AL2 - Surds and exact values

Surds and exact values

A surd is a square root that cannot be written as a fraction, and leaving it as a root is the only way to write its value exactly.

Start with a puzzle. The golden ratio, which Euclid described in his Elements around 300 BC, is $\varphi=\frac{1+\sqrt5}{2}$. Your calculator gives $\varphi=1.6180339887\ldots$. Now press the reciprocal key: $\frac1\varphi=0.6180339887\ldots$. The digits after the decimal point are identical. Is that a lucky coincidence, or must it happen? By the end of this lesson you will prove the answer in a few lines, without a calculator.

First, a concrete case. Remember from GCSE that $\sqrt a$ means the positive square root of $a$, so $\sqrt a\times\sqrt a=a$. Type $\sqrt2$ into your calculator and it shows $1.414213562$. Square that decimal and you get $1.999999998944\ldots$, not 2. The calculator has rounded, and no decimal, however long, squares to exactly 2. The symbol $\sqrt2$ is the only exact way to write this number.

An expression built from surds, such as $3\sqrt5$ or $2+\sqrt3$, is said to be in surd form. When a question asks for an exact answer, this is the form it expects.

Surds have a simple picture. By Pythagoras, the diagonal of a 1 by 1 square is $\sqrt{1^2+1^2}=\sqrt2$, and the diagonal of a 2 by 2 square is $\sqrt{2^2+2^2}=\sqrt8$. Now look at the grid: the long diagonal is made of two short diagonals placed end to end, so $\sqrt8=2\sqrt2$. In the same way, the diagonal of a 3 by 3 square gives $\sqrt{18}=3\sqrt2$.

[DIAGRAM: asset_name: Square diagonals as copies of root 2; asset_slug: surd-diagonals; caption: The diagonals of 1 by 1, 2 by 2 and 3 by 3 squares are 1, 2 and 3 copies of root 2; description: Three squares on a unit grid with sides 1, 2 and 3. Each diagonal is drawn in blue and split at the grid points into unit-square diagonals, each of length root 2. Labels show that 1 squared plus 1 squared is 2 so the diagonal is root 2, 2 squared plus 2 squared is 8 so the diagonal is root 8 which equals 2 root 2, and 3 squared plus 3 squared is 18 so the diagonal is root 18 which equals 3 root 2]

Put the 2 by 2 and 3 by 3 diagonals end to end and you have $2+3=5$ copies of $\sqrt2$, so $\sqrt8+\sqrt{18}=5\sqrt2\approx7.07$. That is not $\sqrt{8+18}=\sqrt{26}\approx5.10$. Roots of a sum behave very differently from sums of roots.

Multiplication is different, and the reason is short. For $a\ge0$ and $b\ge0$, the numbers $\sqrt a\times\sqrt b$ and $\sqrt{ab}$ are both non-negative, and they have the same square: $(\sqrt a\times\sqrt b)^2=(\sqrt a)^2\times(\sqrt b)^2=ab$. Two non-negative numbers with the same square are equal, so $\sqrt a\times\sqrt b=\sqrt{ab}$. That is why $\sqrt8=\sqrt4\times\sqrt2=2\sqrt2$, exactly as the picture shows. The same argument works for division.

You will use this rule all the way through the lesson: to split a surd into a whole number times a smaller surd, and to multiply surds together. It never applies to a sum.

Now test the rule on a student's working.

Simplifying and collecting surds

A surd is in its simplest form when the number under the root has no square factor except 1, and only then can you see which surds can be collected.

To simplify $\sqrt n$, look for the largest square number that divides $n$ (the squares are $4, 9, 16, 25, 36, 49, 64, 81, 100, \ldots$). Split it off using $\sqrt{ab}=\sqrt a\times\sqrt b$, then take its square root. This is the algebra behind the square diagonals in the previous section.

Once surds are simplified, those with the same number under the root are called like surds. They collect exactly like like terms in algebra, with the surd playing the part of the letter: $4\sqrt3+2\sqrt3=6\sqrt3$, just as $4x+2x=6x$. Unlike surds, such as $\sqrt2+\sqrt3$, cannot be combined into one term.

The next example adds one new feature: two surds that look unrelated until you simplify them.

The most common slip at this stage comes in the collecting step.

When a question names a form such as $k\sqrt n$ or $a+b\sqrt c$, finish by checking that the number under each root has no square factor left.

Now try one like Worked example 2, with three surds and a subtraction.

Expanding brackets with surds

Brackets containing surds expand exactly like algebraic brackets; the one extra fact you need is $\sqrt a\times\sqrt a=a$.

To multiply single surds, multiply the numbers outside the roots together and the numbers inside the roots together, then simplify: $2\sqrt3\times5\sqrt6=10\sqrt{18}=10\times3\sqrt2=30\sqrt2$. Whenever a surd meets itself it becomes a whole number, so $\sqrt3\times\sqrt3=3$ and $(4\sqrt5)^2=16\times5=80$. That is what makes surd brackets tidy: the squared surd terms drop out as ordinary numbers.

The next example changes one feature: the surd arrives by substitution, so the work is squaring a bracket.

The error examiners see most often here is in squaring the bracket.

OCR is strict about seeing that expansion when the question says "Determine".

Now try one like Worked example 4, with a minus sign in the value of $x$.

Rationalising the denominator

Rationalising the denominator means rewriting a fraction so that no surd is left on the bottom, without changing its value.

A surd on the bottom is exact but awkward. Compare $\frac{1}{\sqrt2}$ with $\frac{\sqrt2}{2}$: they are the same number, but the second is easy to estimate (half of $1.414$ is $0.707$), easy to combine with other surds, and the form exam answers expect. Before calculators, dividing 1 by $1.414$ by hand was slow, while halving $1.414$ took a moment.

The tool is multiplying the top and bottom of the fraction by the same non-zero number, which is multiplying by 1, so the value never changes. If the denominator is a single surd $\sqrt a$, multiply top and bottom by $\sqrt a$, because $\sqrt a\times\sqrt a=a$.

If the denominator has two terms, such as $2+\sqrt3$, multiplying by $\sqrt3$ does not work: $(2+\sqrt3)\times\sqrt3=2\sqrt3+3$ still contains a surd. You need a partner that cancels the surd completely.

A two-term surd expression times its conjugate is always rational. It is a difference of two squares, so the two cross terms cancel and the squared surd is a whole number.

Here is the same fact as a picture, with $a=3$ and $b=2$. Remove a $\sqrt2$ by $\sqrt2$ square (area 2) from the corner of a 3 by 3 square (area 9), leaving an L-shape of area 7. Cut off strip B, turn it, and place it at the end of piece A: the pieces now make a rectangle $3+\sqrt2$ long and $3-\sqrt2$ high. So $(3+\sqrt2)(3-\sqrt2)=9-2=7$, a whole number.

[DIAGRAM: asset_name: A conjugate pair as an area; asset_slug: conjugate-area; caption: Why 3 plus root 2 times 3 minus root 2 equals 9 minus 2, which is 7; description: Left, a 3 by 3 square with a root 2 by root 2 corner square marked for removal, leaving a blue piece A that is 3 wide and 3 minus root 2 high and a green strip B that is 3 minus root 2 wide and root 2 high, so the area left is 9 minus 2, which is 7. A curved arrow shows strip B turned and moved to the end of piece A. Right, the two pieces form a rectangle 3 plus root 2 wide and 3 minus root 2 high, so its area, 3 plus root 2 times 3 minus root 2, is also 7]

Now for the method itself, starting with the simpler case of a single surd on the bottom.

The next example changes one feature: the denominator has two terms, so you need the conjugate.

Marks are most often lost in the product of the conjugates.

Now try one like Worked example 6, with a plus sign in the denominator.

Exact answers with detailed reasoning

The hardest surd questions combine all of these skills, and OCR often adds the instruction "In this question you must show detailed reasoning".

That instruction means your solution must be a complete analytical method that a reader can follow from start to finish. You may still use your calculator to check, but a calculator value is not evidence. "Show that" is similar: the answer is given, so all the marks are for the steps that reach it.

To add or subtract two fractions with surd denominators, use the product of the two denominators as the common denominator. When the denominators are conjugates, that product is already rational, so there is nothing left to rationalise.

Here is that idea in a detailed reasoning question.

The last example changes one feature: the answer is given, so you must reach it with every step shown.

This settles the puzzle from the start of the lesson. Because $\frac1\varphi$ is exactly $\varphi-1$, the two numbers differ by exactly 1, so they must share every digit after the decimal point. It was never a coincidence: rationalising the denominator proves it.

The way you set out a "show that" matters as much as the algebra.

Now try one like Worked example 8. The number $1+\sqrt2$ is sometimes called the silver ratio.

Quick Recap

• A surd is an exact irrational root. Keep it as a root, not a rounded decimal, whenever an exact answer is needed.
• Simplify with $\sqrt{ab}=\sqrt a\times\sqrt b$ using the largest square factor, and collect only like surds.
• Expand surd brackets like algebra: $\sqrt a\times\sqrt a=a$, and a squared bracket has a middle term.
• Rationalise $\frac{1}{\sqrt a}$ by multiplying top and bottom by $\sqrt a$, and $\frac{1}{a+\sqrt b}$ by the conjugate $a-\sqrt b$, because $(a+\sqrt b)(a-\sqrt b)=a^2-b$.
• "Exact", "Determine", "Show that" and "detailed reasoning" all mean the steps must be written down.

The opening puzzle: $\frac1\varphi=\frac{\sqrt5-1}{2}=\varphi-1$, so $\varphi$ and $\frac1\varphi$ differ by exactly 1 and share every digit after the decimal point.