# Binary Numbers

Binary numbers are representations of numbers using bits--0s and 1s--instead of decimal digits. They are used extensively in mathematics and especially computer science, as bits are easy to create physically using logic gates.

#### Contents

## Definition

A binary number is a number expressed in the binary numeral system (also known as base-2 numeral system), which represents numbers using two digits: 0 and 1.

In contrast to the standard base-10 system, which represents numbers using powers of 10, the place values in binary correspond to powers of 2. Thus the first place (the place just before the decimal) represents \(2^0,\) the second place indicates \(2^1,\) the third \(2^2,\) and so forth. The digits on the right side of the decimal all have a denominator which is a power of 2. So, the first digit after the decimal represents \(2^{-1},\) the second digit \(2^{-2},\) and so on.

## Examples

## What is the representation of \( 1011_2\) using the base-10 system?

Starting from the right, the places each represent \(2^0, 2^1, 2^2,\) and \(2^3\). So \(1011_2= 1\times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 8 + 2 + 1 = 11 \).

## What is the representation of 16 in binary?

Since \( 16 = 2^4 \), we need a single 1 in the place that represents \( 2^4 \).Thus \( 16 = 10000_2 \).

## What is the representation of \(1110_2 + 1001_2\) using the base-10 system?

\(1110_2\) is equal to

\[1\times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 = 8 + 4 + 2 = 14,\]

and \(1001_2\) is equal to

\[1\times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 8 + 1= 9.\]

Thus, the answer is \(14 + 9 = 23\).

Alternatively, it is possible to add the numbers without converting to binary first, using the same arithmetic as for decimal numbers: remembering that \( 1_2+1_2 = 10_2\), the sum is \( 10111_2 = 23_{10}\).

## If 30 is represented in binary, what would the sum of the digits be?

30 is equal to

\[16 + 8 + 4 + 2 = 1\times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 = 11110_2 . \]

Thus, the sum of the digits is \(1+1+1+1+0 = 4.\) \( _\square \)

What is \( \frac13 \) converted to binary?

Solution:Since \( \frac43 =\frac13+1 \), if \[ \frac13 = (0.c_1c_2c_3\ldots)_2, \] then \[ \frac43 = (1.c_1c_2c_3\ldots)_2. \] But \( \frac43 = 4 \cdot \frac13 \), and multiplying by \( 4 = 100_2 \) shifts the binary expansion of a number by two units to the left. So \( 0 = c_1, 1 = c_2, c_1 = c_3, c_2= c_4, \) and so on. This shows that \[ \frac13 = (0.0101\ldots)_2 = (0.{\overline{01}})_2. \] (This corresponds to the geometric series \[ \frac13 = \frac14+\frac1{16}+\frac1{64}+\cdots.) \]