We use numbers all the time but rarely stop to think how they work. In this section we will consider decimal, binary and hexadecimal numbers, we will look at exponential numbers to see how the exponents work under different operations and we will look at scientific and engineering notation (also known as standard form and preferred standard form).

You've grown up with the decimal system. It is so widely used that most people do not realise that it is just one of an infinite number of number systems. The decimal system has ten symbols $0$ to $9$ (dec means ten in Latin). You can add $1$ to $0$ to get $1$, add $1$ to $1$ to get $2$ and so on. If you want to add $1$ to $9$ you have a problem because you have no more symbols left. What do you do?

A solution for this problem was found by Indian mathmaticians in about 650AD when they rediscovered a system used by the ancient Sumerians around 2500BC (more details here). They let the position of a character, as well as the actual character, be significant. So, to add $1$ to $9$ you set the $9$ back to $0$ and carry $1$ into the next column. That may seem obvious to you now, four and a half thousand years later, but it is so clever and it was ground breaking at the time.

Let’s see how that works in binary. In binary you only have two symbols, $0$ and $1$ (bi means two in Latin). Add $1$ to $0$ you get $1$. To add $1$ to $1$ you have the same problem as above. So, you set the first $1$ back to $0$ and carry $1$ into the next column. The result looks like $10$ but it has the value of $2$ in decimal.

How does it work in hexadecimal (hex means six in Latin so hexadecimal means six + ten = sixteen)? In hexadecimal, or hex for short, there are sixteen symbols, $0$ to $9$ followed by $A$ to $F$. $F$ in hex is equal to $15$ in decimal. To add $1$ to $F$ you set the $F$ back to $0$ and carry $1$ to the next column. The result looks like $10$ but this time it has a value of $16$ in decimal.

Subscripts are used to specify the base of a number. $10$, $10_d$ and $10_{10}$ all mean $10$ in decimal. If there is no subscript then you can assume the number is base $10$.

$10_b$ and $10_2$ mean $10$ in binary which is equal to $2$ in decimal.

$10_h$ and $10_{16}$ mean $10$ in hexadecimal which is equal to $16$ in decimal.

In most computers there are 8 binary digits, called bits, in a byte. Half a byte is called a nybble and a 4 bit nybble can have $2^4$ values from $0000$ to $1111$ (in binary) which is $0$ to $15$ (in decimal) or $0$ to $F$ (in hex).

Here is a table of the values:

Decimal | Binary | Hexadecimal | Decimal | Binary | Hexadecimal |
---|---|---|---|---|---|

0 | 0000 | 0 | 8 | 1000 | 8 |

1 | 0001 | 1 | 9 | 1001 | 9 |

2 | 0010 | 2 | 10 | 1010 | A |

3 | 0011 | 3 | 11 | 1011 | B |

4 | 0100 | 4 | 12 | 1100 | C |

5 | 0101 | 5 | 13 | 1101 | D |

6 | 0110 | 6 | 14 | 1110 | E |

7 | 0111 | 7 | 15 | 1111 | F |

A two digit hexadecimal number ranges from $0_h$ to $FF_h$ which is $0$ to $1111\ 1111_b$ in binary and $0$ to $255$ in decimal. In a two digit hexadecimal number the right digit represents units and the left digit represents sixteens.

Converting between binary and hexadecimal is easy, particularly if you treat the binary number in nybbles.

Example 1.1.1

Convert $1010\ 1100$ to hex

In decimal the left digit is $8 + 2 = 10$ which is $A$ in hex.

The right digit is $8 + 4 = 12$ which is $C$ in hex.

So $1010\ 1100_b = AC_h$.

Converting from hex into binary is also easy.

Example 1.1.2

Convert $D6$ to binary

$D_h = 13$ in decimal which is $8 + 4 + 1 = 1101_b$.

$6_h = 6$ in decimal which is $4 + 2 = 0110_b$.

So $D6_h = 1101\ 0110_b$.

The easiest way to convert from decimal into binary is to divide by sixteen (divide by 2 four times) and find the remainder. You can then convert the 16s and the units.

Example 1.1.3

Convert $197$ to binary and hex

Divide by 2 four time ignoring the remainders

$197 \div 2 = 98, 98 \div 2 = 49, 49 \div 2 = 24$ and $24 \div 2 = 12$.

So there are $12$ sixteens in $197$. $12 \times 16 = 192$ so $197 = 12 \times 16 + 5$. That's the hard bit done.

$12 = 8 + 4 = 1100_b = C_h$.

$5 = 4 + 1 = 0101_b = 5_h$.

So $197 = 1100\ 0101_b = C5_h$

and not even a whiff of a calculator.

If you multiply a number by itself you get an exponential number. For example $2 \times 2 = 2^2$. The big $2$ is called the base and the little $2$ is called the exponent. If you multiply $2 \times 2 \times 2$ you get $2^3$. $2$ is the base and $3$ is the exponent.

Each time you multiply a number by itself you increase the exponent by 1 so $2 \times 2 = 2^2$, $2 \times 2 \times 2 = 2^3$ and so on.

$2 = 2^1$ | $2 \times 2 = 2^2$ | $2 \times 2 \times 2 = 2^3$ | $2 \times 2 \times 2 \times 2 = 2^4$ |

$2$ | $4$ | $8$ | $16$ |

Each time we increment the exponent we multiply by the base. What happens if we decrement the exponent?

$2 \times 2 \times 2 = 2^3$ | $2 \times 2 = 2^2$ | $2 = 2^1$ | $2^0 = 1$ |

$8$ | $4$ | $2$ | $1$ |

You can see we are dividing by the base each time we decrement the exponent so it comes as no surprise that $2^0 = 1$. You can probably see that anything raised to the power of $0$ is equal to $1$ except for $0^0$.

$0^0$ is undefined.

What happens if we keep decrementing the exponent?

$2^1$ | $2^0$ | $2^{-1}$ | $2^{-2}$ |

$2$ | $1$ | $1/2$ | $1/4$ |

Which means $2^{-2} = 1/2^2$ or, more generally:

$x^{-a} = 1/x^a $

What is $3^2 \times 3^3$? Well, $3^2 = 9$ and $3^3 = 27$. Now $9 \times 27 = 243$ which is equal to $3^5$ so $3^2 \times 3^3 = 3^5$. To multiply exponential numbers we simply add the exponents. More generally we can write

$x^a \times x^b = x^{a+b}$

Pssssttt - This is how logarithms and slide rules work.

What is $243 \div 9$?

From above we know $243 = 3^5$ and $9 = 3^2$. $243 \div 9 = 27$ and $27 = 3^3$

Writing this as exponential numbers $3^5 \div 3^2 = 3^3$. To divide number $A$ by number $B$ we simply subtract number $B$'s exponent from number $A$'s exponent. More generally we can write

$x^a \div x^b = x^{a-b}$

We know that $x^a \times x^b = x^{a+b}$. What happens in the case of $(x^a)^b$?

First, let’s try this with some easy numbers. Try $(2^3)^2 = (8)^2 = 64 = 2^6$.

As $6 = 3 \times 2$ we can deduce

$(x^a)^b = x^{a \times b}$

$2^3$ is one of the square roots of $2^6$. We could write $2^3 = \sqrt{2^6}$. Using our previous rule we could also write $2^3 = (2^6)^{1/2}$. So $(2^6)^{1/2} = \sqrt{2^6}$ or more generally:

$x^{1/n} = \sqrt[n]{x}$

Look at this table:

$1$ | $10$ | $100$ | $1000$ | $10000$ |

$10^0$ | $10^1$ | $10^2$ | $10^3$ | $10^4$ |

Now look at this table:

$log(1)$ | $log(10)$ | $log(100)$ | $log(1000)$ | $log(10000)$ |

$0$ | $1$ | $2$ | $3$ | $4$ |

Logarithms are exponents. Taking the logarithm of a number is the inverse of raising the number to a power.

$log(1000)=log(10^3)=3$

All the logarithms above have the same base, base 10, so we could have written $log_{10}(1000)=3$ where $1000$ is the argument, $10$ is the base and $3$ is the exponent. Logarithms can have any positive number for the base but the two most common bases are $10$ and Euler's constant $e$. Logarithms to base $10$ are written $log_{10}()$ or, more normally, $log()$. Logarithms to base $e$are written $log_{e}()$ or, more normally, $ln()$.

Look at your calculator. If it has logs it will probably have functions for $10^x$ and $log(x)$ as well as $e^x$ and $ln(x)$.

We know we can write $100$ as $10^2$ and $1000$ as $10^3$. What about the numbers in between? Well, using a calculator or log tables we can see $125=10^{2.0969}$, $150=10^{2.1761}$ and $175=10^{2.2430}$.

If we multiply $100 \times 1000$ we get $100000$ which can be written as $10^5$. If we multiply $125 \times 100$ we get $12500$ which can be written as $10^{4.0969}$.

In the same way we can if we divide $100000$ by $100$ we get $1000$ so $10^5 \div 10^2 = 10^3$. If we divide $125$ by $100$ we get $1.25$ so $10^{2.0969} \div 10^2 = 10^{0.0969}$.

Working with the exponents ($2$, $3$ and $2.0969$) rather than the numbers ($100$, $1000$ and $125$) means we can find products by adding the exponents and we can find quotients by subtracting the exponents.

We know $100=10^2$ so it follows that $log(100)=2$ where $log()$ means logarithm to the base 10.

The $log()$ of a number is the power to which the base, $10$ in this case, needs to be raised to equal the number.

The $log()$ of a number is the power to which the base, $10$ in this case, needs to be raised to equal the number.

Example 1.3.1

Simplify $3log(2x) - 2log(3x)$

$3log(2x)$ | $\ =\ $ | $log(2x)+log(2x)+log(2x)$ |

$\ =\ $ | $log((2x) \times(2x) \times(2x))$ | |

$\ =\ $ | $log(8x^3)$ | |

and | ||

$2log(3x)$ | $\ =\ $ | $log(3x)+log(3x)$ |

$\ =\ $ | $log((3x) \times(3x))$ | |

$\ =\ $ | $log(9x^2)$ | |

so |

$3log(2x) - 2log(3x)$ | $\ =\ $ | $log(8x^3)-log(9x^2)$ |

$\ =\ $ | $log(8x^3/9x^2)$ | |

$\ =\ $ | $log(8x/9)$ |

Example 1.3.2

Simplify $ln(y)+2ln(4x) - ln(8x^5) = 0$

Start by putting the $y$ terms on the left and the $x$ terms on the right

$ln(y)$ | $=$ | $ln(8x^5)-2ln(4x)$ |

$=$ | $ln(8x^5)-ln(16x^2)$ | |

$=$ | $ln(8x^5/16x^2)$ | |

$=$ | $ln(x^3/2)$ | |

We have single logs to the same base on both sides of the equation so we can take inverse logs of both sides giving us: | ||

$y$ | $=$ | $x^3/2$ |

Avogadro's constant is quite a large number. It is $602214000000000000000000$ and represents the number of atoms or molecules in 1 mole of substance. It is not easy to see how many million millions there are.

Plank's constant, by comparison, is quite small. It is $0.000000000000000000000000000000000662607$. It is a a fundamental constant equal to the energy of a quantum of electromagnetic radiation divided by its frequency. How many million millionths are there?

Scientific notation, also known as **standard form**, was devised to help manage very large and very small numbers. In standard form we have a single digit before a decimal point, as many digits after the decimal point as we need and the number is multiplied by $10$ raised to an integer power. The number is called the **mantissa** or the **significand**.

In scientific notation Avogadro's constant would be written $6.02214 \times 10^{23}$. Plank's constant would be written $6.62607 \times 10^{−34}$.

Multiplying numbers in standard form is easy. You multiply the significands and add the exponents. For division you divide the significands and subtract the exponents. For example:

$(3 \times 10^6) \times (2.5 \times 10^3) = 7.5 \times 10^9$

$(7.4 \times 10^5) \div (3 \times 10^3) = 2.47 \times 10^2$

Imagine you have an object that is $500000$ mm long. You would probably not say it was $500000$ mm long, you are more likely to say it was $500$ m or maybe half a kilometre long.

Engineering notation, also known as **preferred standard form**, only uses powers that are factors of $3$.

Significant figures are the figures in a number that carry meaning There are four rules on how to determine how many significant figures there are in a number.

Rules on significant figures

- Non-zero digits are always significant.
- Any zeros between two significant digits are significant.
- Leading zeros are never significant.
- Trailing zeros are only significant in the decimal part of a number.

Let's look at some examples.

Rule 1 | $12.34$ has four non-zero digits so it has four significant figures. |

Rule 2 | $102.304$ has four non-zero digits and two zeros each between significant figures. That makes both the zeros significant so there are six significant figures. |

Rule 3 | $0.003$ has a leading zero before the decimal point and two leading zeros after. They are not significant. There is one non-zero digit so the number has one significant figure. |

Rule 4 | $600$ has one non-zero digit and two trailing zeros so it has one significant figure. |

Rule 4 | $6.2600$ has three non-zero digits and two trailing zeros after the decimal point so it has five significant figures. |