Chapter 1: Real Numbers (Number Systems)

Divide repeatedly by the smallest prime that goes in:

1. $140=2\times 70=2\times 2\times 35=2^2\times 5\times 7$
2. $156=2\times 78=2\times 2\times 39=2^2\times 3\times 13$
3. $3825=3\times 1275=3\times 3\times 425=3^2\times 5\times 85=3^2\times 5^2\times 17$
4. $5005=5\times 1001=5\times 7\times 143=5\times 7\times 11\times 13$
5. $7429=17\times 437=17\times 19\times 23$

(i) 2² × 5 × 7 (ii) 2² × 3 × 13 (iii) 3² × 5² × 17 (iv) 5 × 7 × 11 × 13 (v) 17 × 19 × 23

HCF = product of the smallest power of each common prime; LCM = product of the greatest power of every prime that appears.

(i) $26=2\times 13$, $91=7\times 13$

• HCF $=13$, LCM $=2\times 7\times 13=182$
• Check: $182\times 13=2366$ and $26\times 91=2366$ ✓

(ii) $510=2\times 3\times 5\times 17$, $92=2^2\times 23$

• HCF $=2$, LCM $=2^2\times 3\times 5\times 17\times 23=23460$
• Check: $23460\times 2=46920$ and $510\times 92=46920$ ✓

(iii) $336=2^4\times 3\times 7$, $54=2\times 3^3$

• HCF $=2\times 3=6$, LCM $=2^4\times 3^3\times 7=3024$
• Check: $3024\times 6=18144$ and $336\times 54=18144$ ✓

(i) HCF 13, LCM 182 (ii) HCF 2, LCM 23460 (iii) HCF 6, LCM 3024; in each case LCM × HCF = product of the numbers.

(i) $12=2^2\times 3$, $15=3\times 5$, $21=3\times 7$

HCF $=3$, LCM $=2^2\times 3\times 5\times 7=420$

(ii) 17, 23 and 29 are all prime, so they share no factor.

HCF $=1$, LCM $=17\times 23\times 29=11339$

(iii) $8=2^3$, $9=3^2$, $25=5^2$: no common prime.

HCF $=1$, LCM $=2^3\times 3^2\times 5^2=1800$

(i) HCF 3, LCM 420 (ii) HCF 1, LCM 11339 (iii) HCF 1, LCM 1800

For two positive integers, $\text{HCF}\times \text{LCM}=$ product of the numbers:

$$\text{LCM}=\frac{306\times 657}{9}=34\times 657=22338$$

LCM (306, 657) = 22338

A number ends in 0 only if it is divisible by 10, i.e. its prime factorisation contains both 2 and 5.

$6^n=(2\times 3{)}^n=2^n\times 3^n$. Its only prime factors are 2 and 3; by the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), 5 can never appear.

No: 6ⁿ = 2ⁿ × 3ⁿ has no factor 5, so it can never end in 0.

$$7\times 11\times 13+13=13(7\times 11+1)=13\times 78=1014$$

$$7\times 6\times 5\times 4\times 3\times 2\times 1+5=5(7\times 6\times 4\times 3\times 2\times 1+1)=5\times 1009=5045$$

Each number has a factor other than 1 and itself (13 and 5 respectively), so each is composite.

7 × 11 × 13 + 13 = 13 × 78 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × 1009; both have factors other than 1 and themselves, so both are composite.

They are both at the start again after a time that is a multiple of 18 and of 12; the first such time is the LCM.

$18=2\times 3^2$, $12=2^2\times 3$, so $\text{LCM}=2^2\times 3^2=36$.

(By then Sonia has done 2 rounds and Ravi 3 rounds.)

After 36 minutes.

Suppose, to the contrary, that $\sqrt{5}$ is rational. Then

$$\sqrt{5}=\frac{a}{b}$$

where $a$ and $b$ are integers, $b\ne 0$, and $a$ and $b$ have no common factor other than 1 (we can always cancel common factors).

Squaring: $5b^2=a^2$. So 5 divides $a^2$, and since 5 is prime, 5 divides $a$. Write $a=5c$ for some integer $c$.

Then $5b^2=25c^2$, i.e. $b^2=5c^2$. So 5 divides $b^2$, hence 5 divides $b$.

Now 5 divides both $a$ and $b$, which contradicts the fact that they have no common factor other than 1. This contradiction arose from assuming $\sqrt{5}$ is rational.

Hence √5 is irrational.

Suppose, to the contrary, that $3+2\sqrt{5}$ is rational. Then there are coprime integers $a$, $b$ ($b\ne 0$) with

$$3+2\sqrt{5}=\frac{a}{b}\Rightarrow \sqrt{5}=\frac{a-3b}{2b}$$

Since $a$ and $b$ are integers, $\frac{a-3b}{2b}$ is rational, so $\sqrt{5}$ would be rational.

But $\sqrt{5}$ is irrational (Exercise 1.2, Q1). This contradiction shows our assumption was wrong.

Hence 3 + 2√5 is irrational.

We use the facts that $\sqrt{2}$ and $\sqrt{5}$ are irrational.

(i) Suppose $\frac{1}{\sqrt{2}}$ is rational, say $\frac{1}{\sqrt{2}}=\frac{a}{b}$ with integers $a,b\ne 0$. Then $\sqrt{2}=\frac{b}{a}$, which is rational, a contradiction since $\sqrt{2}$ is irrational. So $\frac{1}{\sqrt{2}}$ is irrational.

(ii) Suppose $7\sqrt{5}$ is rational, say $7\sqrt{5}=\frac{a}{b}$ ($b\ne 0$). Then $\sqrt{5}=\frac{a}{7b}$, which is rational, a contradiction. So $7\sqrt{5}$ is irrational.

(iii) Suppose $6+\sqrt{2}$ is rational, say $6+\sqrt{2}=\frac{a}{b}$ ($b\ne 0$). Then $\sqrt{2}=\frac{a}{b}-6=\frac{a-6b}{b}$, which is rational, a contradiction. So $6+\sqrt{2}$ is irrational.

In each case, assuming the number is rational makes √2 or √5 rational, which is false; so 1/√2, 7√5 and 6 + √2 are irrational.