Is this statement valid for positive $x$ and $y$? If $x>y$, then $\frac{1}{x} < \frac{1}{y}$.

Question

Is this following statement valid (where both $x$ and $y$ are positive)?

If $x>y$, then $\dfrac{1}{x} < \dfrac{1}{y}$.

Answer 1

Counterexample (posted before OP added the condition that $x$ and $y$ are positive):

Let $x=1$ and $y=-1$, the first inequality is $1 > -1$ but the second would become $1 < -1$.

To see that the result holds:

multiply both sides by $\frac1{xy}$.

Answer 2

Because $$\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}>0.$$

Answer 3

It depends on what axioms you were taught.

But the three basics are

0) for any $x,y $ exactly one of the following is true: $x <y $ or $y <x$ or $x=y $.

1) if $x<y $ then for all $c$ we have $x+c <y+c $ and

2) if $x <y $ and $a>0$ the $ax <ay $.

So in a proof by contradiction:

If $\frac 1x \le \frac 1y $ then

$\frac 1xx \le \frac 1y x$

$1\le \frac xy $

$1*y\le \frac xyy $

$y\le x $

Which is a contradiction so $\frac 1x >\frac 1y $