IB Analysis and Approaches HL · Algebra review and mathematical induction · Section 1 of 6

Real numbers, powers and radicals

3 worked examples · By Álvaro López

Key idea

This section covers the numerical tools that the IB takes for granted in every paper: classifying numbers, working with powers and radicals without a calculator (Paper 1) and giving answers to the right degree of accuracy (Paper 2). A mistake here carries over into almost any other topic.

Number sets

Each set exists to solve an operation that the previous one does not allow.

Symbol	Set	What it is for	Examples
$N$	natural numbers: $0, 1, 2, 3, \dots$	counting	$0, 7, 120$
$Z$	integers: $\dots, - 2, - 1, 0, 1, 2, \dots$	always subtracting	$- 5, 0, 18$
$Q$	rationals: $\frac{p}{q}$ with $p, q \in Z$ , $q \neq = 0$	always dividing	$- \frac{3}{4}, 0.125, 5$
$R$	reals: rationals and irrationals	measuring any length	$2, π, - \frac{1}{3}$

In IB notation, $Z^{+} = {1, 2, 3, \dots}$ is the set of positive integers and $Z^{-}$ the negative ones. Each set contains the one before it:

$N \subset Z \subset Q \subset R$

An integer is also rational ( $5 = \frac{5}{1}$ ), but not the other way round. Complex numbers $C$ (needed to solve $x^{2} = - 1$ ) are outside this section.

Be careful with roots: $49 = 7$ is an integer and $\frac{9}{25} = \frac{3}{5}$ is rational. $n$ is irrational only when $n$ is a positive integer that is not a perfect square.

From recurring decimal to fraction

Every recurring decimal is a fraction. The trick is to multiply by a power of 10 that shifts exactly one period, and then subtract, so that the recurring part disappears.

For example, for $x = 0. 18$ the period has 2 digits: $100 x = 18. 18$ , and subtracting $x$ gives $99 x = 18$ , so $x = \frac{18}{99} = \frac{2}{11}$ .

Powers

These two definitions are not arbitrary: they are the only ones that keep the laws below valid. For example, $\frac{a ^{3}}{a ^{3}} = 1$ because a number divided by itself is 1, and by the quotient law it is $a^{3 - 3} = a^{0}$ ; therefore $a^{0}$ must equal $1$ .

Property · Laws of powers

For $a, b \neq = 0$ and exponents $m, n$ :

Law	Formula
Product of the same base	$a^{m} \cdot a^{n} = a^{m + n}$
Quotient of the same base	$\frac{a ^{m}}{a ^{n}} = a^{m - n}$
Power of a power	$(a^{m})^{n} = a^{mn}$
Power of a product	$(ab)^{n} = a^{n} b^{n}$
Power of a quotient	$(\frac{a}{b})^{n} = \frac{a ^{n}}{b ^{n}}$
Negative exponent	$(\frac{a}{b})^{- n} = (\frac{b}{a})^{n}$

The intuition is to count factors: $a^{m} \cdot a^{n}$ has $m$ factors and then $n$ more, $m + n$ in total; $(a^{m})^{n}$ repeats a block of $m$ factors $n$ times, $mn$ in total. These laws also hold for rational and real exponents when the base is positive.

Rational exponents and radicals

For example, $1 6^{3/4} = (416)^{3} = 2^{3} = 8$ and $3 - 8 = - 2$ , but $- 4$ is not a real number. It is better to work out the root first and then raise to the power: the numbers are smaller.

With fractional exponents, the laws of powers give the laws of radicals (with $a, b \geq 0$ if the index is even):

To simplify a radical, take out of the root every factor that is a perfect power: $72 = 36 \cdot 2 = 62$ . You can only add like radicals (same index and same radicand): $35 + 45 = 75$ , which is taking out a common factor. Be careful: $a^{2} = ∣ a ∣$ , not $a$ : $(- 3)^{2} = 3$ .

Rationalising the denominator

Rationalising means removing the radicals from the denominator without changing the value of the fraction, by multiplying the top and bottom by the same quantity. In Paper 1 you are often asked to give the result in exact, rationalised form.

Denominator with one square root: multiply by that root. $\frac{4}{6} = \frac{4 6}{6 6} = \frac{4 6}{6} = \frac{2 6}{3}$ .
Denominator with a sum or difference of roots: multiply by the conjugate (the same binomial with the middle sign changed), because $(p + q) (p - q) = p^{2} - q^{2}$ removes the roots. $\frac{1}{3 - 5} = \frac{3 + 5}{( 3 - 5 ) ( 3 + 5 )} = \frac{3 + 5}{9 - 5} = \frac{3 + 5}{4}$ .

Approximating and rounding

Exact results ( $\frac{1}{3}$ , $2$ ) are approximated when a decimal value is asked for.

Significant figures: count all non-zero digits, zeros between non-zero digits, and trailing zeros to the right of the decimal point. Leading zeros do not count. So $0.00407$ has 3 sf ( $4$ , $0$ and $7$ ) and $5.200$ has 4.

Value	to 3 sf	to 2 dp
$\frac{22}{7} = 3.142857 \dots$	$3.14$	$3.14$
$0.0063549$	$0.00635$	$0.01$
$48750$	$48800$	$48 750.00$

Note that 2 dp and 3 sf are not the same: it depends on where the first non-zero digit is.

Bounds of a rounded value

A rounded value hides the true value. If a quantity has been rounded, the exact value is within half a unit of the last digit kept: between the lower bound (included) and the upper bound (not included).

For example, if a distance is $240$ km rounded to the nearest ten, the true distance $d$ satisfies $235 \leq d < 245$ . If a mass is $0.85$ kg to 2 dp, then $0.845 \leq m < 0.855$ . This is used to bound the error of a calculation or in “greatest possible value” problems.

Scientific notation

It is used for very large or very small quantities: $4200000 = 4.2 \times 1 0^{6}$ and $0.0000017 = 1.7 \times 1 0^{- 6}$ . A positive exponent moves the decimal point to the right; a negative one moves it to the left. Expressions such as $17 \times 1 0^{4}$ or $0.3 \times 1 0^{5}$ are not in standard form, even though they equal $1.7 \times 1 0^{5}$ and $3 \times 1 0^{4}$ .

To operate, separate the mantissas and the powers of 10, apply the laws of powers, and readjust at the end so that the mantissa lies between 1 and 10.

Worked examples

Example 1 Basic

Recurring decimals and classification

a) Express $3. 45$ as an irreducible fraction.

b) Express $0.2 7$ (only the 7 repeats) as an irreducible fraction.

c) Of the numbers $49, - \frac{12}{4}, 18, 0. 45, 2 π$ , how many are rational?

Solution

a) 1. Let $x = 3. 45$ . The period ( $45$ ) has 2 digits, so we multiply by $100$ : $100 x = 345. 45$ .

Subtract: $100 x - x = 345. 45 - 3. 45 = 342$ , so $99 x = 342$ .
$x = \frac{342}{99} = \frac{38}{11}$ (dividing numerator and denominator by $9$ ).

b) 1. Let $x = 0.2 7 = 0.2777 \dots$ Multiply by $10$ and by $100$ so that both have the same tail $7$ after the decimal point: $10 x = 2. 7$ and $100 x = 27. 7$ .

Subtract: $100 x - 10 x = 27. 7 - 2. 7 = 25$ , so $90 x = 25$ .
$x = \frac{25}{90} = \frac{5}{18}$ .

c) Classify each one:

$49 = 7$ : an integer, so rational.
$- \frac{12}{4} = - 3$ : an integer, so rational.
$18 = 32$ : 18 is not a perfect square, so it is irrational.
$0. 45$ : recurring decimal, so rational (it equals $\frac{5}{11}$ ).
$2 π$ : irrational, because $π$ is.

There are 3 rational numbers.

Example 2 Medium

Powers, radicals and rationalising

Without a calculator and in exact form, calculate or simplify:

a) $(\frac{27}{8})^{- 2/3}$

b) $75 + 12 - 27$

c) $\frac{12}{7 - 3}$ , with a rational denominator.

d) $\frac{5 ^{n + 2} - 5 ^{n}}{3 \cdot 5 ^{n + 1}}$ , where $n \in Z$ .

Solution

a) 1. Negative exponent: invert the base, $(\frac{27}{8})^{- 2/3} = (\frac{8}{27})^{2/3}$ .

Root first: $3 \frac{8}{27} = \frac{2}{3}$ .
Then the power: $(\frac{2}{3})^{2} = \frac{4}{9}$ .

b) 1. Take out square factors: $75 = 25 \cdot 3 = 53$ , $12 = 4 \cdot 3 = 23$ , $27 = 9 \cdot 3 = 33$ .

All three are like radicals, so add the coefficients: $5 + 2 - 3 = 4$ . The result is $43$ .

c) 1. Multiply the top and bottom by the conjugate $7 + 3$ .

Denominator: $(7 - 3) (7 + 3) = 7 - 3 = 4$ .
Numerator: $12 (7 + 3)$ . Then $\frac{12 ( 7 + 3 )}{4} = 37 + 33$ .

d) 1. Take out the common factor $5^{n}$ from the numerator: $5^{n + 2} - 5^{n} = 5^{n} (5^{2} - 1) = 24 \cdot 5^{n}$ .

Rewrite the denominator: $3 \cdot 5^{n + 1} = 15 \cdot 5^{n}$ .
Cancel $5^{n}$ (which is not zero): $\frac{24}{15} = \frac{8}{5}$ . The result does not depend on $n$ .

Example 3 Hard

Distance to a star: scientific notation and significant figures

The speed of light is $c = 3.00 \times 1 0^{8}$ m/s. A light year is the distance light travels in a year of 365 days. A star is 4.37 light years away.

a) Calculate the distance of one light year in metres. Give your answer in the form $a \times 1 0^{k}$ , with $1 \leq a < 10$ and $k \in Z$ , rounding $a$ to 3 significant figures.

b) Calculate the distance to the star in metres, also in standard form to 3 sf.

c) Calculate, in seconds, the time it takes the light from that star to reach us, in standard form to 3 sf.

d) A laser has a wavelength of $6.50 \times 1 0^{- 7}$ m, rounded to 3 sf. Find the lower and upper bounds between which the true wavelength $λ$ lies.

Solution

a) 1. Seconds in a year: $365 \cdot 24 \cdot 3600 = 31536000$ .

Distance $= 3.00 \times 1 0^{8} \cdot 31536000 = 9460800000000000$ m.
In standard form: $9.4608 \times 1 0^{15}$ , which to 3 sf is $9.46 \times 1 0^{15}$ m (the next digit, $0$ , is less than 5).

b) 1. Use the unrounded value from a): $4.37 \cdot 9.4608 \times 1 0^{15} = 41.343696 \times 1 0^{15}$ .

Readjust the mantissa: $4.1343696 \times 1 0^{16}$ , which to 3 sf is $4.13 \times 1 0^{16}$ m (the next digit, $4$ , is less than 5).

c) 1. Time is distance divided by speed, and dividing powers of 10 subtracts exponents: $\frac{4.1343696 \times 1 0 ^{16}}{3.00 \times 1 0 ^{8}} = 1.37812 \dots \times 1 0^{8}$ s.

To 3 sf, the next digit is $8 \geq 5$ , so the last digit goes up: $1.38 \times 1 0^{8}$ s (about 4.4 years, consistent with the 4.37 light years).

d) The last digit kept in $6.50 \times 1 0^{- 7}$ is the one in the $1 0^{- 9}$ m place; half a unit at that place is $5 \times 1 0^{- 10}$ m. Then $6.495 \times 1 0^{- 7} \leq λ < 6.505 \times 1 0^{- 7}$ The lower bound is $6.495 \times 1 0^{- 7}$ m and the upper bound is $6.505 \times 1 0^{- 7}$ m.

In the exam: keep the intermediate values unrounded and round only at the end; if you use $9.46 \times 1 0^{15}$ in b), you may get a different digit and lose the accuracy mark. Write the result in the requested form: $41.3 \times 1 0^{15}$ is not in standard form.

Common mistakes

Summary

$N \subset Z \subset Q \subset R$ ; a rational number is a fraction of integers (finite or recurring decimal), and an irrational number has an infinite non-recurring decimal.
A recurring decimal is turned into a fraction by multiplying by powers of 10 and subtracting to eliminate the recurring tail.
$a^{0} = 1$ and $a^{- n} = \frac{1}{a ^{n}}$ ; the laws hold for products, quotients, powers of powers and powers of products or quotients, never for sums.
$a^{m / n} = n a^{m}$ ; work out the root first. An even-index root of a negative number is not real; $a^{2} = ∣ a ∣$ .
Rationalising: multiply by the root or by the conjugate; $(p + q) (p - q) = p^{2} - q^{2}$ .
Rounding: if the next digit is $\geq 5$ , the last digit goes up. Count sf from the first non-zero digit and round only the final result.
A rounded value has bounds half a unit below (included) and above (not included).
Scientific notation: $a \times 1 0^{k}$ with $1 \leq a < 10$ and $k \in Z$ ; readjust the mantissa after operating.

Keep studying

Maths classes in Aldaia and online

Want to work on it together?

Free first consultation: we look at your level and your exam, and build a plan. Álvaro replies on WhatsApp.

Message us on WhatsApp