How to show $0 < \frac{x}{x+y} < 1$ given $0 < x$ and $0 < y$ I am trying to help my student understand why it is true that
$$0 < \frac{x}{x+y} < 1 \impliedby 0 < x \text{ , } 0 < y$$
I so far say that intuitively, $\frac{x}{x}=1$ and since $0<y$ we know 
$$\frac{x}{x+y} \lt 1$$ 
when $y$ is close to $0$.
But when $y$ is much bigger 
$$0<\frac{x}{x+y}$$.
She really has trouble with "intuition" but she does very well with rigorous algebraic proofs.
Is there an "algebraic" approach to this situation?
A would appreciate your help.
 A: If:
$$x,y>0$$
then:
$$x+y>x$$
$$x+y>y$$
and so:
$$\frac x{x+y}<1$$
$$\frac y{x+y}<1$$
also since all terms are greater than $0$, you have a positive number divided by a positive number, which must be positive, and so:
$$\frac x{x+y}>0$$
$$\frac y{x+y}>0$$
A: Since $x$ and $y$ are greater than zero, we can divide by $x + y$ without messing up inequalities.
So if $0 < x < x + y$, which is clear, then $\frac{0}{x + y} < \frac{x}{x+y} < \frac{x+y}{x + y}$, which just states $0 < \frac{x}{x+y} < 1$
A: Similar to one of the answers, but succinctly: since $x, y > 0,$ 
$$0 = \frac{0}{x + y} < \frac{x}{x + y} < \frac{x}{x} = 1$$
A: We can explain by the following simple rules:


*

*Rule 1. Sum of two positive numbers is a positive number. 

*Rule 2. Quotient of two positive numbers is a positive number.

*Rule 3. Subtracting a positive number we get a smaller result. 
Since $x>0,y>0$, $x+y>0$ by Rule 1, and $\displaystyle\frac{x}{x+y}>0$ and 
$\displaystyle\frac{y}{x+y}>0$ by Rule 2. Finally, by Rule 3, we have
$$
\frac{x}{x+y}=\frac{(x+y)-y}{x+y}=1-\frac{y}{x+y}<1.
$$
