Irrational Numbers and Proofs: Quick Revision

A brief note on Irrational Numbers and Proofs. Read it before a test. Try each question at the end before you open its answer.

Proof by contradiction

• A proof shows that a statement is true in every case. Many examples are not a proof.
• One counterexample is enough to show that a statement is false.
• The method: suppose the opposite, reason correctly, reach something impossible, and conclude that the statement is true.

The key fact

• If a prime $p$ divides $a^2$, then $p$ divides $a$.
• It comes from the Fundamental Theorem of Arithmetic: the primes of $a^2$ are exactly the primes of $a$.
• It needs a prime. $9$ divides $6^2 = 36$, but $9$ does not divide $6$.

Why the square root of a prime is irrational

1. Suppose $\sqrt{p} = \dfrac{a}{b}$, in lowest terms.
2. Square: $a^2 = pb^2$. So $p$ divides $a$. Write $a = pc$.
3. Then $b^2 = pc^2$, so $p$ divides $b$.
4. Now $p$ divides both $a$ and $b$, against lowest terms. So $\sqrt{p}$ is irrational.

The square root of any whole number that is not a perfect square is irrational. $\sqrt{49} = 7$ is rational.

Sums and products

• Rational plus irrational is irrational.
• A non-zero rational times an irrational number is irrational.
• Two irrational numbers can give either kind: $\sqrt{3} \times \sqrt{27} = 9$, but $\sqrt{3} \times \sqrt{5} = \sqrt{15}$.

A worked example

Show that $2\sqrt{7} - 4$ is irrational.

1. Suppose the opposite. Let $2\sqrt{7} - 4 = s$, where $s$ is rational.
2. Get the root alone. $2\sqrt{7} = s + 4$, so $\sqrt{7} = \dfrac{s + 4}{2}$.
3. The contradiction. The right side is rational, so $\sqrt{7}$ would be rational. But $7$ is a prime, so $\sqrt{7}$ is irrational.

So $2\sqrt{7} - 4$ is irrational.

Practice with new numbers

A program to try

The program sorts the square roots of $1$ to $30$ into whole numbers and irrational numbers.

Mistakes to avoid

• Proving with a calculator. A decimal shows only a few digits. It cannot prove that a number is irrational.
• Using the key fact for a number that is not a prime. $4$ divides $36$, but not $6$.
• Forgetting lowest terms. The contradiction comes from the common factor.
• Saying that two irrational numbers give an irrational answer. $(5 + \sqrt{2}) + (1 - \sqrt{2}) = 6$.

Check yourself

1. Is $\sqrt{50}$ rational or irrational?

2. In the proof for $\sqrt{13}$, which fact turns "$13$ divides $a^2$" into "$13$ divides $a$"?

3. Prove that $\dfrac{\sqrt{3}}{4}$ is irrational.

4. Give an irrational number $y$ such that $\sqrt{6} + y$ is rational.

5. Asha says: "Every square root is irrational." Give a counterexample.