In adding or subtracting fractions with two or more terms, the resulting denominator is always the product of the denominators of the terms.

Can someone prove or disprove this statement. I'm pretty sure this isn't true, but I don't know how to disprove it. Does a statement like 1/2+1/3+2/3 count?

  • $\begingroup$ Yes that works to disprove it. $\endgroup$ – simonzack Oct 4 '16 at 3:37

It is true if you consider fractions other than the lowest term (or simplest form).

Consider $\frac ab+\frac cd=\frac{ad+bc}{bd}$. Similarly for subtraction.

Clearly, if you reduce to the lowest term, it isn't true. Simply consider something like $\frac 14+\frac 14=\frac 12$; 2 isn't a product of 4 and 4.

  • $\begingroup$ But if you consider fractions not in there lowest terms the sentence is meaningless. the denominator of any fraction is simultaneously equal to, a multiple of, a factor of any other fraction. $\endgroup$ – fleablood Oct 4 '16 at 7:01

It's not always true. When we add or subtract two or more fractions then resulting denominator is the LCM (Least Common Denominator) of the denominators of the terms, being added or subtracted. For example :

$\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}$ here $12$ is the LCM of 3 and 4.

Since the LCM of two or more prime numbers is their product, therefore if two or more fractions which have prime numbers as their denominator are added or subtracted then the resulting denominator will be their product, as we show in above examle.

Consider another example :

$\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}$ Here 4 is the LCM of 2 and 4.

We can see that the resulting denominator is not the product of 2 and 4, but it's the LCM of 2 and 4.

So, it's prooved that it's not necessary that in adding or subtracting two or more fractions, resulting denominator is the product of the denominators of the terms.

  • $\begingroup$ LCM stands for least common multiple, not least common denominator. $\endgroup$ – Daniel Buck Oct 4 '16 at 8:44
  • $\begingroup$ yes, you are right, but i was talking about the least common multiple of the denominators of the terms. $\endgroup$ – user372725 Oct 5 '16 at 4:42
  • $\begingroup$ Sure, the two are equivalent in this situation, i.e., in as much as the LCM of the denominators gives the least common denominator, but the OP just needed that point clarified. $\endgroup$ – Daniel Buck Oct 5 '16 at 8:57

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