9.1 Real Numbers, Surds and Exponents
Key Takeaways
- MAT is the calculator-free afternoon mathematics paper; the official 2023 Intake National Report describes 60 items, while current session instructions control the sitting you actually write.
- The principal square root √a is non-negative; √(a + b) is not equal to √a + √b in general
- Like squared terms combine under one radical: for t > 0, √(25t^2 + 144t^2) = 13t, not 17t and not ±13t
- Exponent laws add or subtract exponents only when the bases match; a^{m/n} is a root, not a^m divided by n
- MAT items are unscaffolded multiple-choice with no school-style hence cue; work the algebra before viewing options
MAT numbers are algebraic, not QL tables
The Mathematics (MAT) paper is the afternoon National Benchmark Tests (NBT) sitting for applicants whose faculty requires mathematics. It is not Quantitative Literacy (QL). QL sits inside the morning Academic and Quantitative Literacy (AQL) paper, uses campus and municipal tables, and prints a formula in the item when one is needed. MAT requires you to carry algebraic structure yourself: properties of real numbers, surds, and exponents, with no calculator on the desk.
Current nbt.ac.za Test Content lists MAT skills that include properties of the real number system and the use of surds and exponents. It does not publish a MAT item count. Do not pace against a 100-question rumour copied from a practice bank. Independent OpenExamPrep MAT practice for this chapter is original teaching. It is not a reprint of confidential National Benchmark Tests Project (NBTP) papers, not a Centre for Educational Assessments (CEA) booklet, and not a claim of official sponsorship.
The 2015 MAT teachers' booklet is useful for format, not for memorising stems. MAT items are unscaffolded multiple-choice: there is no school-style hence that hands you a gradient so you can write a perpendicular line. You decide the method. Calculators are not permitted; numbers are chosen so paper arithmetic is enough. Distractors often encode a popular misconception, so you complete the mathematics before you look at the four options.
Nested sets: natural, integer, rational, real
| Set | Name | Typical members | What it is not |
|---|---|---|---|
| Natural numbers | counting numbers 1, 2, 3, … | 1, 8, 20 | Not −3, not 1/2 |
| Integers | …, −2, −1, 0, 1, 2, … | −4, 0, 15 | Not 3/4, not √2 |
| Rationals | fractions p/q with q ≠ 0, including terminating and repeating decimals | −7, 5/8, 0.125 | Not √2, not π |
| Reals | all rationals and irrationals on the number line | √2, π, −3/5 | Not √(−1) |
Every natural number is an integer. Every integer is rational: write n as n/1. Every rational is real. Irrational numbers are reals that are not rational: √2, √3, π. √4 = 2 is rational; surd notation does not make a value irrational once it simplifies to an integer or a ratio of integers. √(49/25) = 7/5, also rational.
Closure is not a slogan. The sum of two rationals is rational. The product of two irrationals may be rational or irrational: √2 × √8 = √16 = 4, which is rational, while √2 × √3 = √6, which is irrational. Do not treat irrational times irrational equals irrational as a law.
Field properties you actually use:
- Addition and multiplication are commutative and associative
- Multiplication distributes over addition: a(b + c) = ab + ac
- 0 and 1 are identities; every real has an additive inverse, and every non-zero real has a multiplicative inverse
- For any real a, a^2 ≥ 0
- √(a^2) = |a|, not always a. If a = −5, then √25 = 5 = |−5|, not −5
That last identity powers half the surd traps in this section.
Principal square roots and the sum-inside trap
For a real number a ≥ 0, √a denotes the principal (non-negative) square root. √9 = 3, never −3, even though (−3)^2 = 9. The equation x^2 = 9 has two real solutions, x = ±3. The symbol √9 names only the positive one.
√(a + b) is not √a + √b in general. Check with squares you can do by hand: √(9 + 16) = √25 = 5, while √9 + √16 = 3 + 4 = 7. Those are different. The same error appears inside algebra when two like squared terms sit under one radical.
Worked example: original parallel numbers. For t > 0, simplify √(25t^2 + 144t^2).
Inside the radical the terms are like terms: 25t^2 + 144t^2 = 169t^2. Then √(169t^2) = √169 × √(t^2) = 13|t|. The stem says t > 0, so |t| = t, and the value is 13t.
Popular wrong readings:
- Split the radical as a sum of roots: √(25t^2) + √(144t^2) = 5t + 12t = 17t
- Keep both signs because squares have two roots: ±13t
- Combine both mistakes: ±17t
The principal root is positive, and t is already positive, so plus-minus is extra. Independent OpenExamPrep uses 25t^2 + 144t^2 as a new parallel to the familiar sum of two squares under a root misconception. It is not a copy of any booklet stem.
Unlike terms stay unlike. √(9m^2 + 16) is not 3m + 4. You cannot split a radical across a sum. Leave √(9m^2 + 16) unless a later constraint produces a perfect square.
Simplifying surds on paper
A surd is in simple form when the radicand has no perfect-square factor other than 1, a denominator does not hide a surd, and like surds are collected.
Worked example: 3√50 − √18 + √8.
- √50 = √(25 × 2) = 5√2
- √18 = √(9 × 2) = 3√2
- √8 = √(4 × 2) = 2√2
Then 3 × 5√2 − 3√2 + 2√2 = 15√2 − 3√2 + 2√2 = 14√2.
Worked example: rationalise 6 / √12. First √12 = 2√3, so 6 / (2√3) = 3 / √3. Multiply numerator and denominator by √3: 3√3 / 3 = √3.
Worked example: conjugate. 1 / (√5 − √2). Multiply numerator and denominator by √5 + √2. Numerator: √5 + √2. Denominator: 5 − 2 = 3. Result: (√5 + √2)/3.
Worked example: fraction under a root. √(4/9) = 2/3, because both 4 and 9 are perfect squares and the principal root is positive. The value is not 4/3 and not −2/3.
Exponent laws, including fractional and negative
For positive bases, where the expressions exist:
| Law | Form | Paper check |
|---|---|---|
| Product | a^m × a^n = a^{m+n} | 2^3 × 2^5 = 2^8 = 256 |
| Quotient | a^m / a^n = a^{m−n} | 2^7 / 2^3 = 2^4 = 16 |
| Power of a power | (a^m)^n = a^{mn} | (2^3)^4 = 2^{12} |
| Power of a product | (ab)^n = a^n b^n | (2 × 5)^2 = 100 |
| Zero | a^0 = 1 for a ≠ 0 | 7^0 = 1, not 0 |
| Negative | a^{−n} = 1 / a^n | 2^{−3} = 1/8 |
| Fractional | a^{1/n} = n-th root of a; a^{m/n} = (n-th root of a)^m | 8^{2/3} = 2^2 = 4 |
Worked example: mixed powers on one base. (27^{2/3}) × (27^{−1/3}) = 27^{2/3 − 1/3} = 27^{1/3} = 3. Subtracting the bases instead of the exponents lands on 0. Reading 27^{2/3} as (27^2)/3 = 729/3 = 243 treats the fraction as a division of a power, not as a root.
Worked example: rewrite to one base. 4^3 × 8^2. 4 = 2^2 and 8 = 2^3, so (2^2)^3 × (2^3)^2 = 2^6 × 2^6 = 2^{12}. MAT options often keep the power form, so you may stop at 2^{12} rather than expanding 4096 without a calculator.
Worked example: quotient of fractional powers. (16^{3/4}) ÷ (16^{1/4}) = 16^{3/4 − 1/4} = 16^{1/2} = 4. Treating 3/4 and 1/4 as ordinary integers produces 16^2 = 256. Quoting 16^{1/4} = 2 stops one step early. Subtracting the bases 16 − 4 invents 12.
Worked example: algebraic square. (3x)^2 = 9x^2, not 3x^2 and not 6x. The same slip as reading 5^4 as 20.
Sitting an unscaffolded surd item
- Cover the options.
- Simplify using like terms, exponent laws, and the principal root.
- Only then look for your result.
- If you see 17t next to 13t, ask whether you split a radical across a sum.
- If you see ±, ask whether the stem used the √ symbol (principal) or asked for solutions of x^2 = ….
Traps
- √(a + b) treated as √a + √b
- √(k^2) treated as ±k instead of |k|
- Adding unlike surds: √8 + √18 is 2√2 + 3√2 = 5√2, not √26
- a^m × b^n combined as (ab)^{m+n} when bases differ
- a^{2/3} read as a^2 / 3
- Importing a QL formula is given habit into MAT
- Pacing by a 100-item practice-bank rumour instead of the official report's 60-item description and the session clock
On test day, MAT number work is structure first, arithmetic second, with no calculator and no hence. Independent OpenExamPrep teaching here stays on real numbers, surds, and exponents so you can simplify before you browse four tempting wrong forms.
For t > 0, the principal square root of 25t^2 + 144t^2 is equal to:
The expression 3√50 − √18 + √8 simplifies to:
The value of (16^{3/4}) ÷ (16^{1/4}) is: