convert this to base 10 and back to base 2 00001111

Explanation:

Step1: Recall binary - to - decimal formula

The formula to convert a binary number $a_{n}2^{n}+a_{n - 1}2^{n - 1}+\cdots+a_{1}2^{1}+a_{0}2^{0}$, where $a_{i}$ is the $i$-th digit of the binary number. Given the binary number $10001111$.

Step2: Calculate each term

[ \begin{align*} 1\times2^{7}+0\times2^{6}+0\times2^{5}+0\times2^{4}+1\times2^{3}+1\times2^{2}+1\times2^{1}+1\times2^{0}&=128 + 0+0 + 0+8 + 4+2 + 1\ &=143 \end{align*} ]

Step3: Convert decimal to binary (using division method)

Divide 143 by 2 successively and record the remainders. $143\div2 = 71$ remainder $1$ $71\div2=35$ remainder $1$ $35\div2 = 17$ remainder $1$ $17\div2=8$ remainder $1$ $8\div2 = 4$ remainder $0$ $4\div2=2$ remainder $0$ $2\div2 = 1$ remainder $0$ $1\div2=0$ remainder $1$ Reading the remainders from bottom - to - top, we get $10001111$.

Answer:

The binary number $10001111$ in base - 10 is 143, and when converting 143 back to base - 2, we get 10001111.