A rational number is a number which can be written as a fraction where both the numerator and denominator are integers. An irrational number is a number which cannot be written as a fraction of two integers.
We can write any terminating or recurring decimal as a fraction, therefore these are rational numbers. However, decimals which are neither terminating nor recurring are irrational numbers.
Use a calculator to try and work out if the following numbers are rational or irrational. If they're rational, express them in fractional form.
Are the following numbers rational or irrational? $\sqrt{6}$√6, $0.0047$−0.0047, $\sqrt[3]{27}$−^{3}√27 and $\frac{9}{2}$92
Think We want to identify whether these numbers can be expressed as fractions, terminating decimals, recurring decimals, integers, or surds.
Do:
$\sqrt{6}=2.44948974$√6=2.44948974$\ldots$…
This does not simplify to a rational number, so it is irrational.
$0.0047$−0.0047  $=$=  $\frac{0.0047}{1}$−0.00471 
$=$=  $\frac{47}{10000}$−4710000 
This is a fraction so it is rational.
$\sqrt[3]{27}=3$−^{3}√27=−3
This simplifies down to an integer so it is rational.
The square root function reverses the squaring function. Similarly, the cube root function undoes the cubing function.
In algebraic notation, we can write this as $\sqrt{A^2}=A$√A2=A and $\sqrt[3]{B^3}=B$^{3}√B3=B. It is also true that $\left(\sqrt{A}\right)^2=A$(√A)2=A and $\left(\sqrt[3]{B}\right)^3=B$(^{3}√B)3=B.
Square root and cube root expressions can sometimes be written in simpler forms using these facts, together with the other familiar surd rules.
We can use these ideas to simplify expressions like $\sqrt{16}$√16 and $\sqrt{121}$√121, where the numbers in the surd are perfect squares, or expressions like $\sqrt[3]{27}$^{3}√27 and $\sqrt[3]{125}$^{3}√125, where the numbers in the surd are perfect cubes.
If we have a number like $\sqrt{7}$√7 or $\sqrt{29}$√29, we cannot simplify this any more than it already is, as the number in the surd is a prime number, but what if the number in the surd contains a factor that is a perfect square or perfect cube?
For example, an expression like $\sqrt{32}=\sqrt{16\times2}$√32=√16×2. Can we still simplify these types of expressions?
One method to check whether an expression can be simplified is by looking at its prime factors. For example, given the expression $\sqrt{18}$√18, we first look at the prime factors of $18$18 which gives us $3\times3\times2=3^2\times2$3×3×2=32×2.
Now we can write $\sqrt{18}$√18 as $\sqrt{3^2\times2}=\sqrt{3^2}\times\sqrt{2}$√32×2=√32×√2. In this step we have used the important fact that $\sqrt{A\times B}=\sqrt{A}\times\sqrt{B}$√A×B=√A×√B.
The first term in the product $\sqrt{3^2}\times\sqrt{2}$√32×√2 is of the form $\sqrt{A^2}=A$√A2=A, so the fully simplified expression becomes $3\sqrt{2}$3√2.
From this example we can see that if any factor appears two times within a square root, or three times within a cube root, then the expression can be further simplified. This is equivalent to looking for a factor that is a perfect square for square root expressions, or a perfect cube for cube root expressions.
Any positive real number has two square roots, one positive and one negative, but the square root function $\sqrt{x}$√x only gives the positive square root.
This mean that the square root function and the square function are technically not inverses if we consider all real numbers. They are if we only consider the nonnegative numbers.
Simplify $\sqrt{8}$√8.
Think: If we can find any factors of $8$8 that are perfect squares, then we can simplify the expression using the fact that $\sqrt{a^2b}=a\sqrt{b}$√a2b=a√b.
Do: The factors of $8$8 are $1$1, $2$2, $4$4, and $8$8. Let's use the perfect square $4$4 to rewrite the expression and simplify.
$\sqrt{8}$√8  $=$=  $\sqrt{4\times2}$√4×2 
Replace $8$8 with two factors 
$=$=  $\sqrt{4}\times\sqrt{2}$√4×√2 
Use the fact that $\sqrt{A\times B}=\sqrt{A}\times\sqrt{B}$√A×B=√A×√B 

$=$=  $\sqrt{2^2}\times\sqrt{2}$√22×√2 
Rewrite $4$4 as $2^2$22 

$=$=  $2\sqrt{2}$2√2 
Use the fact that $\sqrt{A^2}=A$√A2=A 
Simplify $\sqrt{392}$√392.
Think: Looking at the prime factors gives us $392=2^3\times7^2$392=23×72. Notice that we can write $2^3$23 as $2\times2^2$2×22.
Do: After we substitute $392$392 with its prime factors we can use the rules of manipulation of surds to simplify the expression.
$\sqrt{392}$√392  $=$=  $\sqrt{2\times2^2\times7^2}$√2×22×72 
$=$=  $2\times7\times\sqrt{2}$2×7×√2  
$=$=  $14\sqrt{2}$14√2 
Here are some key identities that we will find useful to simplify expressions involving surds.
$\sqrt{a^2}=a$√a2=a
$\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}$√a×b=√a×√b
And from these two we can see that:
$\sqrt{a^2b}=a\sqrt{b}$√a2b=a√b.
Is $\sqrt[3]{47}$^{3}√47 rational or irrational?
Rational
Irrational
Rational
Irrational
Simplify $\sqrt{180}$√180.
Simplify $\frac{1}{2}\sqrt[3]{8\times6}$12^{3}√8×6.
Perform simple operations with surds.