How Binary Shifts Multiply and Divide by Powers of 2

A left binary shift by $n$ places multiplies an unsigned binary number by $2^n$, provided the result fits in the available bits. A right binary shift by $n$ places divides it by $2^n$, discarding any remainder. This works because each place in a binary number is worth twice as much as the place to its right.

Why shifts work

The place values in binary, reading from right to left, are $1,2,4,8$, and so on. Moving a bit one place left doubles its value; moving it one place right halves its value. Shifting two places has the same effect twice, so it multiplies or divides by $2^2=4$. The number of shift places tells you which power of $2$ to use.

Worked example

Assume we are using an unsigned $8$-bit number, with zeros added to the empty places. Start with $00101101_2$, which is $45$ in decimal:

• Shift left one place: $00101101_2\to 01011010_2$. The result is $90$, or $45\times 2$.
• Shift right two places: $00101101_2\to 00001011_2$. The result is $11$, the whole-number part of $45÷4=11.25$.

On the right shift, bits that move beyond the end are lost. The original number satisfies $45=11\times 4+1$, so the lost bits account for the remainder of $1$. Zeros enter from the left, but they do not preserve that remainder in the stored result.

Bits can also be lost on a left shift. For example, shifting the unsigned $8$-bit number $11001000_2$ left one place stores $10010000_2$, or $144$, rather than $200\times 2=400$. The leading bit has moved beyond the available space: this is overflow.

The exam mistake to avoid is saying that a right shift always gives the exact answer to division. For an unsigned whole number, it gives the whole-number quotient; any remainder is discarded. Also check how many bits are available before assuming a left shift gives the full product.