Chapter 4, Problem 4.7QAP

Interpretation Introduction

(a)

Interpretation:

The given operation should be performed by using binary numbers and after that the final binary number obtained should be converted back to decimal.

Concept introduction:

Rules of Binary addition:

Binary sum of $0$ and $0$ results $0$ and yields $0$ carry.

Binary sum of $0$ and $1$ results $1$ and yields $0$ carry.

Binary sum of $1$ and $1$ results $0$ and yields $1$ carry.

Answer to Problem 4.7QAP

$8_{10}+5_{10}={(1000)}_2+{(101)}_2={(1101)}_2={13}_{10}$

Explanation of Solution

We will convert the decimal to binary as follows:

Divide $8$ by $2$, we will get $4$, remainder will be $0(\text{LSB})$.

Now divide $4$ by $2$, we will get $2$ remainder will be $0$.

Now divide $2$ by $2$, we will get $1(\text{MSB})$ remainder will be $0$.

Hence the binary equivalent of $8$ will be $1000$, it can be written as follows:

$8_{10}={(1000)}_2$

Similarly, we can find the binary equivalent of $5$ as,

$5_{10}={(101)}_2$

The sum is,

$1000$

$+101$

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$1101$

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The decimal equivalent of sum is, ${(1101)}_2={13}_{10}$

Interpretation Introduction

(b)

Interpretation:

The given operation should be performed by using binary numbers and after that the final binary number obtained should be converted back to decimal.

Concept introduction:

Rules of Binary addition:

Binary sum of $0$ and $0$ results in $0$ and yields $0$ carry

Binary sum of $0$ and $1$ results $1$ and yields $0$ carry

Binary sum of $1$ and $1$ results $0$ and yields $1$ carry

Answer to Problem 4.7QAP

${339}_{10}+{25}_{10}={(101010011)}_2+{(11001)}_2={(\text{101101100})}_2={364}_{10}$

Explanation of Solution

we will convert the decimal to binary as follows:

Divide $339$ by $2$, we will get $169$ ,remainder will be $1(\text{LSB})$.

Now divide $169$ by $2$, we will get $84$ remainder will be $1$.

Now divide $84$ by $2$, we will get $42$ remainder will be $0$.

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Now divide $2$ by $2$, we will get $1(\text{MSB)}$ remainder will be $0$.

Hence the binary equivalent of $339$ will be $\text{101010011}$, it can be written as follows:

${339}_{10}={(\text{101010011})}_2$

Similarly we can find the binary equivalent of $25$.

${25}_{10}={(11001)}_2$

$101010011$

$+11001$

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$\text{101101100}$

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${(\text{101101100})}_2={364}_{10}$

Interpretation Introduction

(c)

Interpretation:

The given operation should be performed by using binary numbers and after that the final binary number obtained should be converted back to decimal.

Concept introduction:

Rules of Binary addition:

Binary sum of $0$ and $0$ results $0$ and yields $0$ carry

Binary sum of $0$ and $1$ results $1$ and yields $0$ carry

Binary sum of $1$ and $1$ results $0$ and yields $1$ carry

Answer to Problem 4.7QAP

${47}_{10}+{16}_{10}={(\text{101111})}_2+{(10000)}_2={(\text{111111})}_2={63}_{10}$

Explanation of Solution

we will convert the decimal to binary as follows:

Divide $47$ by $2$, we will get $23$ ,remainder will be $1(\text{LSB})$.

Now divide $23$ by $2$, we will get $11$ remainder will be $1$.

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Now divide $2$ by $2$, we will get $1(\text{MSB)}$ remainder will be $0$.

Hence the binary equivalent of $47$ will be $\text{101111}$, it can be written as follows:

${47}_{10}={(\text{101111})}_2$

Similarly, we can find the binary equivalent of $16.$

${16}_{10}={(10000)}_2$

$101111$

$+10000$

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$111111$

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${(111111)}_2={63}_{10}$

Interpretation Introduction

(d)

Interpretation:

The given operation should be performed by using binary numbers and after that the final binary number obtained should be converted back to decimal.

Concept introduction:

Rules of Binary multiplication:

Binary product of $0$ and $0$ results $0$

Binary product of $0$ and $1$ results $0$

Binary product of $1$ and $1$ results $1$

Answer to Problem 4.7QAP

$2_{10}\times 9_{10}={(1001)}_2\times {(10)}_2={(10010)}_2={18}_{10}$

Explanation of Solution

we will convert the decimal to binary as follows:

Divide $9$ by $2$, we will get $4$ ,remainder will be $1(\text{LSB})$.

Now divide $4$ by $2$, we will get $2$ remainder will be $0$.

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Now divide $2$ by $2$, we will get $1(\text{MSB)}$ remainder will be $0$.

Hence the binary equivalent of $9$ will be $\text{1001}$, it can be written as follows:

$9_{10}={(1001)}_2$

Similarly, we can find the binary equivalent of $2$.

$2_{10}={(10)}_2$

$1001$

$\times 10$

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$0000$

$1001$

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$10010$

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${(10010)}_2={18}_{10}$