Why $x < \frac1a < y$ implies $\frac1y < a < \frac1x$, not $\frac1x < a < \frac1y$? Today, I have a rather simple question.
EDIT: $$ x, y \in [1,\infty) $$
I have the following inequality:
$$ x<\frac{1}{a}<\:y \tag1$$
I want the middle term, which is $\frac{1}{a}$ to be just $ a $,
so it should be this:
$$ \frac{1}{y}<a<\frac{1}{x} \tag2$$
So, the division from (1) to (2) is clear. But what happened to the signs?
Why it is not, instead, this:
$$ \frac{1}{x}<a<\frac{1}{y} $$
Thanks for those who invested time reading the question.
 A: Given $x<\dfrac1a$ with $x>0$ and $a>0$, multiply both sides by $\dfrac ax$ to get $a<\dfrac 1x$.
Do something similar with $\dfrac1a<y\,$ and combine the results to get $\dfrac1y<a<\dfrac1x$.
A: 

*

*If $$p,q>0,$$ then the graph of $\dfrac1x$ shows that $$p<q\implies\frac1p>\frac1q,\tag1$$ since on the $x$-axis $p$ is on the left of $q$ while on the vertical axis $\dfrac1p$ is higher than $\dfrac1q.$
Notice that the converse of $(1)$ is also true.


*Hence, for $$x,a>0,$$ we have $$ x<\frac{1}{a}\iff \frac{1}{x}>a.$$


*In general, when applying a strictly decreasing function to an inequality, the direction of the inequality is flipped. On the other hand, applying a strictly increasing function to an inequality preserves the inequality's direction.
A: Let's start with a general rule:

If $x<y$ and $z > 0$, then $xz < yz$.

In words: if you multiply inequality by a positive number, the signs don't change.
(What happens if you multiply it by a negative number?)
Let's now assume that $x,y>0$. Then, $xy > 0$ and also $\frac 1{xy} > 0$. Therefore, if $x<y$, we can multiply it by $\frac 1{xy}$ to get $\frac{x}{xy} < \frac{y}{xy}$, which simplifies to $\frac 1x > \frac 1y$. Thus, we derived a new rule:

If $x<y$ and $x,y>0$, then $\dfrac 1x > \dfrac 1y$.

In words: taking reciprocals of positive numbers changes the inequality sign.
We can apply this to your problem, because $x,y\geq 1$:
$$x < \frac 1a < y \implies \frac 1x > \frac 1{1/a} > \frac 1y \implies \frac 1x > a > \frac 1y.$$
A: Perhaps a small example will help you:
$$x = 2, y = 3, a = 0.4$$It is easy to see that the first inequality (the original, marked as $(1)$) is satisfied by this triplet.
Now, going as per you, we have:
$$x < \frac 1a < y \color{red}{\implies \frac 1x < a < \frac 1y \implies 0.5<0.4<\frac 13}$$Do you see the flaw?
Let's get to the problem. We know
$$a \times \frac 1a = 1$$
Now, if you increase $a$, then obviously $\frac 1a$ will decrease, otherwise, the product won't remain $1$ (it will increase). So, we understand that if $a$ increases, $\frac 1a$ decreases. This is why if $a<b$, then $\frac 1a > \frac1b$. Using this property, the error is resolved:
$$\color{green}{\frac 1x > a >\frac 1y \implies 0.5>0.4 > 1/3}$$

Another intuition (similar has been already done by Ennar and JW Tanner):
We know that for an inequality, multiplying by a positive number won't change the signs:
$$x<1/a \implies x \cdot \frac ax < \frac1a \cdot \frac ax \implies \color{blue}{a<1/x}$$
$$1/a<y \implies \frac1a \cdot \frac ay < y \cdot \frac ay \implies \color{blue}{1/y<a}$$
Combining the blue inequalities, we get
$$\color{green}{1/x > a > 1/y}$$
