# Binary and Logic

#### Thinker

Joined Jan 9, 2007
62
I've just finished answering these three questions on my Binary and Logic assignment, and i was wondering is someone can check my answers. I'm not good at this subject...i just made an attempt on these question, which i'm not sure if it's right or not.

Here they are:

1. Use 9 bits for mantissa and 3 bits for exponent to represent the following numbers:

(a) 843434343488
(b) 43553.99786763
(c) 0.00000000487
(d) 0.2345899

(a) 843434343488 = 0.843434343
|8|4|3|4|3|4|3|4|3|

(b) 43553.99786763 = 0.4355399786763 x 10^5
|4|3|5|5|4|0|0|7|8|

(c) 0.00000000487 = 0.487 x 10^-8
|0|0|0|0|4|9|0|0|0|

(d) 0.2345899 = 0.2345899 x 10^1
|2|3|4|6|0|0|0|0|1|

2. Use fixed-point notation and 9 bits to multiply and store the following numbers 564.3 and 45.666.

564.3 = 5.643 x 10²
45.666 = 4.5666 x 10¹

3. Use 7 bits for mantissa and 2 bits for exponent and Floating-point notation to simplify and store the following:

(a) 355.898 + 4564.56
(b) 456 – 44.3555

(a) 355.898 + 4564.56 = 4920.458
= 4920458 x 10^4
|4|9|2|0|4|6|0|

(b) 456 – 44.3555 = 411.6445
= 4116445 x 10^3
|4|1|1|6|4|5|0|

#### Papabravo

Joined Feb 24, 2006
14,409
I don't know about the rest of the questions but question number 1 is certainly wrong since you have no exponents, and you are representing the mantissas as 9 DECIMAL digits, and not nine binary digits.

#### Thinker

Joined Jan 9, 2007
62
Thanks Papabravo, do you know right answers to question number 1?

#### Papabravo

Joined Feb 24, 2006
14,409
1(a) cannot be represented with a three bit exponent
1(b) cannot be represented with a three bit exponent
1(c) cannot be represented with a three bit exponent
1(d) 0.2345899 has en exponent of zero and a mantissa of b'001111000'

it would be the equivalent of
( 1/8 + 1/16 + 1/32 + 1/64) * 2^0 = 0.234375 which is as close as you can get with a nine bit mantissa and a three bit exponent. Normalizing the number would allow two more bits in the representation. In that case the exponent would be minus 2 which is still representable in three bits.

#### Thinker

Joined Jan 9, 2007
62
Thanks Papabravo, i'll check it and get back to you.

#### Thinker

Joined Jan 9, 2007
62
1(a) cannot be represented with a three bit exponent
1(b) cannot be represented with a three bit exponent
1(c) cannot be represented with a three bit exponent
1(d) 0.2345899 has en exponent of zero and a mantissa of b'001111000'

it would be the equivalent of
( 1/8 + 1/16 + 1/32 + 1/64) * 2^0 = 0.234375 which is as close as you can get with a nine bit mantissa and a three bit exponent. Normalizing the number would allow two more bits in the representation. In that case the exponent would be minus 2 which is still representable in three bits.
What do you mean by 1 (a), (b) and (c) can't be represented by a three bit exponent, since the question did say it should be?

#### Papabravo

Joined Feb 24, 2006
14,409
There may be a piece of missing information. What base would you like to apply the exponent to. If we choose base 2 then my answer stands. That range of exponents applied to the base 2 is insufficient to represent the indicated numbers.

Does the problem explicitly state which number base to use? If a base other than two is chosen then the nine bits of mantissa for problem 1(d) will need to change.

BTW I've had problems from books and given by instructors that had no solutions. What did you think I meant by my answer. I thought is was clear and unambiguous.

#### Thinker

Joined Jan 9, 2007
62
There may be a piece of missing information. What base would you like to apply the exponent to. If we choose base 2 then my answer stands. That rage of exponents applied to the base 2 is insufficient to represent the indicated numbers.

Does the problem explicitly state which number base to use? If a base other than two is chosen then the nine bits of mantissa for problem 1(d) will need to change.

BTW I've had problems from books and given by instructors that had no solutions. What did you think I meant by may answer. I thought is was clear and unambiguous.
Well i thought your answer for 1 (a), (b) and (c) was very ambiguous! I don't think there is any information, i wrote exactly what the question asked.

I think i have to look into more books to try and get the answer...

#### Papabravo

Joined Feb 24, 2006
14,409
There is no ambiguity in saying that it is impossible to represent the exponents required to express the numbers given in 1(a), 1(b), and 1(c). Furthermore, I challenge you, or your instructor, or any member of the forum to provide an answer using a nine bit mantissa, and a three bit exponent, with 2 as the number base. I got \$100.00 that says you can't do it. So there Mr. "I thought you answer was very ambiguous".