Proving if $n > m$ then $(n-i)/(m-i)<n/m$

Question

How can I prove, using algebra, that if $n > m$, then: $$\frac{n-i}{m-i}<\frac{n}{m}$$

All I can think of is that $n-i>m-i$ because $n>m$, but I can't link that to the above inequality.

Answer 1

The statement if false:

Observe that when $n=4$ and $m=2$, $\frac{n}{m}=2$. On the other hand, for $i=1$, $\frac{n-i}{m-i}=\frac{3}{1}=3$.

If we wanted to try the inequality $n<m$ and see what happens there, then for $n=1$ and $m=2$ with $i=3$, we get $\frac{n}{m}=\frac{1}{2}$ while $\frac{n-i}{m-i}=\frac{-2}{-1}=2$.

Therefore, there needs to be additional assumptions for the inequality to hold.

Answer 2

This is not correct when $n > m> i > 0$

For example, if $n=3, m=2, i=1$ then $$\frac{n-i}{m-i} = \frac{3-1}{2-1} =2 \not < \frac32 = \frac{n}{m}$$

It is correct when $n>m>0$ and $i>m>0$

The case $i=m$ leads to division by $0$

Answer 3

What's true, with the hypothesis $n>m>i>0$, is that, on the contrary, $$\frac nm<\frac{n-i}{m-i}$$ Indeed, since all denominators are positive, it is equivalent to $$n(m-i)<m(n-i)\iff -ni<-mi \Leftrightarrow n>m\quad(\text{simplifying by }{-i)}.$$

Answer 4

The statement isn’t true.

For instance, say $n = 8$, $m = 6$, and $i = 2$. ($n > m > i > 0$)

$$\frac{8-2}{4-2} = \frac{6}{2} = 3$$ $$3 > \frac{8}{6}$$

This can be shown algebraically too.

$$\frac{n-i}{m-i} < \frac{n}{m}$$ $$m(n-i) < n(m-i)$$ $$mn-im < mn-in$$ $$-im < -in \implies \frac{-im}{-i} > \frac{-in}{-i} \implies m > n$$ This contradicts the original statement. Also, you can simply think of the pattern. For instance, assume $i = 1$.

For $m > n$: $$\frac{4}{5} < \frac{5}{6} < \frac{6}{7} < ...$$
As you subtract $i$ from both the numerator and the denominator, the value decreases.

For $n > m$: $$\frac{7}{6} < \frac{6}{5} < \frac{5}{4} < ...$$ As you subtract $i$ from both the numerator and denominator, the value increases. (Once again contradicting your initial statement.)