Question

find the difference.
$1\frac{3}{5}-1\frac{4}{5}=\frac{?}{?}$

Explanation:

Step1: Convert mixed numbers to improper fractions

First, convert \(1\frac{3}{5}\) and \(1\frac{4}{5}\) to improper fractions.
For \(1\frac{3}{5}\), we have \(1\frac{3}{5}=\frac{1\times5 + 3}{5}=\frac{8}{5}\).
For \(1\frac{4}{5}\), we have \(1\frac{4}{5}=\frac{1\times5 + 4}{5}=\frac{9}{5}\).

Step2: Subtract the two fractions

Now, subtract the two improper fractions: \(\frac{8}{5}-\frac{9}{5}=\frac{8 - 9}{5}=\frac{- 1}{5}\). But since we are dealing with the difference (and maybe considering absolute value or the order, but looking at the problem, maybe we made a mistake in order. Wait, the problem is \(1\frac{3}{5}-1\frac{4}{5}\), but if we reverse the order (since \(1\frac{4}{5}>1\frac{3}{5}\)), the difference in terms of positive value? Wait, no, let's check again. Wait, maybe the problem is \(1\frac{3}{5}\) minus \(1\frac{4}{5}\), but that gives a negative. But maybe the problem is written as \(1\frac{3}{5}\) and \(1\frac{4}{5}\) with the subtraction \(1\frac{3}{5}-1\frac{4}{5}\), but let's do the calculation correctly.

Wait, \(1\frac{3}{5}=\frac{5 + 3}{5}=\frac{8}{5}\), \(1\frac{4}{5}=\frac{5+4}{5}=\frac{9}{5}\). Then \(\frac{8}{5}-\frac{9}{5}=\frac{8 - 9}{5}=-\frac{1}{5}\). But maybe the problem is intended to be \(1\frac{4}{5}-1\frac{3}{5}\)? Let's check. If we do \(1\frac{4}{5}-1\frac{3}{5}=\frac{9}{5}-\frac{8}{5}=\frac{1}{5}\). Maybe there was a typo in the order. Assuming the problem is \(1\frac{4}{5}-1\frac{3}{5}\) (since otherwise we get a negative, and the boxes are for positive numerator and denominator). So let's proceed with \(1\frac{4}{5}-1\frac{3}{5}\).

So, converting to improper fractions: \(1\frac{4}{5}=\frac{9}{5}\), \(1\frac{3}{5}=\frac{8}{5}\). Then \(\frac{9}{5}-\frac{8}{5}=\frac{9 - 8}{5}=\frac{1}{5}\).

Answer:

\(\frac{1}{5}\) (So the numerator is 1 and the denominator is 5)