Real Numbers

The set of real numbers is the union of rational and irrational numbers. This chapter formalises everything you already know about numbers — and proves a few results that you will reuse in every later topic.

1. Euclid's Division Lemma

Statement. Given positive integers $a$ and $b$, there exist unique integers $q$ and $r$ satisfying $a=bq+r$, where $0\le r<b$.

In plain English: when you divide $a$ by $b$, you always get a quotient $q$ and a remainder $r$ that is strictly less than the divisor.

Example. Take $a=455$, $b=42$. $455=42\times 10+35,0\le 35<42.$ So $q=10$ and $r=35$.

Euclid's division algorithm (HCF)

To find $\gcd (a,b)$ where $a>b$:

1. Apply the lemma: $a=b{q}_1+r_1$.
2. If $r_1=0$, then $\gcd (a,b)=b$. Done.
3. Otherwise apply the lemma to $b$ and $r_1$, and continue.

The last non-zero remainder is the HCF.

Worked example — HCF(135, 225):

So $\gcd (135,225)=45$.

2. The Fundamental Theorem of Arithmetic

Every composite number can be expressed as a product of primes, and this factorisation is unique apart from the order of the primes.

This is the result behind every HCF/LCM calculation you'll do.

HCF and LCM via prime factorisation.

For two numbers $a,b$:

• $\text{HCF}(a,b)=$ product of the smallest power of each common prime.
• $\text{LCM}(a,b)=$ product of the greatest power of each prime that appears.

Worked example — $a=96$, $b=404$.

$96=2^5\cdot 3,404=2^2\cdot 101.$ $\text{HCF}=2^2=4,\text{LCM}=2^5\cdot 3\cdot 101=9696.$

Sanity check: $\text{HCF}\times \text{LCM}=4\times 9696=38,784=a\cdot b$. ✓

3. Revisiting Irrational Numbers

A number is irrational if it cannot be expressed as $p/q$ for integers $p$ and $q$ with $q\ne 0$.

Classic proof: $\sqrt{2}$ is irrational

Assume for contradiction that $\sqrt{2}=p/q$ in lowest terms. Then $p^2=2q^2$, so $p^2$ is even, so $p$ is even — write $p=2k$. Substituting: $4k^2=2q^2\Rightarrow q^2=2k^2$, so $q$ is also even. But then both $p$ and $q$ are divisible by 2, contradicting "lowest terms". Hence $\sqrt{2}$ is irrational.

The same argument works for $\sqrt{p}$ where $p$ is any prime.

4. Decimal Expansions of Rational Numbers

A rational number $\frac{p}{q}$ (in lowest terms) has a terminating decimal expansion if and only if the prime factorisation of $q$ is of the form $2^n{5}^m$.

Quick check.

Key formulas to memorise

• $\gcd (a,b)\cdot \text{lcm}(a,b)=a\cdot b$ (for two numbers).
• For three numbers $a,b,c$: $a\cdot b\cdot c\ne \gcd \cdot \text{lcm}$ in general — work via primes.
• $a=bq+r$, $0\le r<b$.

Practice (try before checking)

1. Find the HCF of 196 and 38220 using Euclid's algorithm.
2. Show that $5-2\sqrt{3}$ is irrational.
3. Without long division, decide whether $\frac{17}{6250}$ terminates.
4. The HCF of two numbers is 27 and their LCM is 162. If one number is 54, find the other.

Answers

1. 196 (verify by primes: $196=2^2\cdot 7^2$, $38220=2^2\cdot 3\cdot 5\cdot 7^2\cdot 13$, common $=2^2\cdot 7^2=196$).
2. Suppose $5-2\sqrt{3}=r$ is rational. Then $\sqrt{3}=(5-r)/2$, which would make $\sqrt{3}$ rational. Contradiction.
3. $6250=2\cdot 5^5$. The form is $2^n{5}^m$, so the decimal terminates.
4. $a\cdot b=\text{HCF}\cdot \text{LCM}\Rightarrow 54\cdot b=27\cdot 162\Rightarrow b=81$.

What's next

Once Real Numbers is solid, Polynomials builds on the prime factorisation idea but applies it to algebraic expressions — zeroes, the division algorithm and the relationship between coefficients and roots.