Prove there is a rational q so that $|q-\sqrt{2}| < d - c$

Question

Suppose $c < d$. Prove there is a $q \in Q$ so that $|q-\sqrt{2}| < d - c$

I began this proof by setting $c=1, d=2$, and consequently, $q=\sqrt{2}$. However, my professor informed me that proofs need to satisfy general cases, which makes sense.

Then, I created the inequality $-\alpha<q-\sqrt{2}<\alpha$ with $\alpha$ being arbitrary. Because $\sqrt{2}$ is irrational, $i=q-\sqrt{2}$ must be, so we have $-\alpha<i<\alpha$. This process seems typical of similar problems, but I do not know how to complete the proof. I appreciate any and all help!

Answer 1

Hint: Consider the number $x(n)$ represented by the first $n$ digits in the decimal expansion of $\sqrt{2}$. For example, $x(1) = 1$, $x(2) = 1.4$, $x(3) = 1.41$ and so on. Each $x(n)$ is rational since its decimal expansion is finite, and we can see that as $n$ becomes large, $x(n)$ becomes a better and better approximation of $\sqrt{2}$.

To analyze how good these approximations are, note the following inequalities: $$\begin{align*}\sqrt{2} - x(1) = 0.4142\dots & < 1, \\ \sqrt{2} - x(2) = 0.0142\dots & < 1/10, \\ \sqrt{2} - x(3) = 0.0042\dots & < 1/10^2, \end{align*}$$ and so on. Try using this pattern (and be sure to prove it really is a pattern) to show that with sufficiently large $n$, the approximation gets as good as you like (without being equal to $\sqrt{2}$).

Answer 2

Note that the rational numbers are dense in $\mathbb{R}$, meaning that between any two real numbers $a$ and $b$, there is a rational number $c$ where $a < c < b$.

Suppose that c < d and $|q-\sqrt{2}|<d-c$. Note $d$ and $c$ are real numbers, so $d-c$ is also a real number. Denote $d-c$ as $a$. Then \begin{align} & -a < q-\sqrt{2} < a \\ & -a + \sqrt{2} < q < a + \sqrt{2} \end{align} $-a + \sqrt{2}$ and $a + \sqrt{2}$ are real numbers, so we know that there must be a rational between them, and we can let $q$ be such rational number.