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anonymous
 one year ago
Please check my solution: Prove that for all integers n > 0, n3 + 2n is divisible by 3.
anonymous
 one year ago
Please check my solution: Prove that for all integers n > 0, n3 + 2n is divisible by 3.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Prove that for all integers n > 0, n3 + 2n is divisible by 3. Base Case: n=1 (1)^3+2(1) = 3 3 is divisible by 3 Induction Hypothesis P(K+1) (K+1)^3 + 2(K+1) (K+1)(K+1)(K+1)^3 + 2K+2 K^2+K+K+1 = (3K^2 + 1) + 2K+2 (3K^2 + 1)(K + 1) = (3K^3 + 3K^2) + K+1 = 3K^3 + 3K^2 + K + 1 (3K^3 + 3K^2 + K + 1) + 2K + 2 = 3K^3 + 3K^2 + 3K + 3 3K^3 + 3K^2 + 3K + 3 is divisible by 3 QED

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I'm kind of getting lost here: (K+1)(K+1)(K+1)^3 + 2K+2 K^2+K+K+1 = (3K^2 + 1) + 2K+2 (3K^2 + 1)(K + 1) = (3K^3 + 3K^2) + K+1 = 3K^3 + 3K^2 + K + 1

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Use your Binomial Theorem to expand out a cube, \(\large\rm (k+1)^3=k^3+3k^2+3k+3\) No 3 coefficient on the k^3 term.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Woops! +1 on the end, sorry typo

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\rm (k+1)^3=k^3+3k^2+3k+1\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Did you do it the other way? Expanding out two brackets at a time?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just a moment, let me read over this. It might take me a minute to reply I'm not very good at math. Thank you for responding to my question, though.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2The base case looks good :) We're missing an important step though. The `Induction Hypothesis` is actually where we `assume n=k is true`. Then we next apply the `Induction Step`, which you attempted, we just need to clean a few parts up.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The induction hypothesis is basically saying "it's true for all integers," right? And, the induction step is where we do K+1 to show it works for the next integer. Is this the correct idea?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am going to go over this algebra on a sheet of paper. It may take me 45 minutes. If you're lost that means I've done something wrong. Honestly, I haven't done algebra in a long time and I'm kind of fuzzy on my rules. Let me work this out on paper and come to you with concise questions in a few moments.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes :) We SHOWED that it was true for the very first case. Our hypothesis, our assumption, is that this will be true for any number. If we can show it's true for the number after that, we've showed it's true in general. Ok cool.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Can't resist this alternative proof : \(n^3+2n \\= n^3n+3n \\= n(n^21)+3n \\= (n1)(n)(n+1)+3n\) recall the fact that product of any 3 consecutive integers is divisible by 3 to conclude the proof.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(From Above) (K+1)(K+1)(K+1)^3 + 2K+2 K^2+K+K+1 = (3K^2 + 1) + 2K+2 This was pretty sloppy. I was supposed to have: (K + 1)(K + 1)(K + 1) + 2K+2 Now, I have to multiple whatever I can: (K^2 + K + K + 1)(K + 1) + 2K + 2 Simplify and multiple: (K^2 + 2K + 1)(K + 1) + (2K+2) (K^3 + K^2 + 2K^2 + 2K +K + 1) + (2K+2) (K^3 + 3K^2 + 2K + K + 1) + (2K+2) (K^3 + 3K^2 + 3K + 1) + (2K+2) (K^3 + 3K^2 + 5K + 3) Okay, slightly different answer. I tried to make each step clear; but, my eyes get kind of lost looking at these numbers. Sorry, for my poor work.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Ok looks good :) Let's back up a sec though. Stating our `Induction Hypothesis`: We assume this is true for n=k. Which means \(\large\rm \color{orangered}{k^3+2k}\) is assumed to be divisible by 3. Hold onto this we'll need it later!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2In all of your algebra, you actually want to stop here,\[\large\rm (k^3 + 3k^2 + 3k + 1) + (2k+2)\]And group a little differently than you might think.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm k^3+2k+3k^2+3k+3\]So I arranged them a little differently, hopefully that's not confusing.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, you have put your induction hypothesis at the front

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2good good good, we want to see that piece showing up somewhere in our induction step. \[\large\rm \color{orangered}{k^3+2k}+3k^2+3k+3\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Your next step would be to maybe pull a 3 out of each of these terms. So you can say something about it as a group.\[\large\rm \color{orangered}{k^3+2k}+3(k^2+k+1)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2We `assumed` that the orange was divisible by 3. And clearly 3(k^2+k+1) is divisible by 3 (because it's being multiplied by 3). So the whole thing is divisible by 3! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so for these cases I need to do algebra until I get the induction hypothesis. Move that to one side. Factor out the other side.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes. Base Case: Prove true for n=1. Induction Hypothesis will give you `an equation`. Induction Step: Here you are working with a big mess of stuff. You're trying to find `the equation from the induction step` in it, plus some other stuff.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Maybe I'll call it `the equation`. Induction Hypothesis: `the equation`. Induction Step: Big Mess = `the equation` + (stuff that is obviously true)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2The process isn't too bad :) Just gotta get comfortable with the tons and tons of algebra steps I guess

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I do have a few more questions if you don't mind. A) When I wrote the base case I entered 1 because that is the first integer greater than 0. Is that the right reason to be using this base case? B) K+1 is part of the induction step. Does this represent the next integer in the sequence of integers being tested (all, in this case).

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2A) Yes. You'll likely later run into problems which state things like `prove this is true for for n>3`. So you'll have a more awkward starting point of n=4 in that type of problem. But yes, we're only using the counting numbers for n, since these are integers. So n=1 is the next one to go to since it's a `strict inequality`. If they had given us \(\large\rm n\ge0\), then we might want to start at n=0, unless it's incredibly trivial to do so. B) Good, yes. We `assumed` that it would it be true for any number. We `showed` that if that's true, the number after that is always true as well. So ya, we've covered everything by doing that. If this equation holds for any number greater than 0, then it holds for the next number. Since we showed n=1 is true, then by our induction step, we can say that n=2 must be true. (Because we've shown that the next number is always true). We can continue from that point and say that n=3 must be true. And so on... Hopefully that wasn't too long winded of an explanation lol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, that was great. Thank you very much for taking time out of your day to help me and other people. I genuinely appreciate it. I will continue doing some more problems. If you have time and don't mind, then I will attempt a solution and post it to this chat. Otherwise, have a great day and, thanks again! :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2ya I'll be around from time to time. You can use the pager system to get someone's attention, it can be useful. @zepdrix would sent me a notification, makes it easier to find this thread again. Or you can @ any of the other smarty pants in the main lobby. Any of the 99 scores are usually pretty helpful.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix in this chat window? I also see there is an option to send private messages. Whichever you prefer. Anyway, next problem in maybe 1520 min.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes, in this chat window. It's gives me a notification on the sign post icon. I don't like to answer questions in private message >.< I can't use the fancy equation tool there lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix prove that for all integers n>0 (8^n)(3^n) is divisible by 5 Base Case: n=1 (8^1)(3^1)=5 5/5=1 Induction Hypothesis: n=k Induction Step: (8^k+1)(5^k+1) I do not know what to do with these exponents. I have been trying to find what to do; but, the information is not coming easy. Could you point me in the right direction or show me what the first operation would be?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2An exponent rule tells us: \(\large\rm x^{a+b}=x^a\cdot x^b\) We can apply that to our 8 and 3 bases, \(\large\rm 8^{k+1}=8^k\cdot8^1\) \(\large\rm 3^{k+1}=3^k\cdot3^1\) So our expression becomes:\[\large\rm 8^{k+1}3^{k+1}\quad=\quad 8^k\cdot83^k\cdot3\]From that point.. hmm lemme think about it a sec, it's gonna be a tricky step I think.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Recall from your base case that 83=5. Manipulate that to get \(\large\rm \color{orangered}{3=(85)}\) I think we'll have to use this in our Induction Step:\[\large\rm 8^k\cdot83^k\cdot\color{orangered}{3}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Think you can work with that? :) Or still too confusing?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am thinking about it. I need to get 8^k3^k out of this. Is 8^k3^k = 85?? I understand what you are saying by manipulating 3=(85). Perhaps it is necessary. I have no idea what I'm doing with this problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix I am thinking about it. I need to get 8^k3^k out of this. Is 8^k3^k = 85?? I understand what you are saying by manipulating 3=(85). Perhaps it is necessary. I have no idea what I'm doing with this problem.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I colored them orange to show that we're going to make a substitution.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So from here,\[\large\rm 8^k\cdot83^k\cdot\color{orangered}{3}\]We'll rewrite 3 as (85),\[\large\rm 8^k\cdot83^k\cdot\color{orangered}{(85)}\]This algebra step is going to be a little tricky. Distribute the 3^k to each term in the brackets.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 8^k\cdot83^k\cdot8+3^k\cdot5\]Something like that, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Are you familiar with this notation or no?  that bar?  means "divides" If I say `x+y is divisible by 5`, another way to say it is, `5 divides x+y` Mathematically it looks like this \(\large\rm 5x+y\) Have you been introduced to that? :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am familiar, now :P I would have never got this problem. I suppose you can manipulate the base case for the int 8, too.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2To be honest, I didn't remember how to do this problem either. I had to google it lol. Kind of a weird little step in the middle there.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Do you see how to finish that problem up though? :o \[\large\rm 8^k\cdot83^k\cdot8+3^k\cdot5\]From that point?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.08k⋅8−3k⋅(8−5) = 8k * 8  24k  15k No? 8k⋅8−3k⋅8+3k⋅5 < How did this happen??

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oh boy what happened there +_+ Hmmm

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix 8^k⋅8−3^k⋅8+3^k⋅5 from here it's pretty easy to put the induction hypothesis in front. 8^k3^k*8*8+3^k*5; but, I'm still left with this 3^K which I don't know what to do with. I mean, even if 3^k was set to 0 or 1 that side would be divisible by 5. I'm not sure how to proceed.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Woops, the way you got your `induction hypothesis` looks a little sloppy. Lemme write it out and make sure this makes sense to you,\[\large\rm \color{royalblue}{8^k\cdot83^k\cdot8}+3^k\cdot5\]We're going to factor an 8 out of each blue term,\[\large\rm \color{royalblue}{8(8^k3^k)}+3^k\cdot5\]Since our Induction hypothesis tells us that \(\large\rm 5~~8^k3^k\), that implies that \(\large\rm 5~~8(8^k3^k)\) Good, so you can make use of your induction hypothesis to deal with that part.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Try to get comfortable with `what divisibility means`. If I have a number \(\large\rm 2n\), where n is an integer, then I can confidently say that it's an even number, or in other words, it's a multiple of 2. Which also means it's divisible by 2! (Pssst: Because a 2 is multiplying the number).

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2If I have \(\large\rm 5\cdot stuff\), I can confidently say that the `stuff` is a multiple of 5, and is therefore divisible by 5.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So as your last step, maybe you just say something like: Since our Induction hypothesis tells us that \(\large\rm 5~~8^k3^k\), this implies that \(\large\rm 5~~8(8^k3^k)\). And further, since \(\large\rm 5~~5\cdot3^k\), we have that \(\large\rm 5~~8(8^k3^k)+5\cdot3^k\), proving our induction step.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2The first chunk is divisible by 5, the second chunk of stuff is divisible by 5, so their sum is divisible by 5.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2And then wrap it up all fancy at the end with something like: Therefore, by the Principle of Mathematical Induction, \(\large\rm \text{ For all integers }n>0,~~(8^n)(3^n)\text{ is divisible by }5 \).

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Unless that seems to wordy :) lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2To further nail the point about divisibility though... If something is divisible by 5, it means it can be written as a multiple of 5. Example: \(\large\rm 5~~x+y\) means that \(\large\rm x+y=5k\). Where k is some integer. In our problem: \(\large\rm 5~~8^k3^k\) means that \(\large\rm 8^k3^k=5\ell\) where \(\large\rm \ell\) is some integer. If I multiply both sides by 8,\[\large\rm 8(8^k3^k)=8\cdot5\ell\]If we write the right side like this,\[\large\rm 8(8^k3^k)=5\cdot(8\ell)\]Do you see how the right side is of the form 5*(stuff)? So it's still divisible by 5! :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Ok that was a lot to chew on :U I'll simmer down

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you're totally fine man. i've been here trying to learn these algebra skills all day. i just forgot about factoring... your explanations are crystal clear and really helpful. simmer as much as you want because I have a test on tuesday :/
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