0
$\begingroup$

Which one is less than others?

$\frac{3}{5} , \frac{2}{3} , \frac{6}{13} , \frac{23}{38}$

Yes the answer is $\frac{6}{13}$ but the real question is this:

I've a 12 years old brother and he just asked me this question so he is kid and I need to explain this with the most simple way.

First I thought using common Denominator but it's hard and time consuming then I find out that if we multiply all numbers by 2, all numbers would be greater than their Denominator except for $\frac{6}{13}$ so this most be less than others.

Is there another way to explain this?

$\endgroup$
  • 5
    $\begingroup$ Just observe that $\;\frac 6{13}\;$ is the only fraction in which the denominator is greater than twice the numerator, so the fraction has to be less than $\;\frac12\;$ . In all the other three cases this doesn't happen. $\endgroup$ – Timbuc Dec 31 '14 at 13:25
  • $\begingroup$ @Timbuc. Good catch ! Happy New Year ! $\endgroup$ – Claude Leibovici Dec 31 '14 at 13:26
  • $\begingroup$ Same to you, Claude mon ami: Bonne Année 2015! $\endgroup$ – Timbuc Dec 31 '14 at 13:29
  • $\begingroup$ Your 12 years old brother needs to know the general method, because the next time he is asked a similar question it might not be as easy as this one. $\endgroup$ – barak manos Dec 31 '14 at 13:58
  • $\begingroup$ Do the easy comparisons first, then you just need to compare with the least fraction you've found so far. No need to put them all over the same common denominator. $\endgroup$ – Mark Bennet Dec 31 '14 at 14:48
2
$\begingroup$

The easiest way to see whether $\frac ab<\frac cd$ (where I assume that $b,d>0$) is to test whether $ad<bc$. This is really what one does when bringing fractions to the common denominator $bd$, but one need not actually change any fraction when going further to compare it to other fractions.

There may be a common denominator smaller than $bd$, but in general it is not worth while to go looking for it if you just want to compare two fractions.

Concretely to look for the minimal element one could determine that $\frac35<\frac23$ since $3\times3<5\times2$, then that $\frac35>\frac6{13}$ since $3\times13>5\times 6$, and finally that $\frac6{13}<\frac{23}{38}$ since $6\times38<13\times23$, showing that $\frac6{13}$ is the smallest of the four fractions. The last inequality is actually fairly gross, and the computation can be somewhat simplified (at least for mental calculation) by using an intermediate fraction: $\frac6{13}<\frac12<\frac{23}{38}$ since both $6\times2<13\times1$ and $1\times38<2\times23$ (using $\frac23$ would work as well). But of course searching a simple intermediate fraction is not guaranteed to be successful.

$\endgroup$
1
$\begingroup$

Bake four cakes and cut peaces in an accurate way.

$\endgroup$
-1
$\begingroup$

Note that you don't have to find the least common denominator, just any common denominator.

So you can just multiply each numerator by all of the other denominators, and compare the results. (These would be the numerators corresponding to the denominator that would be the product of all of the separate denominators.)

In your case, you would compare the numbers $$3\cdot(3\cdot 13\cdot 38)=4446$$ $$2\cdot(5\cdot 13\cdot 38)=4940$$ $$6\cdot (5\cdot 3\cdot 38)=3420$$ $$23\cdot (5\cdot 3\cdot 13)=4485$$

The smallest is the third, whic corresponds to the original fraction $\frac{6}{13}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.