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IB Maths: Applications & Interpretation: Number and algebra

Standard form, error, sequences, compound interest, loans. 15 subtopics, 60 key terms, full free lesson below.

This topic is free in the app too. Interactive diagrams, worked examples, check questions and exam-style practice with mark schemes. Open “Number and algebra” in FortyFive →

In this topic: 1.1 · 1.2 · 1.3 · 1.4 · 1.5 · 1.6 · 1.7 · 1.8 · 1.9 · 1.10 · 1.11 · 1.12 · 1.13 · 1.14 · 1.15

1.1Standard form

Key terms

Standard form: a × 10ᵏ with 1 ≤ a < 10 and k an integer.
Positive power: The number is 10 or more: 4.5 × 10⁶ = 4 500 000.
Negative power: The number is between 0 and 1: 3.2 × 10⁻³ = 0.0032.
GDC entry: Use the EE / ×10ˣ key, never type ×10^ as two separate operations.

Science, economics and computing questions use numbers that are far too big or small to write out. Standard form makes them easy to read and compare.

\[a\times10^{k},\qquad 1\le a\lt10,\ k\in\mathbb{Z}\]

To multiply, multiply the \(a\)-values and add the powers. To divide, divide the \(a\)-values and subtract the powers.
Always finish by adjusting so that \(1\le a\lt10\): \(12\times10^{3}=1.2\times10^{4}\).
Your GDC may show 4.5E6. That is calculator notation: in your answer write \(4.5\times10^{6}\).

Worked example

The mass of the Earth is \(5.97\times10^{24}\) kg and the mass of the Moon is \(7.35\times10^{22}\) kg. How many times heavier is the Earth?

Divide: \(\dfrac{5.97\times10^{24}}{7.35\times10^{22}}=\dfrac{5.97}{7.35}\times10^{2}\).
\(\dfrac{5.97}{7.35}\approx0.8122\), so the answer is \(0.8122\times10^{2}\).
Adjust: about 81.2 times (\(8.12\times10^{1}\)).

Exam tip: Copying “4.5E6” from your GDC into your answer loses the mark. Write 4.5 × 10⁶.

1.2Arithmetic sequences and series

Key terms

Arithmetic sequence: Each term is the previous term plus a fixed amount.
Common difference d: d = uₙ₊₁ − uₙ. Negative d means the sequence decreases.
Series: The sum of the terms of a sequence.
Simple interest: Interest on the original amount only, so the balance grows arithmetically.

\[u_n=u_1+(n-1)d\]

\[S_n=\frac{n}{2}\big(2u_1+(n-1)d\big)=\frac{n}{2}\,(u_1+u_n)\]

Real contexts: seats in rows, a salary that rises by a fixed amount each year, savings with fixed deposits, simple interest.
Sigma notation: \(\displaystyle\sum_{k=1}^{n}u_k=u_1+u_2+\dots+u_n\). Your GDC can evaluate it directly.
To find \(n\) when you know a term, solve \(u_1+(n-1)d=u_n\). To find \(n\) from a sum, use the GDC solver or a table.
Check that \(n\) is a positive whole number: a context answer of \(n=14.3\) means “the 15th term” or “14 complete years”, depending on the question.

Worked example

A theatre has 25 rows. The first row has 20 seats and each row has 3 more seats than the row in front. How many seats are there in total?

Arithmetic: \(u_1=20,\ d=3,\ n=25\).
Last row: \(u_{25}=20+24\times3=92\) seats.
Total: \(S_{25}=\dfrac{25}{2}(20+92)=12.5\times112=\mathbf{1400}\) seats.

Exam tip: Write down u₁, d and n before you substitute. Most lost marks come from using n instead of n − 1.

1.3Geometric sequences and series

Key terms

Geometric sequence: Each term is the previous term multiplied by a fixed number.
Common ratio r: r = uₙ₊₁ ÷ uₙ. A 5% increase means r = 1.05; a 5% decrease means r = 0.95.
Growth factor: The multiplier applied each period, e.g. 1.05.
Decay: 0 < r < 1: the terms shrink.

\[u_n=u_1\,r^{\,n-1}\]

\[S_n=\frac{u_1\left(r^{n}-1\right)}{r-1}=\frac{u_1\left(1-r^{n}\right)}{1-r},\qquad r\ne1\]

Percentage change \(\to\) ratio: increase by \(p\%\) gives \(r=1+\frac{p}{100}\); decrease by \(p\%\) gives \(r=1-\frac{p}{100}\).
Contexts: population growth, spread of a virus, bacteria, depreciation, bounce heights.
To find \(n\), use the GDC solver or a table of values; exact answers can use logarithms.
Watch the counting: if the population is \(P_0\) at the start, after \(n\) years it is \(P_0r^{n}\) (that is the \((n+1)\)th term).

Worked example

A town has 12 000 people and the population grows by 5% per year. Estimate the population after 10 years.

The growth factor is \(r=1.05\).
After 10 years: \(12\,000\times1.05^{10}\).
\(=19\,546.7\ldots\approx\mathbf{19\,500}\) people (3 s.f.).

Exam tip: In context questions, decide first whether “after n years” means multiplying by rⁿ or by rⁿ⁻¹. Write one line explaining your choice.

1.4Financial applications: compound interest and depreciation

Key terms

Present value PV: The amount invested or borrowed now.
Future value FV: The amount after interest has been added.
Compounding: Interest is added to the balance, so you earn interest on interest.
Depreciation: The value of an asset falls by a fixed percentage each year.
Inflation: Prices rise over time, so money loses buying power.

\[FV=PV\times\left(1+\frac{r}{100k}\right)^{kn}\]

\[V=V_0\left(1-\frac{r}{100}\right)^{n}\]

\(r\) = nominal annual rate (%), \(k\) = number of compounding periods per year (monthly \(k=12\), quarterly \(k=4\)), \(n\) = years.
GDC finance (TVM) solver: \(N=kn\), \(I\%=r\), \(PV\) = amount invested (enter as negative: money leaves you), \(PMT=0\), \(P/Y=C/Y=k\). Solve for \(FV\).
Inflation: to find the real value, use a real rate of about (interest rate − inflation rate), or divide by \((1+i)^{n}\).
Money answers: 2 decimal places, or to the nearest cent/penny.

Worked example

€3000 is invested at 2.4% per year, compounded monthly. Find its value after 6 years.

\(PV=3000,\ r=2.4,\ k=12,\ n=6\).
\(FV=3000\left(1+\frac{2.4}{1200}\right)^{72}=3000\times1.002^{72}\).
\(FV=\mathbf{\text{€}3464.15}\). TVM: N = 72, I% = 2.4, PV = −3000, PMT = 0, P/Y = C/Y = 12.

Exam tip: Write your TVM entries in the answer (N = …, I% = …, PV = …). They earn method marks even if you mistype one value.

1.5Laws of exponents and introduction to logarithms

Key terms

Exponent: The power a number is raised to: in 2⁵, the exponent is 5.
Logarithm: log_a b is the power you raise a to in order to get b.
log x: Means log₁₀ x (common logarithm).
ln x: Means logₑ x, where e ≈ 2.718 (natural logarithm).

\[a^{m}\times a^{n}=a^{m+n}\qquad \frac{a^{m}}{a^{n}}=a^{m-n}\qquad \left(a^{m}\right)^{n}=a^{mn}\]

\[a^{0}=1\qquad a^{-n}=\frac{1}{a^{n}}\qquad (ab)^{n}=a^{n}b^{n}\]

\[a^{x}=b\iff x=\log_a b\]

In AI SL the exponents are integers. Rational exponents such as \(a^{\frac12}\) are AHL (1.10).
Logarithms answer “what power?”: \(\log_{10}1000=3\) because \(10^{3}=1000\).
To solve \(a^{x}=b\), use \(x=\log_a b\) on the GDC, or the equation solver, or the intersection of two graphs.

Worked example

Solve (a) \(5\times2^{x}=80\) and (b) \(10^{x}=0.002\).

(a) Divide by 5: \(2^{x}=16=2^{4}\), so \(x=\mathbf{4}\).
(b) \(x=\log_{10}0.002\).
\(x=-2.698\ldots\approx\mathbf{-2.70}\) (3 s.f.).

Exam tip: Before reaching for logs, check whether both sides can be written as powers of the same base. It is quicker and gives an exact answer.

1.6Approximation, bounds and percentage error

Key terms

Decimal places (d.p.): Digits after the decimal point.
Significant figures (s.f.): Digits counted from the first non-zero digit.
Upper and lower bounds: The range of true values that round to a given measurement.
Percentage error: How far an approximation is from the exact value, as a % of the exact value.
Estimation: A rough value found by rounding first, used to check that an answer is sensible.

\[\varepsilon=\left|\frac{v_A-v_E}{v_E}\right|\times100\%\]

\(v_A\) = approximate value, \(v_E\) = exact value. Always divide by the exact value.
Bounds: a length of 12 cm to the nearest cm lies in \(11.5\le L\lt12.5\). To 1 d.p., 6.3 lies in \(6.25\le x\lt6.35\).
Bounds of a calculation: maximum of a product = product of the upper bounds; maximum of a difference = upper bound − lower bound.
Unless told otherwise, IB answers are exact or to 3 significant figures.

Worked example

A rectangle is measured as 8 m by 5 m, each to the nearest metre. Find the upper and lower bounds of its area.

Length: \(7.5\le L\lt8.5\). Width: \(4.5\le W\lt5.5\).
Upper bound: \(8.5\times5.5=\mathbf{46.75\ m^{2}}\).
Lower bound: \(7.5\times4.5=\mathbf{33.75\ m^{2}}\).

Exam tip: Keep full values in your GDC during a calculation and round only the final answer. Early rounding is a classic lost accuracy mark.

1.7Amortization and annuities

Key terms

Loan: Money borrowed (PV) that is repaid with regular payments (PMT).
Amortization: Paying off a loan with equal regular payments that cover interest and part of the debt.
Annuity: A series of equal regular payments into or out of an account.
Outstanding balance: What is still owed after some payments.

These questions are solved with the finance (TVM) solver on your GDC. You need to know what each entry means and which signs to use.

\(N\) = total number of payments, \(I\%\) = annual rate, \(PV\) = present value, \(PMT\) = regular payment, \(FV\) = future value, \(P/Y\) = payments per year, \(C/Y\) = compounding periods per year.
Sign rule: money you receive is positive, money you pay is negative. For a loan, \(PV\gt0\) (you receive it) and \(PMT\lt0\). A loan that is fully repaid has \(FV=0\).
Total interest paid = \(N\times|PMT|-PV\).
For savings (an annuity), \(PV=0\) or negative, \(PMT\lt0\) and you solve for \(FV\).
Early payments are mostly interest; later payments are mostly repayment of the debt.

Worked example

Sam borrows €15 000 at 5% per year, compounded monthly, and repays it in equal monthly payments over 4 years. Find the monthly payment and the total interest.

TVM: \(N=48,\ I\%=5,\ PV=15\,000,\ FV=0,\ P/Y=C/Y=12\).
Solve: \(PMT=-345.44\), so Sam pays €345.44 per month.
Total paid \(=48\times345.44=\text{€}16\,581.12\) (using the unrounded payment: €16 581.09).
Interest \(\approx16\,581.09-15\,000=\mathbf{\text{€}1581.09}\).

Exam tip: State the TVM values you used, including P/Y and C/Y. A wrong sign gives an error or a negative answer: say which way the money flows.

1.8Systems of linear equations and polynomial equations

Key terms

System of linear equations: Two or three equations that must all be true at the same time.
Unknowns: The variables you solve for (up to three in AI).
Polynomial equation: An equation such as ax³ + bx² + cx + d = 0.
Root: A solution of an equation: an x-intercept of y = f(x).

Use the GDC simultaneous equation solver for systems with 2 or 3 unknowns, and the polynomial root finder for polynomial equations.
Most marks are for setting up the equations from words: define each variable clearly, then write one equation per piece of information.
Write the equations in the same order of variables before entering them, with every variable on the left: \(ax+by+cz=d\).
A polynomial of degree \(n\) has at most \(n\) real roots. Check that each root makes sense in the context (e.g. a negative length is rejected).

Worked example

Three adult and two child tickets cost €34. Two adult and five child tickets cost €41. Find the price of each ticket.

Let \(a\) = adult price and \(c\) = child price (€).
\(3a+2c=34\) and \(2a+5c=41\).
GDC solver (or elimination): \(11c=55\), so \(c=5\) and \(a=8\).
Adult ticket €8, child ticket €5.

Exam tip: Define your variables in words (“let x be the number of …”). Without that line you can lose a mark even with the right numbers.

1.9Laws of logarithms HL

Key terms

Product law: log(xy) = log x + log y
Quotient law: log(x/y) = log x − log y
Power law: log(xᵐ) = m log x
Change of base: log_a x = log_b x ÷ log_b a

\[\log_a(xy)=\log_a x+\log_a y\qquad \log_a\!\left(\frac{x}{y}\right)=\log_a x-\log_a y\]

\[\log_a\!\left(x^{m}\right)=m\log_a x\qquad \log_a x=\frac{\log_b x}{\log_b a}\]

To solve \(a^{kx}=b\): take logs of both sides and use the power law to bring \(x\) down.
There is no law for \(\log(x+y)\).
Log arguments must be positive: check every solution.

Worked example

Solve \(3^{2x}=20\).

Take \(\ln\) of both sides: \(2x\ln3=\ln20\).
\(x=\dfrac{\ln20}{2\ln3}\).
\(x\approx\mathbf{1.36}\).

Exam tip: Show the log step (“taking ln of both sides”). A bare GDC answer with no method can score only the final mark.

1.10Simplifying expressions with rational exponents HL

Key terms

Rational exponent: An exponent that is a fraction, such as ½ or ⅔.
nth root: a^(1/n) = ⁿ√a
a^(m/n): (ⁿ√a)ᵐ: take the root first, then the power.

\[a^{\frac1n}=\sqrt[n]{a}\qquad a^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^{m}=\sqrt[n]{a^{m}}\]

\[a^{-\frac{m}{n}}=\frac{1}{a^{\frac{m}{n}}}\]

All the integer laws of exponents still work with fractions.
Taking the root first keeps the numbers small: \(8^{\frac23}=2^{2}=4\).
Writing roots as powers (\(\sqrt{x}=x^{\frac12}\)) is essential before differentiating or integrating.

Worked example

Simplify \(\left(27x^{6}\right)^{\frac23}\).

\(27^{\frac23}=\left(\sqrt[3]{27}\right)^{2}=3^{2}=9\).
\(\left(x^{6}\right)^{\frac23}=x^{4}\).
Answer: \(\mathbf{9x^{4}}\).

Exam tip: Negative exponent means reciprocal, not negative number: 16^(−3/4) is positive.

1.11Sum of an infinite geometric sequence HL

Key terms

Convergent series: Its sum approaches a finite limit.
Divergent series: Its sum grows without limit (or keeps oscillating).
Sum to infinity S∞: The limit of Sₙ as n → ∞.

\[S_\infty=\frac{u_1}{1-r},\qquad |r|\lt1\]

If \(|r|\ge1\) there is no sum to infinity.
Contexts: total distance travelled by a bouncing ball, total effect of repeated drug doses, long-run savings.
For a bouncing ball, every bounce is travelled up and down, so double the rebound series.

Worked example

A ball is dropped from 10 m. After each bounce it rises to 60% of its previous height. Find the total vertical distance travelled.

First fall: 10 m.
Rebounds: \(6+3.6+\dots\), each travelled up and down: \(2\times\dfrac{6}{1-0.6}=2\times15=30\) m.
Total \(=10+30=\mathbf{40\ m}\).

Exam tip: Always state |r| < 1 before using S∞. It is often worth a reasoning mark.

1.12Complex numbers: Cartesian form HL

Key terms

i: The imaginary unit, with i² = −1.
Cartesian form: z = a + bi, with real part a and imaginary part b.
Complex conjugate: z* = a − bi
Argand diagram: Real part on the horizontal axis, imaginary part on the vertical axis.

\[i^{2}=-1,\qquad z=a+bi,\qquad z^{*}=a-bi\]

Add and subtract by collecting real and imaginary parts; multiply by expanding brackets and replacing \(i^{2}\) with \(-1\).
To divide, multiply the top and bottom by the conjugate of the denominator.
A quadratic with negative discriminant has two complex roots \(a\pm bi\) (a conjugate pair).
Your GDC works in complex mode: use it to check your algebra.

Worked example

Write \(\dfrac{3+i}{1-i}\) in the form \(a+bi\).

Multiply top and bottom by \(1+i\): \(\dfrac{(3+i)(1+i)}{(1-i)(1+i)}\).
Top: \(3+3i+i+i^{2}=2+4i\). Bottom: \(1-i^{2}=2\).
Answer: \(\mathbf{1+2i}\).

Exam tip: Do not leave i² in your final answer. Replace it with −1 every time.

1.13Complex numbers: polar and Euler form HL

Key terms

Modulus |z|: The distance from 0 on the Argand diagram: √(a² + b²).
Argument arg z: The angle from the positive real axis, usually in −π < θ ≤ π.
Polar form: z = r(cos θ + i sin θ) = r cis θ
Euler form: z = re^(iθ)

\[z=r(\cos\theta+i\sin\theta)=r\,\mathrm{cis}\,\theta=re^{i\theta}\]

\[|z_1z_2|=|z_1||z_2|,\quad \arg(z_1z_2)=\arg z_1+\arg z_2,\quad \left(re^{i\theta}\right)^{n}=r^{n}e^{in\theta}\]

Find \(\theta\) with a sketch: \(\tan\theta=\frac{b}{a}\), then check the quadrant.
Multiplying by \(re^{i\theta}\) is a rotation by \(\theta\) and an enlargement by \(r\).
AI application: two sinusoids with the same frequency, such as voltages in AC circuits, can be added as complex numbers. The modulus of the sum is the new amplitude and the argument is the phase shift.

Worked example

Write \(z=1+i\sqrt3\) in Euler form, and find \(z^{6}\).

\(|z|=\sqrt{1+3}=2\), and \(\arg z=\arctan\sqrt3=\frac{\pi}{3}\) (first quadrant).
\(z=2e^{i\pi/3}\).
\(z^{6}=2^{6}e^{i\,6\pi/3}=64e^{2\pi i}=\mathbf{64}\).

Exam tip: Set your GDC to radians for Euler form, and give arguments in radians unless the question uses degrees.

1.14Matrices HL

Key terms

Order m × n: m rows and n columns.
Identity matrix I: 1s on the leading diagonal, 0s elsewhere: AI = IA = A.
Determinant: For [[a, b], [c, d]], det = ad − bc.
Inverse A⁻¹: AA⁻¹ = I. It exists only if det A ≠ 0.

\[A=\begin{pmatrix}a&b\\c&d\end{pmatrix},\quad \det A=ad-bc,\quad A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\]

\(AB\) exists only if the number of columns of \(A\) equals the number of rows of \(B\). In general \(AB\ne BA\).
A system of linear equations can be written \(AX=B\), so \(X=A^{-1}B\) (multiply on the left).
For 3 × 3 matrices, use the GDC for determinants and inverses.

Worked example

Use a matrix method to solve \(2x+3y=8\) and \(x-y=-1\).

\(A=\begin{pmatrix}2&3\\1&-1\end{pmatrix}\), \(\det A=-2-3=-5\).
\(A^{-1}=-\frac15\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}\).
\(X=A^{-1}\begin{pmatrix}8\\-1\end{pmatrix}=-\frac15\begin{pmatrix}-5\\-10\end{pmatrix}=\begin{pmatrix}1\\2\end{pmatrix}\), so \(\mathbf{x=1,\ y=2}\).

Exam tip: Write X = A⁻¹B, not BA⁻¹. Order matters for matrices.

1.15Eigenvalues and eigenvectors HL

Key terms

Eigenvector: A non-zero vector v whose direction is unchanged by A: Av = λv.
Eigenvalue λ: The scale factor along an eigenvector.
Characteristic equation: det(A − λI) = 0
Diagonalisation: A = PDP⁻¹, with eigenvectors in P and eigenvalues in D.

\[A\mathbf{v}=\lambda\mathbf{v}\qquad \det(A-\lambda I)=0\]

\[A=PDP^{-1}\ \Rightarrow\ A^{n}=PD^{n}P^{-1}\]

For a \(2\times2\) matrix: \(\lambda^{2}-(\text{trace})\lambda+\det A=0\).
Each eigenvalue gives an eigenvector from \((A-\lambda I)\mathbf v=\mathbf 0\).
\(D^{n}\) is easy: raise each diagonal entry to the power \(n\). This gives long-term behaviour, e.g. of Markov chains (4.19) and coupled systems (5.17).

Worked example

Find the eigenvalues and eigenvectors of \(A=\begin{pmatrix}4&1\\2&3\end{pmatrix}\).

\(\lambda^{2}-7\lambda+10=0\Rightarrow(\lambda-2)(\lambda-5)=0\), so \(\lambda=2,\ 5\).
\(\lambda=5\): \(-x+y=0\), eigenvector \(\begin{pmatrix}1\\1\end{pmatrix}\).
\(\lambda=2\): \(2x+y=0\), eigenvector \(\begin{pmatrix}1\\-2\end{pmatrix}\).

Exam tip: Check an eigenvector by multiplying: A × v should give λ × v. It takes ten seconds and catches sign errors.

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