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How do you know if a fraction is in lowest terms

  1. Jan 21, 2015 #1
    when I am adding, subtracting, multiplying and dividing fractions and I get an answer is there a way I can check to make sure my answer is in lowest terms ?
     
  2. jcsd
  3. Jan 21, 2015 #2

    Mentallic

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    The easiest cheat method would be to use a calculator. Most scientific calculators let you alternate between decimal and fractions, and the fractional form is always shown in the lowest terms.

    If you need to do it by hand however, then break the numerator and denominator up into their prime factors. If any factor is common in both, then you can cancel them. If all factors are different, then you have the lowest term.

    For example:

    [tex]\frac{10}{15}=\frac{2\times 5}{3\times 5}=\frac{2}{3}[/tex]
    since you cancel the factors of 5.

    [tex]\frac{48}{88}=\frac{2^4\times 3}{2^3\times 11}=\frac{2\times 3}{11}=\frac{6}{11}[/tex]
    where we cancelled a common factor of [itex]2^3=8[/itex].

    In some cases, it's quite obvious that we don't have things in the lowest terms. In the first example, it should be obvious that 5 goes into both 10 and 15, hence you can quickly divide both numbers by 5. In the second, since both numbers are even, divide by 2. Then since your results are both still even, divide by 2 again, etc.

    Of course, breaking each number into its prime factors is a tedious process, and only serves as more work than you need. If you come across two fairly large numbers and are unsure if they have common prime factors, then start testing each prime. Try dividing the numerator by each prime 2,3,5,7,11,13,17,19,23,29,31,37...
    up until the square root of the number you're trying to break down. For example, if you're trying to find out if

    [tex]\frac{1231}{2341}[/tex]

    have common factors, then start by trying to divide 1231 by each prime, up until you reach [itex]\sqrt{1231}\approx 35[/itex], so each prime up to and including 31. If you find that 1231 is divisible by at least one of those primes, then test to see if the other number is also divisible by those. If not, then the last check is to see if 2341 is a multiple of 1231, or vice versa if the numerator is larger. If all of these fail, then they have no common factors.
     
    Last edited: Jan 21, 2015
  4. Jan 21, 2015 #3

    Mark44

    Staff: Mentor

    The short answer is: the fraction is in lowest terms if there are no factors other than 1 that appear in both the numerator and denominator.

    For example ##\frac{3}{6}## is not reduced, since 3 is a factor of both numerator and denominator.
    ##\frac{3}{8}## is reduced - there are no factors of the numerator that are also factors of the denominator.
     
  5. Jan 21, 2015 #4

    mathman

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    If the numbers are really small, by inspection. However in general no, except by brute force. Example: 91/143.
     
  6. Jan 21, 2015 #5
    Calculate the greatest common divisor of the numerator and the denominator using euclids algorithm.
    http://en.wikipedia.org/wiki/Euclidean_algorithm
    It that's equal to 1, the fractions are in lowest terms, if it's not divide the numerator and the denominator by the greatest common divisor,
     
  7. Jan 22, 2015 #6

    HallsofIvy

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    Or simply factor the numerator and denominator to prime factors. In mathman's "91/143" example, 91 is not divisible by 2, 3, or 5 but 91= 7(13) and 13 is also prime. 143 is not divisible by 2, 3, 5, or 7 but 143= 11(13). Since 91 and 143 have prime factor 13 in common, the fraction 91/143= (7(13))/(11(13)) is not "reduced to lowest terms" we can reduce further by cancelling the "13" in both numerator and denominator to get 7/11 which is "reduced to lowest terms.

    To apply willem2's suggestion, n91 divides into 143 once with remainder 143- 91= 52. 52 divides into 91 once with remainder 91- 52= 39. 39 divides into 52 once with remainder 52- 39= 13. And 13 divides into 39 exactly three times with no remainder: 39= 3(13). That immediately tells us that the two numbers, 143 and 91 have common factor 13.

    To see that this is true, since 39= 3(13), 52- 39= 52- 3(13)= 13 so 53= 4(13). Then 91- 52= 91- 4(13)= 3(13) so 91= 7(13). And then 143- 91= 143- 7(13)= 4(13) so 143= 11(13).
     
  8. Jan 22, 2015 #7

    mathman

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    I think you proved my point. As the numbers get larger lowest terms is less than obvious.
     
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