E1.17 - Rationalising denominators (Extended)

E1.17 - Rationalising denominators (Extended)

Can the exact number 13−1\frac{1}{\sqrt3-1} be written without a square root underneath the fraction? E1.17 develops a way to change a fraction's appearance while keeping its value. You will rationalise single-term and two-term denominators, including a negative rational term.

Single surd denominator

The opening fraction has an awkward denominator, but begin with a simpler one: 63\frac{6}{\sqrt3}. You already know that 3×3=3\sqrt3\times\sqrt3=3. Multiplying the denominator by another 3\sqrt3 would therefore make it rational.

Rationalising the denominator

To rationalise a denominator is to write an equivalent fraction with a rational denominator. The numerator may still contain a surd. The original denominator must be non-zero.

Multiplying only the denominator would change the value. Multiply the numerator by the same non-zero number: 33=1\frac{\sqrt3}{\sqrt3}=1, so this changes the form without changing the number.

Worked example 1: one surd underneath

Question. Rationalise the denominator of 63\frac{6}{\sqrt3}.

Plan. Multiply top and bottom by 3\sqrt3, then simplify the fraction.

63=633×3multiply by 1=633square the root=23divide the coefficient by 3\begin{aligned} \frac{6}{\sqrt3}&=\frac{6\sqrt3}{\sqrt3\times\sqrt3} &&\text{multiply by 1}\\ &=\frac{6\sqrt3}{3} &&\text{square the root}\\ &=2\sqrt3 &&\text{divide the coefficient by 3} \end{aligned}

Answer. 232\sqrt3.

Check. Multiply the answer by the original denominator: 23×3=62\sqrt3\times\sqrt3=6, the original numerator.

The next denominator has a coefficient as well as a root. Only the root needs a partner; the rational coefficient can stay.

Worked example 2: a coefficient in the denominator

Question. Rationalise the denominator of 623\frac{6}{2\sqrt3}.

Plan. Multiply both parts of the fraction by 3\sqrt3, then cancel the common numerical factor.

623=6323×3same multiplier top and bottom=636denominator is 2×3=3simplify fully\begin{aligned} \frac{6}{2\sqrt3}&=\frac{6\sqrt3}{2\sqrt3\times\sqrt3} &&\text{same multiplier top and bottom}\\ &=\frac{6\sqrt3}{6} &&\text{denominator is }2\times3\\ &=\sqrt3 &&\text{simplify fully} \end{aligned}

Answer. 3\sqrt3.

Check. 3×23=2×3=6\sqrt3\times2\sqrt3=2\times3=6, recovering the numerator. Doubling the first example's denominator has halved its answer.

The two checks use multiplication rather than repeating the rationalising steps.

Watch out

63≠63\frac{6}{\sqrt3}\ne\frac{6}{3}. Changing just the denominator changes the number. The correct multiplication gives 633\frac{6\sqrt3}{3}, including the root in the numerator.

Try the same method with a different coefficient and root.

Conjugate products

Multiplying 2+32+\sqrt3 by 3\sqrt3 gives 23+32\sqrt3+3, which still contains a surd. A two-term denominator needs a different partner: change the sign between its terms.

Conjugate expressions

For the real two-term surd expressions used here, a conjugate pair has the same terms with the joining sign changed. For example, 4+54+\sqrt5 and 4−54-\sqrt5 are conjugates.

When multiplying two sums, multiply every term of the first by every term of the second. In a conjugate product, the two middle products cancel.

Worked example 3: the cancellation

Question. Work out (4+5)(4−5)(4+\sqrt5)(4-\sqrt5).

Plan. Write all four products, then combine the opposite surd terms.

(4+5)(4−5)=16−45+45−5four products=16−5middle terms cancel=11\begin{aligned} (4+\sqrt5)(4-\sqrt5)&=16-4\sqrt5+4\sqrt5-5 &&\text{four products}\\ &=16-5 &&\text{middle terms cancel}\\ &=11 \end{aligned}

Answer. 1111.

Check. (4+5)(4−5)(4+\sqrt5)(4-\sqrt5) is about 6.236×1.7646.236\times1.764, which is about 11. Both factors are positive, so the positive answer makes sense.

Now change only the coefficient of the surd. That coefficient is squared along with the root.

Worked example 4: the whole term is squared

Question. Work out (4+25)(4−25)(4+2\sqrt5)(4-2\sqrt5).

Plan. Multiply the four pairs and cancel the middle terms.

(4+25)(4−25)=16−85+85−20=16−20since (25)2=4×5=−4\begin{aligned} (4+2\sqrt5)(4-2\sqrt5)&=16-8\sqrt5+8\sqrt5-20\\ &=16-20 &&\text{since }(2\sqrt5)^2=4\times5\\ &=-4 \end{aligned}

Answer. −4-4.

Check. 25>42\sqrt5>4, so one factor is positive and the other is negative. A negative product is required; approximately 8.472×(−0.472)=−4.008.472\times(-0.472)=-4.00.

The same cancellation works for general real numbers.

Difference of squares

(A+B)(A−B)=A2−B2(A+B)(A-B)=A^2-B^2

The middle products −AB-AB and +AB+AB cancel. For a rationalising multiplier, the conjugate must also be non-zero.

Not on the supplied formula lists: learn it.

Square the entire surd term when you use this result.

Watch out

(25)2=20(2\sqrt5)^2=20, not 10. Both copies of 2 are multiplied, as well as both copies of 5\sqrt5.

Use the cancellation to get an exact answer.

Two-term denominators

A conjugate removes the surd from a two-term denominator because its product is a difference of squares. Keep the denominator's exact signs until that product has been worked out.

For instance, the conjugate of 3−1\sqrt3-1 is 3+1\sqrt3+1. Writing the same denominator as −1+3-1+\sqrt3 does not change its value; 1+31+\sqrt3 is still a useful partner, giving (3−1)(3+1)=2(\sqrt3-1)(\sqrt3+1)=2.

Worked example 5: a positive rational term

Question. Rationalise the denominator of 11+3\frac1{1+\sqrt3}.

Plan. Multiply top and bottom by 1−31-\sqrt3, then remove a negative denominator.

11+3=1−3(1+3)(1−3)=1−31−3difference of squares=1−3−2=3−12change both numerator signs\begin{aligned} \frac1{1+\sqrt3}&=\frac{1-\sqrt3}{(1+\sqrt3)(1-\sqrt3)}\\ &=\frac{1-\sqrt3}{1-3} &&\text{difference of squares}\\ &=\frac{1-\sqrt3}{-2}\\ &=\frac{\sqrt3-1}{2} &&\text{change both numerator signs} \end{aligned}

Answer. 3−12\frac{\sqrt3-1}{2}.

Check. Multiplying by the original denominator gives (3−1)(1+3)2=3−12=1\frac{(\sqrt3-1)(1+\sqrt3)}2=\frac{3-1}2=1.

The next example changes only the sign of the rational term. Reordering the addition makes the conjugate easy to see.

Worked example 6: a negative rational term

Question. Rationalise the denominator of 1−1+3\frac1{-1+\sqrt3}.

Plan. Rewrite the denominator as 3−1\sqrt3-1 and multiply top and bottom by 3+1\sqrt3+1.

1−1+3=13−1same terms, reordered=3+1(3−1)(3+1)=3+13−1middle products cancel=1+32\begin{aligned} \frac1{-1+\sqrt3}&=\frac1{\sqrt3-1} &&\text{same terms, reordered}\\ &=\frac{\sqrt3+1}{(\sqrt3-1)(\sqrt3+1)}\\ &=\frac{\sqrt3+1}{3-1} &&\text{middle products cancel}\\ &=\frac{1+\sqrt3}{2} \end{aligned}

Answer. 1+32\frac{1+\sqrt3}{2}.

Check. 1+32(3−1)=3−12=1\frac{1+\sqrt3}{2}(\sqrt3-1)=\frac{3-1}{2}=1. Also the denominator 3−1\sqrt3-1 is between 0 and 1, so its reciprocal must be greater than 1; the answer is about 1.366.

This resolves the opening question with an exact expression. No decimal approximation is needed in the answer.

Watch out

A conjugate changes the sign between the terms, not the sign of the whole expression. The negative of 3−1\sqrt3-1 is 1−31-\sqrt3; multiplying by it does not produce the cancellation you want.

Choose the partner before expanding anything.

Quick Recap

  • Multiply the numerator and denominator by the same non-zero expression.
  • A single root becomes rational when multiplied by itself.
  • Conjugates give a difference of squares; square coefficients as well as roots.
  • Keep exact values and simplify any common factors.

The opening number is 13−1=1+32\frac1{\sqrt3-1}=\frac{1+\sqrt3}{2}.