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Math Help - Help Check my work on a few Simple problems?

  1. #1
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    Help Check my work on a few Simple problems?

    If anyone could help check my work on these problems i would be grateful

    Try to be picky like if you were the grader.

    Help Check my work on a few Simple problems?-2.jpg
    Help Check my work on a few Simple problems?-3.jpg
    Help Check my work on a few Simple problems?-4.jpg



    Here is a pdf

    mathtest.pdf
    Last edited by outkast32; April 23rd 2014 at 07:54 AM. Reason: added pics
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  2. #2
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    Re: Help Check my work on a few Simple problems?

    (1) looks good.

    With (2) you start off "since 5^n- 1 is divisible by 4" but that is what you are trying to prove. You have to start by proving it is true for n= 1 (which is easy) and then say "If, for some k, 5^k- 1 is divisible by 4" or "suppose 5^k- 1 is divisible by 4". The difference is that (1) I am saying "if" rather than "since" and (2) I am using "k" rather than "n". That is not crucially important but makes it clear that I am not assuming what I want to prove.

    For (3), I personally don't like using "formulas" like that. Here is what I would do:
    Assume a population of 10000 people. 4%, 400, use drugs and 96%, 9600, do not. There is a test for drug use which has a 3% false positive rate and a 2% false negative rate. So of the 9600 who do not use drugs, 97%, 9312, will test negative and 3%, 288, will test positive. Of the 400 who use drugs, 98%, 392, test positive and 2%, 8, test negative.

    So there are a total of 288+ 392= 680 who test positive, 392 of whom are really drug users. So the probability that a person who tests positive really is a drug user is [tex]\frac{392}{680}= .5765... or 57.65% as you have.

    Similarly, there are a total of 9312+ 8= 9320 who test negative, 9312 of whom do not use drugs. So the probability that a person who tests negative is not a drug user is \frac{9312}{9320}= 0.9991 or 99.9% as you have.
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  3. #3
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    Re: Help Check my work on a few Simple problems?

    Is this better for #2? i made some changes you mentioned:

    Help Check my work on a few Simple problems?-q3.jpg


    And for # 3 is a matter of opinion on using the formulas or not, right? - your way seems to make sense too, but its the same answer either way right, so it should be fine?
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  4. #4
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    Re: Help Check my work on a few Simple problems?

    Quote Originally Posted by outkast32 View Post
    Is this better for #2? i made some changes you mentioned:

    Click image for larger version. 

Name:	q3.JPG 
Views:	3 
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ID:	30748


    And for # 3 is a matter of opinion on using the formulas or not, right? - your way seems to make sense too, but its the same answer either way right, so it should be fine?
    Yes. That does it for problem 2. Few minor points. First, in next to last line, you mean that 5^(k + 1) - 1 is divisible by 4, not that 4 is divisible by 5^(k + 1) - 1.

    Second, and this is a personal preference about style, I find most intuitive this format for proofs by weak mathematical induction.

    $Prove\ n \in \mathbb Z^+ \implies 5^n - 1\ is\ evenly\ divisible\ by\ 4.$

    Step 1: $n = 1 \implies 5^1 - 1 = 5 - 1 = 4 = 4 * 1 \implies 5^n\ is\ evenly\ divisible\ by\ 4.$

    Not part of the proof: This means that there certainly are one or more positive integers such that if n is one of those integers 5^n - 1 is evenly divisible by 4. Choose an arbitrary one of those integers and call it k. This then, at least to me, completely justifies the induction hypothesis as being more than an arbitrary assumption. Back to the proof.

    Step 2: $k\ is\ a\ positive\ integer\ such\ that\ 5^k - 1\ is\ evenly\ divisible\ by\ 4 \implies m\ is\ a\ positive\ integer\ such\ that\ 4m = 5^k - 1.$

    Your logic works like a charm, but I like to get out of the way what I need to know about k right up front.

    $5^{(k + 1)} - 1 = (5^{(k + 1)} - 1) - 4 + 4 = (5^{(k + 1)} - 5) + 4 = 5(5^k - 1) + 4 = 5(4m) + 4 = 4(5m + 1).$

    $But\ m\ is\ a\ positive\ integer \implies 5m + 1\ is\ a\ positive\ integer \implies 5^{(k + 1)} - 1\ is\ evenly\ divisible\ by\ 4.$

    $\therefore n \in \mathbb Z^+ \implies 5^n - 1\ is\ evenly\ divisible\ by\ 4.$

    Now of course if your teacher prefers another format, use that.
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  5. #5
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    Re: Help Check my work on a few Simple problems?

    Okay for step 1 did you mean n=1 => 5^1 -1 = 5 - 1 = 4 = 4*1 => 5^n -1 instead of just 5n is evenly divisible by 4?
    Does it matter if we use that base case or would the base case of n = 0, 5^0 -1 = 1 -1 = 0 and 4 is divisible by 4 work better?

    Okay so by your format, this looks good??: http://i.imgur.com/BrYvcee.jpg

    Thank you for sticking with me on this by the way
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  6. #6
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    Re: Help Check my work on a few Simple problems?

    Quote Originally Posted by outkast32 View Post
    Okay for step 1 did you mean n=1 => 5^1 -1 = 5 - 1 = 4 = 4*1 => 5^n -1 instead of just 5n is evenly divisible by 4?
    Does it matter if we use that base case or would the base case of n = 0, 5^0 -1 = 1 -1 = 0 and 4 is divisible by 4 work better?

    Okay so by your format, this looks good??: http://i.imgur.com/BrYvcee.jpg

    Thank you for sticking with me on this by the way
    First, yes I did mean to say in step 1: $n = 1 \implies a\ bunch\ of\ steps \implies 5^n - 1\ is\ evenly\ divisible\ by\ 4.$

    Sorry about that.

    Second, a few very minor things about your new style proof. (You did ask for fussy: the way an obnoxious grader might look at it.)

    A. Be aware that many say $Assume\ \exists\ a\ positive\ integer\ k\ such\ that\ ...$ instead of $k\ is\ an\ arbitrary\ positive\ integer\ such\ that\ ...$

    I personally think saying that is an assumption is silly because in step 1 you have already proved that at least one such positive integer exists, but make sure your teacher agrees.

    B. I think the word "arbitrary" is important in my style. You know only that k has the properties common to all positive integers and the property that you are trying to prove is general. So k could be any of the positive integers for which the property in question is true. In your proof, you said "k is a positive integer such that ..." I prefer "k is an arbitrary positive integer such that."

    C. I am pretty sure a fussy grader would object to your "and m is a positive integer ..." because, for those who think that you are making an assumption about k, you SEEM to be making an extra assumption about m. In fact, the existence of m is a consequence of what you have already affirmed about k. If 5^k - 1 is evenly divisible by 4, then, by the meaning of the phrase "evenly divisible by 4," there exists a positive integer m such that 4m = 5^k - 1. So I would say

    $5^k - 1\ is\ evenly\ divisible\ by\ four \implies \exists\ m \in \mathbb Z^+\ such\ that\ 4m = 5^k - 1.$

    Again, this way of showing a proof by weak mathematical induction strikes me personally as being intuitive, but your teacher may not like it because it is a little atypical. So check with your teacher about what he or she likes.
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