Define rationalising the denominator .

Rationalising the denominator changes a fraction with surds in its denominator into an equivalent fraction whose denominator is rational, with any surds moved to the numerator.

True or False? Rationalising the denominator changes the value of a fraction.

False. You multiply the numerator and denominator by the same quantity, which is the same as multiplying by 1, so you get an equivalent fraction .

A LevelMathsEdexcelPureFlashcardsAlgebra & FunctionsLaws of Indices & Surds

Laws of Indices & Surds (Edexcel A Level Maths: Pure): Flashcards

Exam code: 9MA0

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FrontLaws of Indices
Define index.

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Cards in this collection (23)

Define index.
In $a^{n}$ , the index is $n$ , the number that the base $a$ is raised to.
It is also called the power or the exponent.
How do you simplify $a^{m} \times a^{n}$ and $a^{m} \div a^{n}$ ?
Multiplying adds the indices: $a^{m} \times a^{n} = a^{m + n}$
Dividing subtracts them: $a^{m} \div a^{n} = a^{m - n}$
Complete the index law by filling in the missing index: ${(a^{m})}^{n} = a^{______}$
${(a^{m})}^{n} = a^{m n}$
Raising a power to another power multiplies the indices.
True or False?
$a^{0} = 0$
False.
$a^{0} = 1$ for any non-zero $a$ . This follows from $a^{m} \div a^{m} = a^{m - m} = a^{0}$ , and any number divided by itself is 1.
What does a negative index mean?
It means the reciprocal of the positive power: $a^{- m} = \frac{1}{a^{m}}$
In $a^{\frac{m}{n}}$ , the denominator $n$ tells you which $______$ to take.
In $a^{\frac{m}{n}}$ , the denominator $n$ tells you which root to take, and the numerator $m$ is the power: $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$
How can you simplify an expression involving powers whose terms have different bases?
Rewrite the terms so that they share a base, then use the index laws.
For example $9^{4} = {(3^{2})}^{4} = 3^{8}$ , so $9^{4} \div 3^{7} = 3^{8 - 7} = 3$
How do you apply a power to a product, such as ${(\frac{1}{27} x^{3})}^{\frac{1}{3}}$ ?
Apply the power to each factor separately, using ${(a b)}^{n} = a^{n} b^{n}$
Here that gives ${(\frac{1}{27})}^{\frac{1}{3}} {(x^{3})}^{\frac{1}{3}} = \frac{1}{3} x$
Define surd.
If $n$ is a positive integer that is not a square number, then $\sqrt{n}$ is a surd. Any rational multiple, such as $5 \sqrt{2}$ , is also a surd.
Surds are irrational numbers.
Why is $\sqrt{16}$ not a surd, when $\sqrt{13}$ is?
$16$ is a square number, so $\sqrt{16} = 4$ , which is rational.
$13$ is not a square number, so $\sqrt{13}$ is irrational, and therefore a surd.
Simplify by looking for a square factor: $\sqrt{720} = \sqrt{______\times 5} =______\sqrt{5}$
$\sqrt{720} = \sqrt{144 \times 5} = \sqrt{144} \times \sqrt{5} = 12 \sqrt{5}$
Look for a factor that is a square number, so that its root is a whole number.
What are the two rules for multiplying and dividing surds?
$\sqrt{a b} = \sqrt{a} \times \sqrt{b}$
$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$
What is the reciprocal of $\sqrt{\frac{a}{b}}$ ?
The fraction under the root flips:
$\frac{1}{\sqrt{\frac{a}{b}}} = \frac{\sqrt{b}}{\sqrt{a}} = \sqrt{\frac{b}{a}}$
True or False?
$\sqrt{a} + \sqrt{b} = \sqrt{a + b}$
False.
You cannot add or subtract under the surd. The multiplication and division rules work, but there is no matching rule for addition or subtraction.
How do you simplify $5 \sqrt{2} - 2 \sqrt{5} - 8 \sqrt{2} + 3 \sqrt{5}$ ?
Collect like terms, treating $\sqrt{2}$ and $\sqrt{5}$ like different letters in algebra:
$\begin{array}{rcl} 5 \sqrt{2} - 2 \sqrt{5} - 8 \sqrt{2} + 3 \sqrt{5} & = & (5 - 8) \sqrt{2} + (- 2 + 3) \sqrt{5} \\ = & - 3 \sqrt{2} + \sqrt{5} \end{array}$
Leaving an answer as $5 \sqrt{2}$ rather than $7.071067812$ keeps it in $______$ form.
Leaving an answer as $5 \sqrt{2}$ rather than $7.071067812$ keeps it in exact form.
This matters when the value has to be carried through into later working.
Define rationalising the denominator.
Rationalising the denominator changes a fraction with surds in its denominator into an equivalent fraction whose denominator is rational, with any surds moved to the numerator.
How do you rationalise a denominator of the form $\sqrt{a}$ ?
Multiply the numerator and the denominator by $\sqrt{a}$ , since $\sqrt{a} \times \sqrt{a} = a$
True or False?
Rationalising the denominator changes the value of a fraction.
False.
You multiply the numerator and denominator by the same quantity, which is the same as multiplying by 1, so you get an equivalent fraction.
How do you rationalise a denominator of the form $\sqrt{a} + \sqrt{b}$ ?
Multiply the numerator and denominator by $\sqrt{a} - \sqrt{b}$ , changing the sign between the terms.
For a denominator $\sqrt{a} - \sqrt{b}$ , multiply by $\sqrt{a} + \sqrt{b}$ instead.
To rationalise $\frac{1}{5 - \sqrt{3}}$ , multiply the numerator and denominator by $______$ .
To rationalise $\frac{1}{5 - \sqrt{3}}$ , multiply the numerator and denominator by $5 + \sqrt{3}$ .
Either term can be an integer rather than a surd; just change the sign between them.
Why does multiplying by the conjugate of a denominator such as $5 - \sqrt{3}$ rationalise it?
It creates a difference of squares, and squaring removes the surds:
$(5 - \sqrt{3}) (5 + \sqrt{3}) = 25 - 3 = 22$
True or False?
If an answer must be in the form $p + q \sqrt{3}$ where $p$ and $q$ are rational, then $p$ and $q$ must be integers.
False.
Rational numbers include fractions and negatives, so $\frac{1}{2} - \frac{1}{2} \sqrt{3}$ is a valid answer in that form.

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