Number System - Binary, Octal, Hex

The string $101$ means five in binary, sixty-five in octal, and one hundred one in decimal — same three symbols, three different values. Once you see that the base controls everything, binary, octal, and hexadecimal stop looking mysterious.

When you use this conversion method

These place-value systems differ only in base: binary is base $2$, octal is base $8$, hexadecimal is base $16$. You reach for them whenever you need to read or convert between bases — binary is the language of digital systems (switches have two states), while octal and hexadecimal are compact ways to write long binary strings. The math idea stands on its own: in any positional system each place is a power of the base, so the value depends on the base and the place.

The procedure, step by step

1. Identify the base. Check how many digit symbols the system allows and what each place means:

• Binary uses only $0$ and $1$.
• Octal uses $0$ through $7$.
• Hexadecimal uses $16$ symbols: $0$ through $9$, then $A$ through $F$ for values $10$ through $15$.

One hex place counts in powers of $16$, so a hex digit holds more information than a binary digit.

2. Read place values. In base $b$, each position is a power of $b$: $b^0, b^1, b^2, \ldots$. In base $10$ the places are $1, 10, 100, \ldots$; in base $2$ they are $1, 2, 4, 8, 16, \ldots$.

3. Expand or regroup. Convert by writing the value as a sum of powers of the base, or by grouping binary digits — sets of $3$ for octal, sets of $4$ for hexadecimal.

4. Verify the value. Confirm that all forms represent the same base-$10$ quantity. A number does not change value when you rewrite its base; only the representation changes. Different bases are like different languages for the same amount.

Full worked example: write $45$ in binary, octal, and hex

Start in base $10$ and expand into powers of $2$:

So binary has $1$s in the $2^5$, $2^3$, $2^2$, and $2^0$ places:

Regroup for octal. Since $8 = 2^3$, group binary digits in threes from the right:

Regroup for hexadecimal. Since $16 = 2^4$, group in fours from the right, adding leading zeros:

Verify — all four forms are the same quantity:

Where the steps break down, and how to check

• Step 1, using a digit the base does not allow. $2$ cannot appear in binary, and $8$ cannot appear in octal. Scan your digits against the allowed set first.
• Step 2, forgetting the base sets the place values. $101$ does not mean the same thing in base $2$, base $8$, and base $10$.
• Step 3, grouping binary digits incorrectly. Always group from the right, and add leading zeros to complete a group.

Self-check the whole conversion by expanding each form back to base $10$ — they must all match (here, $45$).

Try a conversion

Convert $58_{10}$ into binary, octal, and hexadecimal. First write it as a sum of powers of $2$, then group the binary digits into threes and fours for the other two forms — and expand back to base $10$ to confirm you landed on $58$ every time.