Answer
See explanation
Work Step by Step
To show that between any two real numbers \(a\) and \(b\) (with \(a < b\)) there is an **irrational** number, one standard argument goes like this:
1. **Set up the gap**: Let \(c = b - a > 0\). We want to place an irrational number strictly between \(a\) and \(b\).
2. **Choose a small irrational increment**: For example, pick a positive integer \(n\) large enough so that
\[
\frac{\sqrt{2}}{n} \;<\; c.
\]
(Such an \(n\) always exists because \(\sqrt{2}/n\) can be made arbitrarily small by choosing \(n\) large.)
3. **Construct the candidate**: Define
\[
x \;=\; a \;+\;\frac{\sqrt{2}}{n}.
\]
- Since \(\sqrt{2}/n\) is irrational, and \(a\) is a real number, the sum \(a + \sqrt{2}/n\) is **irrational** unless \(a\) itself was specifically chosen to “cancel out” \(\sqrt{2}/n\) (which it cannot, in general).
- In any usual scenario, if \(a\) is rational, then adding any nonzero irrational number yields an irrational. If \(a\) is already irrational, adding a rational multiple of \(\sqrt{2}\) will not accidentally become rational.
4. **Check it lies between \(a\) and \(b\)**:
Since \(\sqrt{2}/n < c\), we get
\[
a
\;<\;
a + \frac{\sqrt{2}}{n}
\;<\;
a + c
\;=\;
b.
\]
Hence \(x = a + \tfrac{\sqrt{2}}{n}\) is an **irrational** number satisfying
\[
a \;<\; x \;<\; b.
\]
Because this construction works for any \(a < b\), we conclude the **irrationals** are **dense** on the real number line.
