Question

If \(a\) and \(b\) are positive and \(a

Short answer

Yes, if \(a\) and \(b\) are positive and \(a<b\), then \(\frac{1}{b}<\frac{1}{a}\) is true. As shown in the example, where \(a=2\) and \(b=3\), we can see that \(\frac{1}{3}<\frac{1}{2}\).

Step-by-step solution

Understanding what happens when a number is divided by a larger number

When a number is divided by a larger number, the result is a smaller quantity. For example, for any positive number \(p\), \( \frac{p}{2} > \frac{p}{3}\) because 2 is less than 3. Here, although \(p\) is the same, the result depends on whether \(p\) is divided by 2 or 3.

Applying this understanding to our problem

In our problem, \(a<b\), which means \(b\) is larger than \(a\). So if we divide 1 (a positive number) by \(b\) and \(a\), we should get a smaller result when we divide it by \(b\) than when we divide it by \(a\). So \(\frac{1}{b}<\frac{1}{a}\).

Providing an example

Let's use an example to illustrate this. Let's say \(a=2\) and \(b=3\). \(a<b\) and both are positive numbers. Then if we divide 1 by \(b\) and \(a\) we get \(\frac{1}{3} = 0.33\) and \(\frac{1}{2} = 0.5\) respectively. Clearly, \(\frac{1}{3}<\frac{1}{2}\) which shows \(\frac{1}{b}<\frac{1}{a}\) when \(a\) and \(b\) are positive and \(a<b\).