Number Systems for Math Olympiad: Complete Preparation Guide

Why This Matters

Number systems form the theoretical foundation of mathematics, and Olympiad papers test this foundation rigorously. From classifying real numbers to working with irrational numbers and surds, this topic demands deep understanding.

For Class 9 students, competition problems test your ability to rationalize denominators, represent irrationals on the number line, and apply laws of exponents to rational and irrational numbers.

Best Strategy

Common Pitfalls

Mistakes:

• $\pi$ is NOT $\frac{22}{7}$ — $\frac{22}{7}$ is an approximation. $\pi$ is irrational.
• Sum/product of irrationals — $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$ (irrational), but $\sqrt{2} \times \sqrt{2} = 2$ (rational!).
• Rationalization — Multiply by conjugate: $\frac{1}{a+\sqrt{b}} \times \frac{a-\sqrt{b}}{a-\sqrt{b}}$.
• $0.999... = 1$ — This is true! The repeating decimal equals exactly 1.

Practice Questions

Try these!

Question 1

Rationalize: $\frac{1}{\sqrt{5}+\sqrt{3}}$

Solution: $= \frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} = \frac{\sqrt{5}-\sqrt{3}}{5-3} = \frac{\sqrt{5}-\sqrt{3}}{2}$

Question 2

Simplify: $2^{3/2} \times 2^{1/2}$

Solution: $= 2^{3/2 + 1/2} = 2^2 = 4$

Question 3

Is $\sqrt{2} + \sqrt{3}$ rational or irrational?

Solution: Irrational. Proof by contradiction: if $\sqrt{2}+\sqrt{3} = r$ (rational), then $\sqrt{3} = r - \sqrt{2}$. Squaring: $3 = r^2 - 2r\sqrt{2} + 2$, so $\sqrt{2} = \frac{r^2-1}{2r}$ (rational). Contradiction!

How SparkEd Helps