x\]\[(k+1)\cdot \heartsuit = k \cdot \heartsuit \oplus \heartsuit\]\[(k+1)\cdot \spadesuit = k \cdot \spadesuit \oplus\spadesuit \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Oh  never mind then...

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Finally, you can crack my skull to put the logic in. Thanks for being patient to me.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Anyway, if you ever receive these questions on a multiplechoice test, spend a minute thinking if these operators are related to any that we already know. That makes things easier.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0Again, thank you so much.:)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Which is why it is useful to know that these operators are just addition and multiplication. lol :P No problem!
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