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Comparing 2 Fraction based variables

  1. Jun 14, 2016 #1
    1. The problem statement, all variables and given/known data
    c & d are positive integers &
    1/c = 1 + 1/d
    Now tell which is greater c or d?

    2. Relevant equations
    1/c = 1 + 1/d

    3. The attempt at a solution
    1/c = (d+1)/d

    c = d/ (d+1)
    c = 1 + d

    Thus if c =10
    therefore d = 9
    But this is not a correct answer. Some body please guide me.

    Zulfi.
     
  2. jcsd
  3. Jun 14, 2016 #2

    Nidum

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    Hint : Which is smaller 1/c or 1/d ?
     
  4. Jun 14, 2016 #3
    You could also just sub in values and make inferences from that
     
  5. Jun 14, 2016 #4

    Charles Link

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    For c, d positive integers, 1/c is always less than 1 unless c=1. I think the statement of the problem is incorrect that c and d are positive integers. There is no solution. c=1 implies d=+infinity. If the problem statement had said positive reals, then in one case c<1 and d>1. In another case, c<1 and d<1 and c<d, so that c<d if c, d are positive reals.
     
    Last edited: Jun 14, 2016
  6. Jun 14, 2016 #5

    Ray Vickson

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    As pointed out in #4, the equation is impossible if c and d are positive integers; however, if they are just positive real numbers, then the equation says that 1/c is a bigger number than 1/d. What does that say about c and d themselves?
     
  7. Jun 14, 2016 #6
    Hi,
    Thanks for your help. Actually it says that " c & d are positive".

    <Which is smaller 1/c or 1/d ?>
    I think it depends on values. They can be equal.

    <
    In another case, c<1 and d<1 and c<d, so that c<d if c, d are positive reals.>
    If c & d are positive reals this means that c & d can be integers also.
    What's wrong with my solution. I have proved that c is greater. However the answer says that d is greater.

    Zulfi.
     
  8. Jun 14, 2016 #7

    Charles Link

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    Your last line of post #1 c=1+d is incorrect. Meanwhile, since 1/c=1+1/d, this says the right side of the equation is greater than 1. That means c must be less than 1 (to make the left side greater than 1) and therefore c can not be a positive integer.
     
  9. Jun 15, 2016 #8

    Nidum

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    I guessed that the use of integer was a typo - the question did not make sense otherwise .

    Now think about this :

    If x is greater than y then is 1/x greater or less than 1/y ? Put some numbers in if it helps - say 10 for x and 5 for y .
     
  10. Jun 15, 2016 #9
    Hi,
    Thanks for your response. I got my mistake. I am now able to understand book's solution.
    According to the book,
    c = d/ (d+1)
    Since d is positive, so d+1> d. if d= 7 then 7/(7+1) (which is a fraction & less than d). So d > d/(d+1). But c=d/(d+1). So d>c.

    Now your question:
    "If x is greater than y then is 1/x greater or less than 1/y ? Put some numbers in if it helps - say 10 for x and 5 for y" .

    i.e. x> y so 1/x < 1/y. Inequalities reverse. 10> 5 so 1/10 < 1/5.

    Thanks for help.

    Zulfi.
     
  11. Dec 3, 2016 #10
  12. Dec 3, 2016 #11

    Mark44

    Staff: Mentor

    Because it is incorrect. See below.
    The reciprocal of ##1 + \frac 1 d## is NOT 1 + d.
    Here is an example using numbers.
    $$\frac 1 {\frac 1 3} = 1 + \frac 1 {\frac 1 2}$$
    Does it follow that ##\frac 1 3 = 1 + \frac 1 2##?


     
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