Answer to Question #147172 in Excel for Ashhar

Question

Convert the following decimal numbers into equivalent binary numbers and then convert the binary answer back into equivalent decimal Show all the steps of conversions

479

8735

Solve the given Boolean Expression by using truth table,

Answer 1

To convert a given numbers to a binary system, we will use the Euclidean algorithm.

(more information: https://en.wikipedia.org/wiki/Euclidean_division)

More specifically, we will divide by 2 and we will be interested in the remainders.

Note : With a separate picture, I will show how this division looks like if you do it "in a column".

1) $479$

1 case : $479_{10}\to x_2$

$$479\div 2=239\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 239\div 2=119\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 119\div 2=59\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 59\div 2=29\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 29\div 2=14\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 14\div 2=7\left(remainder\,\,\,\boxed{0}\right)\\[0.3cm] 7\div 2=3\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 3\div 2=\boxed{1}\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm]$$

It remains to write down the results obtained: writes down the numbers in reverse order, that is, it starts with the quotient from the last step, and then writes down only the remainders.

Conclusion,

$$\boxed{479_{10}\to(111011111)_2}$$

When converting from a binary number to a decimal number, we do "as if the opposite": we multiply "1" or "0" by "2" to the degree that corresponds to the (position-1) of the specified "0" or "1".

2 case : $(111011111)_2\to x_{10}$

$$(111011111)_2=1\cdot 2^{9-1}+1\cdot2^{8-1}+1\cdot2^{7-1}+0\cdot2^{6-1}+1\cdot2^{5-1}+\\[0.3cm] +1\cdot2^{4-1}+1\cdot2^{3-1}+1\cdot2^{2-1}+1\cdot2^{1-1}=\\[0.3cm] =2^8+2^7+2^6+2^4+0+2^3+2^2+2^1+2^0=\\[0.3cm] =256+128+64+16+8+4+2+1=\\[0.3cm] =384+80+15=464+15=479$$

2) $8735$

1 case : $8735_{10}\to x_2$

$$8735\div 2=4367\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 4367\div 2=2183\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 2183\div 2=1091\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 1091\div 2=545\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 545\div 2=272\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 272\div 2=136\left(remainder\,\,\,\boxed{0}\right)\\[0.3cm] 136\div 2=68\left(remainder\,\,\,\boxed{0}\right)\\[0.3cm] 68\div 2=34\left(remainder\,\,\,\boxed{0}\right)\\[0.3cm] 34\div 2=17\left(remainder\,\,\,\boxed{0}\right)\\[0.3cm] 17\div 2=8\left(remainder\,\,\,\boxed{1}\right)\\[0.3cm] 8\div 2=4\left(remainder\,\,\,\boxed{0}\right)\\[0.3cm] 4\div 2=2\left(remainder\,\,\,\boxed{0}\right)\\[0.3cm] 2\div 2=\boxed{1}\left(remainder\,\,\,\boxed{0}\right)\\[0.3cm]$$

Conclison,

$$8735_{10}\to(01001000011111)_2$$

Note : the first "0" is not taken into account.

$$\boxed{8735_{10}\to(1001000011111)_2}$$

2 case : $(1001000011111)_2\to x_{10}$

$$(1001000011111)_2=1\cdot2^{13-}+1\cdot2^{10-1}+1\cdot2^{5-1}+\\[0.3cm] +1\cdot2^{4-1}+1\cdot2^{3-1}+1\cdot2^{2-1}+1\cdot2^{1-1}=\\[0.3cm] =2^{13}+2^9+2^4+2^3+2^2+2^1+2^0=\\[0.3cm] =8192+512+16+8+4+2+1=8735$$