Loser66
  • Loser66
Please, explain me.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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Loser66
  • Loser66
I understand 3, but 2, how??
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ParthKohli
  • ParthKohli
Hey, you can probably use induction for proving II.
Loser66
  • Loser66
The given information doesn't include \(n* (x \oplus y)\), hence we can't define the product of it.

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ParthKohli
  • 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
  • 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
  • Loser66
We don't have \(k(x\oplus y) = kx\oplus ky\)
ParthKohli
  • ParthKohli
But that's the inductive step!
Loser66
  • Loser66
But first off, the operation must be defined, right?
ParthKohli
  • ParthKohli
It has been defined recursively...
ParthKohli
  • ParthKohli
\[(k+1)x = kx \oplus x\]
ParthKohli
  • ParthKohli
Hey, you could observe that the operators are + and * though, right?
Loser66
  • Loser66
ok, you can say if \(k (x\oplus y)\), then \((k+1) (x\oplus y)= k(x\oplus y)+(x\oplus y)\)
ParthKohli
  • ParthKohli
That \(+\) should be \(\oplus\). Now \(k(x\oplus y)=kx \oplus ky\).
Loser66
  • Loser66
hihihi... I don't see anything allows me to distribute like that.
ParthKohli
  • ParthKohli
That is the inductive step, man.
ParthKohli
  • ParthKohli
We have assumed that \(k(x\oplus y) = kx \oplus ky\). We have to show that the same holds for \(k+1\)
Loser66
  • Loser66
Ok, I temporarily "pretend" that I agree with you on that step, then??
ParthKohli
  • 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
  • ParthKohli
Anyway, I gotta go. So you see how this induction works, right?
Loser66
  • 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
  • Loser66
I don't know what given condition allows me pick like terms to add up. :(
ParthKohli
  • ParthKohli
Hey, that is the definition of \(\cdot\).\[(k+1)x = kx\oplus x\]given in your question
ParthKohli
  • ParthKohli
That \(\color{red}+\) sign should be \(\oplus\).
ParthKohli
  • ParthKohli
OK, let's start from scratch. Sorry for being a little casual the last time.
Loser66
  • Loser66
Do steps please.
Loser66
  • Loser66
Question: is it a Set theory problem??
ParthKohli
  • 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
  • Loser66
I know induction.
zzr0ck3r
  • zzr0ck3r
me too, hes a jerk sometimes
zzr0ck3r
  • 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
  • Loser66
@zzr0ck3r who is your "he" ??
zzr0ck3r
  • zzr0ck3r
fun
zzr0ck3r
  • zzr0ck3r
induction is my he
Loser66
  • Loser66
ok, :) good luck (for your final)
zzr0ck3r
  • zzr0ck3r
I am giving it, not taking it :)
ParthKohli
  • 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
  • ParthKohli
We've thus proven \(P(k+1)\).
ParthKohli
  • ParthKohli
Try proving all those properties for rational numbers as an exercise.
Loser66
  • Loser66
the line " definition of dot" !!!
Loser66
  • Loser66
oh, got it. hihihih
Loser66
  • Loser66
Thanks a ton.............
ParthKohli
  • 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
  • ParthKohli
Oh - never mind then...
Loser66
  • Loser66
Finally, you can crack my skull to put the logic in. Thanks for being patient to me.
ParthKohli
  • 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
  • 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
  • Loser66
Again, thank you so much.:)
ParthKohli
  • 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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