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NSWMaths AdvancedSyllabus dot point

What are the real numbers, and how do we simplify surds, rationalise denominators and apply the index laws without leaving a surd in the wrong place or a slip in the arithmetic?

Work with the real number system and surds: simplify surds, add, multiply and expand surdic expressions, rationalise single- and binomial-surd denominators, and apply the index laws to expressions with integer indices

A focused answer to the Year 11 Maths Advanced number groundwork: the sets N, Z, Q and R, why a surd is irrational, simplifying surds by taking out square factors, surd arithmetic and binomial expansions, rationalising single-term and binomial-surd denominators with the conjugate, and the index laws for integer indices, with worked examples and original practice questions.

Generated by Claude Opus 4.817 min answer

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What this dot point is asking

Before you can graph, differentiate or solve a quadratic with confidence, you have to be at home with the numbers themselves. NESA expects you to know the real number system, to recognise when a root is a surd (an irrational number that cannot be written as a fraction), and to manipulate surds fluently: simplify them, add and multiply them, expand brackets that contain them, and clear them out of a denominator. You also need the index laws cold, because powers are the language the rest of the course is written in. The methods are short, but they recur in almost every later topic, so the goal is speed without slips, and understanding why each rule holds so you can rebuild it under pressure.

The answer

The real number system: N, Z, Q and R

Mathematics builds its numbers up in layers, each one larger than the last:

  • Natural (counting) numbers N\mathbb{N}: 1,2,3,4,1, 2, 3, 4, \dots (and often 00). The numbers you count with.
  • Integers Z\mathbb{Z}: the naturals together with zero and the negatives, ,2,1,0,1,2,\dots, -2, -1, 0, 1, 2, \dots
  • Rational numbers Q\mathbb{Q}: every number that can be written as a fraction ab\dfrac{a}{b} of two integers (with b0b \neq 0). Equivalently, every terminating or recurring decimal. For example 34=0.75\dfrac{3}{4} = 0.75 and 311=0.27\dfrac{3}{11} = 0.\overline{27} are rational.
  • Real numbers R\mathbb{R}: all the points on the number line. The reals that are not rational are the irrational numbers, such as 2\sqrt{2}, 3\sqrt{3} and π\pi. An irrational number has a decimal that never terminates and never recurs.

Each set sits inside the next: NZQR\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}. The picture below places a sample number in the smallest set that contains it, so you can see at a glance where any given number lives, and that the irrationals are exactly the reals left over once you remove the fractions.

The number systems nestedFour nested boxes showing the counting numbers inside the integers inside the rationals inside the real numbers, with example numbers placed in the smallest set that contains each.R (real)Q (rational)Z (integers)N (counting)170−3¾−2/5√2π√3-3-2-101234√2πEvery real number is a point on the line; the rationals leave gaps that the surds fill.

A subtle point worth keeping: the rationals are dense, meaning between any two of them there is always another, yet they still leave gaps. The number 2\sqrt{2} sits in one of those gaps. It is the length of the diagonal of a unit square, so it certainly names a point on the line, but no fraction equals it exactly. That is what "irrational" means, and it is why we keep such numbers in exact form (2\sqrt{2}, not 1.411.41) until a decimal is actually required.

What a surd is, and why 2\sqrt{2} is irrational

A surd is a root such as 2\sqrt{2}, 53\sqrt[3]{5} or 7\sqrt{7} that is not itself a rational number. The test is whether the root comes out exactly: 9=3\sqrt{9} = 3 and 83=2\sqrt[3]{8} = 2 are not surds because they simplify to integers, whereas 2\sqrt{2} is a surd because it does not. Notice that x\sqrt{x} is the positive square root only: although 2525 has two square roots, 55 and 5-5, the symbol 25\sqrt{25} denotes just 55, and we write 25-\sqrt{25} for the negative one.

That surds really are irrational is not obvious, and a famous proof by contradiction shows it. The argument is worth seeing once because it explains why surds cannot be tidied into fractions; the worked example "Prove that 3\sqrt{3} is irrational" below runs it in full.

Simplifying a surd: take out the largest square factor

A surd is in simplest form when the number under the root sign has no square factor bigger than 11. To simplify, pull out the largest perfect square (4,9,16,25,36,4, 9, 16, 25, 36, \dots) that divides the number, using the law ab=ab\sqrt{ab} = \sqrt{a}\,\sqrt{b} for non-negative aa and bb:

72=36×2=362=62.\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36}\,\sqrt{2} = 6\sqrt{2}.

Always take out the largest square you can spot, or you will have to simplify twice. If you only notice 72=4×18=218\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18}, you are not finished, because 18=9×218 = 9 \times 2 still has the square factor 99. The two laws that do all the work are

ab=abandab=ab(a,b0, b0),\sqrt{a}\,\sqrt{b} = \sqrt{ab} \qquad\text{and}\qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \quad (a, b \ge 0,\ b \neq 0),

and the squaring facts a2=a\sqrt{a^2} = a and (a)2=a(\sqrt{a})^2 = a. There is no corresponding law for a sum: a+b\sqrt{a + b} is not a+b\sqrt{a} + \sqrt{b}, a trap worth fixing in your memory now.

Surd arithmetic: collecting, multiplying and expanding

Once surds are simplified, the ordinary rules of algebra apply, with 2\sqrt{2}, 3\sqrt{3}, and so on treated like the pronumerals xx, yy.

  • Adding and subtracting works only for like surds (the same number under the root). Simplify first so the like surds become visible, then add the coefficients: 50+18=52+32=82\sqrt{50} + \sqrt{18} = 5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}. Unlike surds, such as 2+3\sqrt{2} + \sqrt{3}, cannot be combined at all.
  • Multiplying uses ab=ab\sqrt{a}\,\sqrt{b} = \sqrt{ab}, then check whether the result simplifies further: 26×510=1060=10×215=20152\sqrt{6} \times 5\sqrt{10} = 10\sqrt{60} = 10 \times 2\sqrt{15} = 20\sqrt{15}.
  • Expanding brackets is exactly the algebra of the previous page, with the same three special expansions. In particular (a+b)(ab)=ab(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b, a difference of squares that turns two surds into a plain integer. For example (7+3)(73)=73=4(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3}) = 7 - 3 = 4. That single identity is the engine behind rationalising a binomial denominator.

Rationalising the denominator

It is conventional, and almost always required for full marks, to leave no surd in a denominator. Removing it is called rationalising, and there are two cases.

A single-term denominator (a surd or a multiple of one). Multiply the top and bottom by that surd. Because aa=a\sqrt{a}\,\sqrt{a} = a, the surd in the denominator becomes rational:

63=63×33=633=23.\frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}.

A two-term (binomial) denominator such as 5+1\sqrt{5} + 1 or 2362\sqrt{3} - \sqrt{6}. Multiply the top and bottom by the conjugate, which is the same expression with the middle sign reversed. The denominator then becomes a difference of squares and the surds vanish. The stepped figures below carry 45+1\dfrac{4}{\sqrt{5} + 1} through the whole process.

Rationalise a binomial denominator, stage by stage

Stage 1, spot the conjugate. The denominator is 5+1\sqrt{5} + 1. Write down its conjugate, 51\sqrt{5} - 1 (same terms, opposite middle sign). Multiplying these two will give (5)212(\sqrt{5})^2 - 1^2, a difference of squares with no surd, so the plan is to multiply top and bottom by 51\sqrt{5} - 1.

Spot the conjugateThe denominator is root 5 plus 1. Its conjugate flips the sign to root 5 minus 1.Stage 1Spot the conjugateThe denominator is √5 + 1. Its conjugate flips the sign: √5 − 1.4√5 + 1×√5 − 1√5 − 1Plan: multiply top and bottom by √5 − 1.

Stage 2, multiply by 11. Multiply the fraction by 5151\dfrac{\sqrt{5} - 1}{\sqrt{5} - 1}. That quotient equals 11, so the value of the fraction is unchanged; only its appearance changes. The numerator becomes 4(51)4(\sqrt{5} - 1) and the denominator becomes (5+1)(51)(\sqrt{5} + 1)(\sqrt{5} - 1).

Multiply by 1Multiply by root 5 minus 1 over root 5 minus 1, which equals 1, so the value is unchanged.Stage 2Multiply by 1Multiply by (√5 − 1)/(√5 − 1). That equals 1, so the value is unchanged.4 (√5 − 1)(√5 + 1)(√5 − 1)Top and bottom both gain the factor √5 − 1.

Stage 3, difference of squares. Expand the denominator with (A+B)(AB)=A2B2(A + B)(A - B) = A^2 - B^2. Here A=5A = \sqrt{5} and B=1B = 1, so the denominator is (5)212=51=4(\sqrt{5})^2 - 1^2 = 5 - 1 = 4. The surd is gone: the denominator is now the integer 44.

Difference of squaresThe bottom is A plus B times A minus B, equals A squared minus B squared, equals root 5 squared minus 1 squared, equals 5 minus 1, equals 4. The surd is gone.Stage 3Difference of squaresThe bottom is (A+B)(A−B) = A² − B² = (√5)² − 1² = 5 − 1 = 4.4 (√5 − 1)4(√5)² = 5, so the denominator becomes the integer 4.

Stage 4, cancel and finish. The numerator 4(51)4(\sqrt{5} - 1) and the denominator 44 share the factor 44. Cancel it to reach the answer 51\sqrt{5} - 1, which has no surd in any denominator and so is fully rationalised.

Cancel and finishCancel the common factor 4 from top and bottom to leave root 5 minus 1.Stage 4Cancel and finishCancel the common factor 4 from top and bottom.√5 − 11Answer: √5 − 1 (no surd in any denominator).

The index laws (integer indices)

A power ana^n is repeated multiplication, and five laws let you combine powers of the same base. They are the rules you will use to differentiate, to handle exponentials, and to tidy almost any algebraic expression:

  • Multiplying adds indices: am×an=am+na^m \times a^n = a^{m+n}.
  • Dividing subtracts indices: aman=amn\dfrac{a^m}{a^n} = a^{m-n}.
  • Power of a power multiplies indices: (am)n=amn(a^m)^n = a^{mn}.
  • Power of a product distributes: (ab)n=anbn(ab)^n = a^n b^n.
  • Power of a quotient distributes: (ab)n=anbn\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}.

When an expression has several factors, work through it systematically: the numbers first, then each pronumeral in turn. For instance (20x7y3)÷(4x5y3)=5x2y0=5x2(20x^7y^3) \div (4x^5y^3) = 5x^{2}y^{0} = 5x^2, since 20÷4=520 \div 4 = 5, 75=27 - 5 = 2 and 33=03 - 3 = 0 (and any non-zero base to the power 00 is 11). A common slip is to add the coefficients instead of multiplying them, or to apply a law across different bases: a3×b2a^3 \times b^2 does not simplify, because the bases differ.

How exam questions ask about surds and indices

These skills underpin many longer questions, and a handful of command words tell you exactly which one is wanted:

  • "Simplify" a surd means take out every square factor and collect like surds. State the answer in simplest surd form (no square factor left under the root, no like surds left uncombined).
  • "Express in simplest exact form" or "leave your answer in surd form" is a signal not to give a decimal: keep 30\sqrt{30}, do not write 5.485.48.
  • "Rationalise the denominator" means clear the surd from the bottom: multiply by the surd for a single term, or by the conjugate for a two-term denominator, then use the difference of squares.
  • "Show that" a surd expression equals a given value usually means rationalise or expand until both sides match; write every line so the marker can follow the manipulation.
  • "Simplify, leaving your answer with positive indices" is an index-law question: combine the powers, then make sure no negative or zero index is left in the final form.
  • Pythagoras or geometry problems often produce a surd answer (d2=30d^2 = 30 gives d=30d = \sqrt{30}); take the positive root for a length, and simplify the surd.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksSimplify 200\sqrt{200}, and hence simplify 45+20\sqrt{45} + \sqrt{20}.
Show worked solution →

Take out the largest square factor of 200200. The squares up to 200200 are 4,9,16,25,36,49,64,81,100,121,144,169,1964, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, and 100100 divides 200200. So

200=100×2=1002=102.\sqrt{200} = \sqrt{100 \times 2} = \sqrt{100}\,\sqrt{2} = 10\sqrt{2}.

Simplify each surd in the second expression. 45=9×5=35\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} and 20=4×5=25\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}.

Collect like surds. Both are multiples of 5\sqrt{5}, so add the coefficients:

45+20=35+25=55.\sqrt{45} + \sqrt{20} = 3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}.

Check numerically. 45+206.7082+4.4721=11.1803\sqrt{45} + \sqrt{20} \approx 6.7082 + 4.4721 = 11.1803, and 5511.18035\sqrt{5} \approx 11.1803. They agree.

foundation2 marksRationalise the denominator of 126\dfrac{12}{\sqrt{6}}.
Show worked solution →

Multiply top and bottom by the surd in the denominator. To clear 6\sqrt{6}, multiply by 66\dfrac{\sqrt{6}}{\sqrt{6}}, which equals 11 and so does not change the value:

126=126×66=1266.\frac{12}{\sqrt{6}} = \frac{12}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{12\sqrt{6}}{6}.

Simplify the rational coefficient. 126=2\dfrac{12}{6} = 2, so

1266=26.\frac{12\sqrt{6}}{6} = 2\sqrt{6}.

Check numerically. 126122.4495=4.8990\dfrac{12}{\sqrt{6}} \approx \dfrac{12}{2.4495} = 4.8990, and 264.89902\sqrt{6} \approx 4.8990. They agree.

core2 marksExpand and simplify (325)(1+5)(3 - 2\sqrt{5})(1 + \sqrt{5}).
Show worked solution →

Expand every pair of terms. Multiply each term in the first bracket by each term in the second:

(325)(1+5)=3+3525255.(3 - 2\sqrt{5})(1 + \sqrt{5}) = 3 + 3\sqrt{5} - 2\sqrt{5} - 2\sqrt{5}\cdot\sqrt{5}.

Simplify the surd product. 55=5\sqrt{5}\cdot\sqrt{5} = 5, so 255=2×5=10-2\sqrt{5}\cdot\sqrt{5} = -2 \times 5 = -10.

Collect rational parts and surd parts separately. Rational: 310=73 - 10 = -7. Surd: 3525=53\sqrt{5} - 2\sqrt{5} = \sqrt{5}. So

(325)(1+5)=7+5.(3 - 2\sqrt{5})(1 + \sqrt{5}) = -7 + \sqrt{5}.

Check numerically. (325)(1+5)(34.4721)(1+2.2361)=(1.4721)(3.2361)=4.7639(3 - 2\sqrt{5})(1 + \sqrt{5}) \approx (3 - 4.4721)(1 + 2.2361) = (-1.4721)(3.2361) = -4.7639, and 7+57+2.2361=4.7639-7 + \sqrt{5} \approx -7 + 2.2361 = -4.7639. They agree.

core3 marksSimplify (2a3b)2×3ab4a4b2\dfrac{(2a^3b)^2 \times 3ab^4}{a^4b^2}, where aa and bb are non-zero, leaving your answer with positive indices.
Show worked solution →

Apply the power to the bracket first. Raising a product to a power raises each factor: (2a3b)2=22a3×2b1×2=4a6b2(2a^3b)^2 = 2^2 a^{3\times 2} b^{1 \times 2} = 4a^6b^2.

Multiply out the numerator. Multiply 4a6b24a^6b^2 by 3ab43ab^4: multiply the numbers (4×3=124 \times 3 = 12) and add the indices of each letter (aa: 6+1=76 + 1 = 7; bb: 2+4=62 + 4 = 6):

(2a3b)2×3ab4=12a7b6.(2a^3b)^2 \times 3ab^4 = 12a^7b^6.

Divide by the denominator. To divide powers of the same base, subtract indices (aa: 74=37 - 4 = 3; bb: 62=46 - 2 = 4):

12a7b6a4b2=12a74b62=12a3b4.\frac{12a^7b^6}{a^4b^2} = 12a^{7-4}b^{6-2} = 12a^3b^4.

Every index is already positive, so the simplified expression is 12a3b412a^3b^4.

exam3 marksRationalise the denominator of 275\dfrac{2}{\sqrt{7} - \sqrt{5}}, expressing your answer in simplest exact form.
Show worked solution →

Identify the conjugate of the denominator. The denominator 75\sqrt{7} - \sqrt{5} is a two-term (binomial) surd. Its conjugate keeps the terms but flips the middle sign: 7+5\sqrt{7} + \sqrt{5}. Multiplying a binomial surd by its conjugate gives a difference of squares, which removes the surds.

Multiply top and bottom by the conjugate.

275×7+57+5=2(7+5)(75)(7+5).\frac{2}{\sqrt{7} - \sqrt{5}} \times \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}} = \frac{2(\sqrt{7} + \sqrt{5})}{(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5})}.

Expand the denominator as a difference of squares. (7)2(5)2=75=2(\sqrt{7})^2 - (\sqrt{5})^2 = 7 - 5 = 2:

2(7+5)75=2(7+5)2.\frac{2(\sqrt{7} + \sqrt{5})}{7 - 5} = \frac{2(\sqrt{7} + \sqrt{5})}{2}.

Cancel the common factor 22.

2(7+5)2=7+5.\frac{2(\sqrt{7} + \sqrt{5})}{2} = \sqrt{7} + \sqrt{5}.

Check numerically. 27522.64582.2361=20.4097=4.8818\dfrac{2}{\sqrt{7} - \sqrt{5}} \approx \dfrac{2}{2.6458 - 2.2361} = \dfrac{2}{0.4097} = 4.8818, and 7+52.6458+2.2361=4.8818\sqrt{7} + \sqrt{5} \approx 2.6458 + 2.2361 = 4.8818. They agree.

exam4 marksA rectangular garden bed at a community garden has length 232\sqrt{3} metres and width 323\sqrt{2} metres. Find, in simplest exact (surd) form, (a) the exact area of the bed, and (b) the exact length of its diagonal.
Show worked solution →

(a) Multiply length by width for the area. Multiply the numbers and the surds separately, then simplify:

A=23×32=(2×3)(3×2)=66 m2.A = 2\sqrt{3} \times 3\sqrt{2} = (2 \times 3)(\sqrt{3} \times \sqrt{2}) = 6\sqrt{6} \ \text{m}^2.

Here 3×2=6\sqrt{3}\times\sqrt{2} = \sqrt{6}, and 66 has no square factor, so 666\sqrt{6} is fully simplified. Numerically A6×2.4495=14.697 m2A \approx 6 \times 2.4495 = 14.697\ \text{m}^2.

(b) Use Pythagoras' theorem for the diagonal. The diagonal dd of a rectangle satisfies d2=length2+width2d^2 = \text{length}^2 + \text{width}^2. Square each side, remembering (km)2=k2m(k\sqrt{m})^2 = k^2 m:

d2=(23)2+(32)2=(4×3)+(9×2)=12+18=30.d^2 = (2\sqrt{3})^2 + (3\sqrt{2})^2 = (4 \times 3) + (9 \times 2) = 12 + 18 = 30.

Take the positive square root (a length is positive):

d=30 m.d = \sqrt{30} \ \text{m}.

Since 30=2×3×530 = 2 \times 3 \times 5 has no square factor, 30\sqrt{30} is already in simplest form. Numerically d5.477 md \approx 5.477\ \text{m}.

Answers. Area =66 m2= 6\sqrt{6}\ \text{m}^2; diagonal =30 m= \sqrt{30}\ \text{m}.

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