# IP to Binary

### What is an Octet?

Octet is a grouping of eight binary bits, same as a byte. But the word octet is used when in networking so not to confuse with storage (byte). An IPv4 address is made up with four octets separated with a dot. In total an IP address comprises 32 binary bits. In binary the range of a octet starts at 00000000 up to 11111111. In Decimal a octet can a value from 0 up to 255.

### How to convert a IP to a binary?

There may be other methods out there that you could use to convert an IP address to binary. But there are two tried and tested methods: division and table method. In this article we will discuss how to convert an IP address using the division method. Which one you chose depends on you but they will have the same outcome if the methodology is applied correctly.

#### Converting IP to Binary Using Division

To convert an IP address of 193.168.67.33 using division you simply continually divide the octet by two until the quotation (answer of the division) reaches zero. During each division iteration the remainder should be noted. An IPv4 address consists of four octets, each separated with a dot between them. These octets will be converted separately to one another. Start with the first octet which is 193 and divide it by 2 which results in 193/2 = 96 with a remainder of 1. So the next step would be noting the remainder which is 1 and then taking the quotation from the first step which is 96 then divide that by two. Continue to repeat each step again until the quotation is zero, Then reverse the order list of eight remainders that was noting during the division to reverse the binary equivalent of the integer. See the steps below and compare your answers:

1. 193 / 2 = 96 remainder 1
2. 96 / 2 = 48 remainder 0
3. 48 / 2 = 24 remainder 0
4. 24 / 2 = 12 remainder 0
5. 12 / 2 = 6 remainder 0
6. 6 / 2 = 3 remainder 0
7. 3 / 2 = 1 remainder 1
8. 1 / 2 = 0 remainder 1

When you reverse the order of the remainders you get 11000001, so 193 is 11000001 in binary. Repeat the same steps for the next octet which is 168 and repeat for octet 67 and 33. In the end your answer should exactly match “11000001.10101000.01000011.00100001”

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