a. To prove: “Adding one to nine’s complement of a number is equal to ten’s complement of the same number”. Proof: Formula for ten’s complement to compute the number: Negative ( I ) = 10 k − I (1) In the above formula, “k” represents the number of digits. “I” represents the number to be calculated. Formula for nine’s complement to compute the number: Negative[I]=9999 .... − I In the above formula, “I” represents the number to be calculated. “9999….” represents all 9s in which the number of 9s should be equal to the number of digits in a given number “I”. Adding one to Nine’s complement formula to calculate the numbers: Negative[I]=(9999 .... − I)+1 (2) Comparing Equation (1) and Equation (2), 10 k − I=(9999 .... − I)+1 10 k − I=9999 .... − I+1 10 k − I + I =9999

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Answer 1

Expert Solution & Answer

Chapter 3, Problem 39E

Explanation of Solution

a.

To prove: “Adding one to nine’s complement of a number is equal to ten’s complement of the same number”.

Proof:

Formula for ten’s complement to compute the number:

$Negative(I) = 10k − I$ (1)

In the above formula,

“k” represents the number of digits.

“I” represents the number to be calculated.

Formula for nine’s complement to compute the number:

$Negative[I]=9999....−I$

In the above formula,

“I” represents the number to be calculated.

“9999….” represents all 9s in which the number of 9s should be equal to the number of digits in a given number “I”.

Adding one to Nine’s complement formula to calculate the numbers:

$Negative[I]=(9999....−I)+1$ (2)

Comparing Equation (1) and Equation (2),

$10k −I=(9999....−I)+110k−I=9999....−I+110k−I+I=9999$

Explanation of Solution

b.

Calculating ten’s complements using nine’s complement plus one formula for “-35768”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999....−I)+1$

$Negative[35768]=(99999−35768)+1=64231+1=64232$

Therefore, ten’s complement using nine’s complement plus one formula for “-35768” is “64232”.

Calculating ten’s complements using nine’s complement plus one formula for “-444455”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999...$

Explanation of Solution

c.

Nine’s complement plus one method is easier.

Justification:

Borrowing of values are needed for subtraction in direct ten’s complement method but in nine’ complement plus one method, borrowing of values are not needed for subtraction...

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Answer 2

Expert Solution & Answer

Chapter 3, Problem 39E

Explanation of Solution

a.

To prove: “Adding one to nine’s complement of a number is equal to ten’s complement of the same number”.

Proof:

Formula for ten’s complement to compute the number:

$Negative(I) = 10k − I$ (1)

In the above formula,

“k” represents the number of digits.

“I” represents the number to be calculated.

Formula for nine’s complement to compute the number:

$Negative[I]=9999....−I$

In the above formula,

“I” represents the number to be calculated.

“9999….” represents all 9s in which the number of 9s should be equal to the number of digits in a given number “I”.

Adding one to Nine’s complement formula to calculate the numbers:

$Negative[I]=(9999....−I)+1$ (2)

Comparing Equation (1) and Equation (2),

$10k −I=(9999....−I)+110k−I=9999....−I+110k−I+I=9999$

Explanation of Solution

b.

Calculating ten’s complements using nine’s complement plus one formula for “-35768”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999....−I)+1$

$Negative[35768]=(99999−35768)+1=64231+1=64232$

Therefore, ten’s complement using nine’s complement plus one formula for “-35768” is “64232”.

Calculating ten’s complements using nine’s complement plus one formula for “-444455”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999...$

Explanation of Solution

c.

Nine’s complement plus one method is easier.

Justification:

Borrowing of values are needed for subtraction in direct ten’s complement method but in nine’ complement plus one method, borrowing of values are not needed for subtraction...

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 3

Expert Solution & Answer

Chapter 3, Problem 39E

Explanation of Solution

a.

To prove: “Adding one to nine’s complement of a number is equal to ten’s complement of the same number”.

Proof:

Formula for ten’s complement to compute the number:

$Negative(I) = 10k − I$ (1)

In the above formula,

“k” represents the number of digits.

“I” represents the number to be calculated.

Formula for nine’s complement to compute the number:

$Negative[I]=9999....−I$

In the above formula,

“I” represents the number to be calculated.

“9999….” represents all 9s in which the number of 9s should be equal to the number of digits in a given number “I”.

Adding one to Nine’s complement formula to calculate the numbers:

$Negative[I]=(9999....−I)+1$ (2)

Comparing Equation (1) and Equation (2),

$10k −I=(9999....−I)+110k−I=9999....−I+110k−I+I=9999$

Explanation of Solution

b.

Calculating ten’s complements using nine’s complement plus one formula for “-35768”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999....−I)+1$

$Negative[35768]=(99999−35768)+1=64231+1=64232$

Therefore, ten’s complement using nine’s complement plus one formula for “-35768” is “64232”.

Calculating ten’s complements using nine’s complement plus one formula for “-444455”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999...$

Explanation of Solution

c.

Nine’s complement plus one method is easier.

Justification:

Borrowing of values are needed for subtraction in direct ten’s complement method but in nine’ complement plus one method, borrowing of values are not needed for subtraction...

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 4

Expert Solution & Answer

Chapter 3, Problem 39E

Explanation of Solution

a.

To prove: “Adding one to nine’s complement of a number is equal to ten’s complement of the same number”.

Proof:

Formula for ten’s complement to compute the number:

$Negative(I) = 10k − I$ (1)

In the above formula,

“k” represents the number of digits.

“I” represents the number to be calculated.

Formula for nine’s complement to compute the number:

$Negative[I]=9999....−I$

In the above formula,

“I” represents the number to be calculated.

“9999….” represents all 9s in which the number of 9s should be equal to the number of digits in a given number “I”.

Adding one to Nine’s complement formula to calculate the numbers:

$Negative[I]=(9999....−I)+1$ (2)

Comparing Equation (1) and Equation (2),

$10k −I=(9999....−I)+110k−I=9999....−I+110k−I+I=9999$

Explanation of Solution

b.

Calculating ten’s complements using nine’s complement plus one formula for “-35768”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999....−I)+1$

$Negative[35768]=(99999−35768)+1=64231+1=64232$

Therefore, ten’s complement using nine’s complement plus one formula for “-35768” is “64232”.

Calculating ten’s complements using nine’s complement plus one formula for “-444455”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999...$

Explanation of Solution

c.

Nine’s complement plus one method is easier.

Justification:

Borrowing of values are needed for subtraction in direct ten’s complement method but in nine’ complement plus one method, borrowing of values are not needed for subtraction...

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 3, Problem 39E

Explanation of Solution

a.

To prove: “Adding one to nine’s complement of a number is equal to ten’s complement of the same number”.

Proof:

Formula for ten’s complement to compute the number:

$Negative(I) = 10k − I$ (1)

In the above formula,

“k” represents the number of digits.

“I” represents the number to be calculated.

Formula for nine’s complement to compute the number:

$Negative[I]=9999....−I$

In the above formula,

“I” represents the number to be calculated.

“9999….” represents all 9s in which the number of 9s should be equal to the number of digits in a given number “I”.

Adding one to Nine’s complement formula to calculate the numbers:

$Negative[I]=(9999....−I)+1$ (2)

Comparing Equation (1) and Equation (2),

$10k −I=(9999....−I)+110k−I=9999....−I+110k−I+I=9999$

Explanation of Solution

b.

Calculating ten’s complements using nine’s complement plus one formula for “-35768”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999....−I)+1$

$Negative[35768]=(99999−35768)+1=64231+1=64232$

Therefore, ten’s complement using nine’s complement plus one formula for “-35768” is “64232”.

Calculating ten’s complements using nine’s complement plus one formula for “-444455”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999...$

Explanation of Solution

c.

Nine’s complement plus one method is easier.

Justification:

Borrowing of values are needed for subtraction in direct ten’s complement method but in nine’ complement plus one method, borrowing of values are not needed for subtraction...

Answer 8

Chapter 3, Problem 39E

Explanation of Solution

a.

To prove: “Adding one to nine’s complement of a number is equal to ten’s complement of the same number”.

Proof:

Formula for ten’s complement to compute the number:

$Negative(I) = 10k − I$ (1)

In the above formula,

“k” represents the number of digits.

“I” represents the number to be calculated.

Formula for nine’s complement to compute the number:

$Negative[I]=9999....−I$

In the above formula,

“I” represents the number to be calculated.

“9999….” represents all 9s in which the number of 9s should be equal to the number of digits in a given number “I”.

Adding one to Nine’s complement formula to calculate the numbers:

$Negative[I]=(9999....−I)+1$ (2)

Comparing Equation (1) and Equation (2),

$10k −I=(9999....−I)+110k−I=9999....−I+110k−I+I=9999$

Answer 9

Chapter 3, Problem 39E

Explanation of Solution

a.

To prove: “Adding one to nine’s complement of a number is equal to ten’s complement of the same number”.

Proof:

Formula for ten’s complement to compute the number:

$Negative(I) = 10k − I$ (1)

In the above formula,

“k” represents the number of digits.

“I” represents the number to be calculated.

Formula for nine’s complement to compute the number:

$Negative[I]=9999....−I$

In the above formula,

“I” represents the number to be calculated.

“9999….” represents all 9s in which the number of 9s should be equal to the number of digits in a given number “I”.

Adding one to Nine’s complement formula to calculate the numbers:

$Negative[I]=(9999....−I)+1$ (2)

Comparing Equation (1) and Equation (2),

$10k −I=(9999....−I)+110k−I=9999....−I+110k−I+I=9999$

Answer 10

Explanation of Solution

b.

Calculating ten’s complements using nine’s complement plus one formula for “-35768”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999....−I)+1$

$Negative[35768]=(99999−35768)+1=64231+1=64232$

Therefore, ten’s complement using nine’s complement plus one formula for “-35768” is “64232”.

Calculating ten’s complements using nine’s complement plus one formula for “-444455”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999...$

Answer 11

Explanation of Solution

b.

Calculating ten’s complements using nine’s complement plus one formula for “-35768”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999....−I)+1$

$Negative[35768]=(99999−35768)+1=64231+1=64232$

Therefore, ten’s complement using nine’s complement plus one formula for “-35768” is “64232”.

Calculating ten’s complements using nine’s complement plus one formula for “-444455”:

Nine’s complement plus one formula to compute the numbers:

$Negative[I]=(9999...$

Answer 12

Explanation of Solution

c.

Nine’s complement plus one method is easier.

Justification:

Borrowing of values are needed for subtraction in direct ten’s complement method but in nine’ complement plus one method, borrowing of values are not needed for subtraction...

Answer 13

Explanation of Solution

c.

Nine’s complement plus one method is easier.

Justification:

Borrowing of values are needed for subtraction in direct ten’s complement method but in nine’ complement plus one method, borrowing of values are not needed for subtraction...

Answer 14

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Answer 15

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Answer 16

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Answer 17

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Answer 18

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Answer 19

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Answer 20

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Answer 21

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Answer 22

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Answer 23

Ch. 3 - Prob. 1E Ch. 3 - Prob. 2E Ch. 3 - Prob. 3E Ch. 3 - Prob. 4E Ch. 3 - Prob. 5E Ch. 3 - Prob. 6E Ch. 3 - Prob. 7E Ch. 3 - Prob. 8E Ch. 3 - Prob. 9E Ch. 3 - Prob. 10E

Explanation of Solution

Explanation of Solution

Explanation of Solution

c. Nine’s complement plus one method is easier. Justification: Borrowing of values are needed for subtraction in direct ten’s complement method but in nine’ complement plus one method, borrowing of values are not needed for subtraction...

Chapter 3 Solutions

c.

Nine’s complement plus one method is easier.

Justification:

Borrowing of values are needed for subtraction in direct ten’s complement method but in nine’ complement plus one method, borrowing of values are not needed for subtraction...