Is This True?

Simple counter example:

$\Large\dfrac{1}{-3} < \dfrac{1}{2} \iff -3 > 2 \rightarrow \text{ false}$

The statement is true only if both $a,b$ are either positive or negative.

We could easily answer the question by picking one example like what Nihar Mahajan did:

$\frac{1}{-3}<\frac{1}{2} \iff -3 > 2$

which is false .

However, we can also answer the question by remembering the general principle of inequality: If both sides are multiplied by a negative number, then the symbol must be changed.

For example, if $x>y$ , then $-x<-y$ .

Now, how do we go from $a>b$ to $\frac{1}{a}<\frac{1}{b}$ ? It only takes one step:

$a>b$
Multiply both sides with $\frac{1}{ab}$
$\frac{1}{b} ... \frac{1}{a}$

The problem stated that $a$ and $b$ are non zero, but it didn't state whether both have the same sign (both positive or both negative). So, $\frac{1}{ab}$ could be positive or negative. If it is positive, $\frac{1}{b} > \frac{1}{a}$ . If it is negative, $\frac{1}{b} < \frac{1}{a}$ . No conclusion can be made.

Let us take 1 &-1

1>-1

But 1/1<1/(-1) is not true.

If we assume the inequality holds always, then we can proof by contradiction. We assume a $a > b, \frac{1}{a} < \frac{1}{b}$ , though we are allowed to multiply both sides by the same factor. Hence if we multiply by ab, we get $\frac{ab}{a} < \frac{ab}{b} = b < a$

Hence a contradiction to our assumption the a >b.

not true at all once B is negative, just having a = 1 and b = -1 makes it false... such a silly question.

A simple counter exemple: a = 1 and b = -1. a>b. 1/1<1/(-1) = 1<-1 is false.

if a = 1/4 and b=-1/8 a > b
but if we make it 1/a=4 and 1/b=-8 then a is still greater than b

Use two negative numbers it doesn't work!!

1/A >1/B =A>B ONLY IF A and B are both positive the case is reversed if sign of anyone of them changes [1/(-4)<1/(-1) does not mean (-4>-1) here (-1>-4)}

$a$ and $b$ should be 1, so it's not necessarily true

If a and b are numbers between 0 and 1 then its not possible.