Topic 1: Number and algebra | IB Maths AI

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🎯What you need to be able to do

Work with numbers in the form $ a \times 10^{k} $, and never write calculator notation in an answer.
Use the $n$th term and sum formulae for arithmetic and geometric sequences, including sigma notation.
Apply geometric sequences to compound interest, depreciation and real value under inflation.
Round to a given number of decimal places or significant figures, state upper and lower bounds, and calculate percentage error.
Judge whether an answer is reasonable, and estimate before calculating.
Use the financial package on your calculator for amortization and annuities.
Solve systems of linear equations and polynomial equations with technology.
Use the laws of exponents with integer exponents, and evaluate logarithms base 10 and $ \mathrm{e} $.
AHL Apply the laws of logarithms and simplify rational exponents.
AHL Sum an infinite geometric series.
AHL Work with complex numbers in Cartesian, polar and exponential form, including adding sinusoids.
AHL Use matrices: algebra, determinants, inverses and solving systems.
AHL Find eigenvalues and eigenvectors, and use them to compute powers of a $ 2 \times 2 $ matrix.

📚The mathematics

1.1 Standard form

A number in standard form is $ a \times 10^{k} $ with $ 1 \le a < 10 $ and $k$ an integer. Your calculator may display 5.2E30, but that is calculator notation and it is not accepted in an answer. Write $ 5.2 \times 10^{30} $. This applies to every paper and it is free marks lost.

1.2 Arithmetic sequences and series

$ u_n = u_1 + (n-1)d $

$ S_n = \dfrac{n}{2}\left(2u_1 + (n-1)d\right) $

$ S_n = \dfrac{n}{2}\left(u_1 + u_n\right) $

The $ (n-1) $ is the classic slip: the tenth term involves nine steps. Given two terms, subtract the equations rather than guessing — from $ u_7 = 25 $ and $ u_{12} = 45 $ you get $ 5d = 20 $ immediately.

Arithmetic models describe anything growing by a fixed amount: simple interest, a salary rising by a set sum each year, seats increasing by a constant number per row. If the growth is by a fixed percentage, it is geometric instead, and choosing wrongly between the two is the most consequential decision in this topic.

1.3 Geometric sequences and series

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

$ S_n = \dfrac{u_1(r^{n}-1)}{r-1} $

$ S_n = \dfrac{u_1(1-r^{n})}{1-r} $

To find $r$ from two given terms, divide rather than subtract: $ \dfrac{u_6}{u_3} = r^{3} $. Geometric sequences model anything with a constant percentage change — population growth, radioactive decay, salary rises of 3% a year, the spread of a disease in its early phase.

1.4 Financial applications

Compound interest is a geometric sequence. With $PV$ invested at a nominal annual rate of $r\%$ compounded $k$ times per year for $n$ years:

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

Half-yearly is $ k = 2 $, quarterly $ k = 4 $, monthly $ k = 12 $. Depreciation is the same formula with a negative rate — an asset losing 15% a year is multiplied by $ 0.85 $ annually.

To find the real value of an investment, adjust the growth for inflation by dividing rather than subtracting: a 6% nominal return with 4% inflation gives a real growth factor of $ \dfrac{1.06}{1.04} $, about 1.92% — close to, but not equal to, the 2% that subtraction suggests. Over twenty years that difference is visible.

1.7 Amortization and annuities

An annuity pays out a regular sum from an invested lump; amortization pays off a loan with regular repayments. Both are done with the finance solver (TVM) on your calculator, and the syllabus expects that rather than a formula.

The sign convention is where these questions are won or lost. Money flowing away from you is negative and money flowing towards you is positive. Take a loan: the present value is positive (the bank gives you money) and the payment is negative (you pay it back). Save into an annuity: the payment is negative and the future value comes out positive. Get it backwards and the solver returns a confident, wrong number rather than an error. Also check $ P/Y $ and $ C/Y $, and that payments are at the end of the period.

1.6 Approximation, bounds and error

Round to a stated number of decimal places or significant figures, and choose a sensible accuracy from the data you were given. If $ x = 4.1 $ to one decimal place then $ 4.05 \le x < 4.15 $ — those are the bounds, and note the asymmetry of the inequality signs.

\[ \text{percentage error} = \left|\frac{v_A - v_E}{v_E}\right| \times 100\% \]

where $ v_A $ is the approximate value and $ v_E $ the exact one. The exact value goes on the bottom — dividing by the approximation instead is the standard error, and it gives a different answer. Estimation matters too: before trusting a calculator output, ask whether the magnitude is plausible. Lengths cannot be negative, probabilities cannot exceed 1, and a population cannot be 4.7 people.

1.5 Exponents and logarithms, and 1.8 solving with technology

The laws of exponents with integer exponents: $ a^{m}a^{n} = a^{m+n} $, $ \dfrac{a^{m}}{a^{n}} = a^{m-n} $, $ (a^{m})^{n} = a^{mn} $, $ a^{-n} = \dfrac{1}{a^{n}} $. A logarithm is an exponent: $ a^{x} = b $ means exactly $ \log_a b = x $, for $ b > 0 $, and $ \ln x $ is $ \log_{\mathrm{e}} x $. At standard level you evaluate logarithms numerically with technology; the manipulation laws are HL.

You are also expected to solve systems of up to three linear equations in three unknowns, and polynomial equations, using technology — no particular method is required, and in examinations a system will always have a unique solution. Learn the words though: the solutions of $ f(x) = 0 $ are the roots of the equation and the zeros of the function.

AHL 1.9 & 1.10 Logarithm laws and rational exponents

$ \log_a xy = \log_a x + \log_a y $

$ \log_a \dfrac{x}{y} = \log_a x - \log_a y $

$ \log_a x^{m} = m\log_a x $

In examinations the base will be 10 or $ \mathrm{e} $. Beware the false laws: $ \log(x+y) \ne \log x + \log y $, and $ \dfrac{\log x}{\log y} \ne \log\dfrac{x}{y} $. These laws are what make it possible to linearise exponential data in Topic 2, which is their main use in AI.

Rational exponents: $ a^{1/n} = \sqrt[n]{a} $ and $ a^{m/n} = \sqrt[n]{a^{m}} $, so $ 32^{3/5} = \left(\sqrt[5]{32}\right)^{3} = 2^{3} = 8 $. Take the root first — the numbers stay small.

AHL 1.11 Infinite geometric series

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

State the convergence condition explicitly; it is often a mark of its own. This connects to fractals in Topic 3 and to Markov chains in Topic 4.

AHL 1.12 & 1.13 Complex numbers

Define $ \mathrm{i} $ by $ \mathrm{i}^{2} = -1 $. In Cartesian form $ z = a + b\mathrm{i} $, with conjugate $ z^{*} = a - b\mathrm{i} $ and modulus $ |z| = \sqrt{a^{2}+b^{2}} $, plotted on an Argand diagram. A quadratic with real coefficients and $ b^{2} - 4ac < 0 $ has a conjugate pair of complex roots.

Polar form is $ z = r(\cos\theta + \mathrm{i}\sin\theta) = r\,\mathrm{cis}\,\theta $, and exponential (Euler) form is $ z = r\mathrm{e}^{\mathrm{i}\theta} $. Multiplying multiplies the moduli and adds the arguments; geometrically, multiplication is a rotation combined with a stretch.

The application AI cares about is adding sinusoids of the same frequency: two alternating voltages with different phase shifts add to a single sinusoid of the same frequency, and representing each as a complex number turns an awkward trigonometric sum into a straightforward vector addition. That is why complex numbers appear in this course at all — the electrical engineering context is the point, not a decoration.

AHL 1.14 Matrices

A matrix has order rows $ \times $ columns. Addition and scalar multiplication are element-wise. Multiplication is defined only when the inner dimensions agree, and it is not commutative — $ AB \ne BA $ in general, which is the single most important structural fact about matrices and the one most often forgotten under time pressure. It is associative and distributive.

For a $ 2 \times 2 $ matrix $ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $, $ \det A = ad - bc $ and

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

Swap the leading diagonal, negate the other, divide by the determinant. A matrix with $ \det A = 0 $ is singular and has no inverse. Larger matrices are handled with technology. A system of linear equations written $ A\mathbf{x} = \mathbf{b} $ is then solved by $ \mathbf{x} = A^{-1}\mathbf{b} $, and applications include coding and decoding messages.

AHL 1.15 Eigenvalues and eigenvectors

An eigenvector of $M$ is a direction the matrix does not turn: $ M\mathbf{v} = \lambda\mathbf{v} $, where the scalar $ \lambda $ is the eigenvalue telling you how much that direction is stretched. Find the eigenvalues from the characteristic equation $ \det(M - \lambda I) = 0 $, then substitute each back to find its eigenvector.

With distinct real eigenvalues you can diagonalise, and that makes high powers cheap:

\[ M^{n} = PD^{n}P^{-1} \]

where $P$ has the eigenvectors as its columns and $D$ is the diagonal matrix of eigenvalues — and raising a diagonal matrix to a power just means raising each entry. This is the machinery behind long-run population models and Markov chains, where the eigenvalue equal to 1 corresponds to the steady state.

✏️Worked example

Maya invests $6500 in an account paying a nominal annual rate of 4.8% compounded quarterly. (a) Find the value of the investment after 5 years, to the nearest dollar. (b) Find the number of complete years before the investment first exceeds $9000. (c) Over the same 5 years inflation averaged 2.9% per year. Find the real value of the investment in today’s money, and comment.

(a) Quarterly means $ k = 4 $, so the quarterly rate is $ \dfrac{4.8}{4} = 1.2\% $ and there are $ 4 \times 5 = 20 $ compounding periods:

\[ FV = 6500\left(1 + \frac{4.8}{400}\right)^{20} = 6500(1.012)^{20} = \$8248 \]

(b) Solve $ 6500(1.012)^{4n} > 9000 $, so $ (1.012)^{4n} > 1.3846 $. Using technology — either the finance solver or the graph of the left-hand side — $ 4n > 27.28 $, so $ n > 6.82 $. The question asks for complete years, so the answer is 7 years.

(c) Divide by the inflation factor rather than subtracting the rate:

\[ \text{real value} = \frac{8248}{(1.029)^{5}} = \frac{8248}{1.1537} = \$7150 \]

So although the account grew by about $1750 in nominal terms, in purchasing power it gained only about $650. The investment did beat inflation, but by far less than the headline figure suggests.

Check it. The real growth factor is $ \dfrac{1.048}{1.029} \approx 1.0185 $, so roughly 1.85% real growth per year. Over 5 years that is about 9.6%, and 9.6% of $6500 is about $625 — consistent with the $650 found above. Two routes agreeing is worth thirty seconds.

Rounding at the wrong moment, and rounding the wrong way. In (b), 6.82 years rounds up to 7 because the question asks when the value is first exceeded — normal rounding rules do not apply to a threshold. And in (c), using the rounded $8248 rather than the stored full-accuracy value can shift the final answer by a dollar or two; keep the unrounded figure in your calculator and round only at the very end.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A theatre has 22 seats in the front row, and each subsequent row has 3 more seats than the one in front. There are 18 rows. Find the number of seats in the last row and the total seating capacity.

Arithmetic, since the increase is a fixed amount. $ u_1 = 22 $, $ d = 3 $, $ n = 18 $. Last row: $ u_{18} = 22 + 17(3) = 73 $ seats. Total: $ S_{18} = \tfrac{18}{2}(22 + 73) = 9(95) = 855 $ seats. Using $ u_1 + 18(3) = 76 $ is the classic error — there are only 17 gaps between 18 rows.

2. A machine bought for €45 000 depreciates by 18% per year. Find its value after 4 years, and the first year in which its value falls below €15 000.

Each year the value is multiplied by $ 0.82 $. After 4 years: $ 45000(0.82)^{4} = $ €20 349. For the threshold, solve $ 45000(0.82)^{n} < 15000 $, so $ (0.82)^{n} < 0.3333 $. By technology $ n > 5.54 $, so the value first falls below €15 000 during the 6th year. (Checking: $ 45000(0.82)^{6} = $ €13 682, which is indeed below.)

3. A student measures the length of a room as 5.8 m. The true length is 5.94 m. Find the percentage error, and state the bounds within which the true length lies if the measurement is correct to 1 decimal place.

Percentage error $ = \left|\dfrac{5.8 - 5.94}{5.94}\right| \times 100 = \dfrac{0.14}{5.94} \times 100 = 2.36\% $. Note the exact value 5.94 is the denominator. For the bounds, a measurement of 5.8 to 1 d.p. means $ 5.75 \le x < 5.85 $. The true value 5.94 lies outside those bounds, so the measurement is not merely imprecise — it is inconsistent with a correct reading to 1 d.p.

4. A loan of $18 000 is taken out at 7.2% per year compounded monthly, to be repaid in equal monthly instalments over 5 years. Find the monthly repayment and the total interest paid.

Use the finance solver: $ n = 60 $ payments, $ I\% = 7.2 $, $ PV = 18000 $, $ FV = 0 $, $ P/Y = C/Y = 12 $. This gives $ PMT = -\$358.19 $, so the monthly repayment is $358.19. Total paid $ = 60 \times 358.19 = \$21\,491 $, so the interest is $ 21491 - 18000 = \$3491 $. The $PV$ is positive and the $PMT$ negative because the money comes to you and then flows away — entering both as positive gives a nonsensical result.

5. AHL Find the inverse of $ A = \begin{pmatrix} 3 & 5 \\ 1 & 4 \end{pmatrix} $ and use it to solve $ 3x + 5y = 11 $, $ x + 4y = 9 $.

$ \det A = (3)(4) - (5)(1) = 7 $, so $ A^{-1} = \dfrac{1}{7}\begin{pmatrix} 4 & -5 \\ -1 & 3 \end{pmatrix} $. The system is $ A\mathbf{x} = \begin{pmatrix} 11 \\ 9 \end{pmatrix} $, so $ \mathbf{x} = A^{-1}\begin{pmatrix} 11 \\ 9 \end{pmatrix} = \dfrac{1}{7}\begin{pmatrix} 44 - 45 \\ -11 + 27 \end{pmatrix} = \dfrac{1}{7}\begin{pmatrix} -1 \\ 16 \end{pmatrix} $. So $ x = -\tfrac{1}{7} $ and $ y = \tfrac{16}{7} $. Check in the second equation: $ -\tfrac{1}{7} + \tfrac{64}{7} = \tfrac{63}{7} = 9 $. Correct.

6. AHL Find the eigenvalues of $ M = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} $ and an eigenvector for each.

The characteristic equation is $ \det(M - \lambda I) = (4-\lambda)(3-\lambda) - 2 = 0 $, which expands to $ \lambda^{2} - 7\lambda + 10 = 0 $, so $ (\lambda - 5)(\lambda - 2) = 0 $ and $ \lambda = 5 $ or $ \lambda = 2 $.
For $ \lambda = 5 $: $ (4-5)x + y = 0 $ gives $ y = x $, so $ \begin{pmatrix} 1 \\ 1 \end{pmatrix} $ is an eigenvector.
For $ \lambda = 2 $: $ (4-2)x + y = 0 $ gives $ y = -2x $, so $ \begin{pmatrix} 1 \\ -2 \end{pmatrix} $ works.
Any non-zero multiple of either is equally valid — an eigenvector specifies a direction, not a length, so an answer differing from the markscheme by a scalar factor is still right.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

3Blue1Brown — Essence of Linear Algebra, especially the episode on eigenvectors
Khan Academy — sequences, series and financial mathematics
Your GDC’s own manual — genuinely worth twenty minutes on the finance and matrix menus