Is it necessary for the MSB of the Mantissa to be non zero in Normalized Form ?

Consider the following $32\text{-bit}$ floating-point representation scheme as shown in the format below. A value is specified by $3$ fields, a one bit sign field (with $0$ for positive and $1$ for negative values), a $24 \;\text{bit}$ fraction field (with the binary point is at the left end of the fraction bits), and a $7\;\text{bit}$ exponent field (in $\text{excess-}64$ signed integer representation, with $16$ is the base of exponentiation). The sign bit is the most significant bit.

- It is required to represent the decimal value $- 7.5$ as a normalized floating point number in the given format. Derive the values of the various fields. Express your final answer in the hexadecimal.
- What is the largest value that can be represented using this format? Express your answer as the nearest power of $10$.

### 8 Comments

Sir,

But then when you converted **111.1 * 16^0** to **0.0111*16^1 **

**0.0111*16^1 **, is not in Normalized Form ( and the question says we want it in Normalized Form)

the defination of Normalised Number -

**A number in Scientific Notation with no leading 0s is called a Normalised Number: 1.0 × 10**

^{-8 }

where am I going wrong ?

Re-writing the same answer as that of Arjun Sir (correct me if I am wrong anywhere)

The Floating point representation given in the question is not that of IEEE-754. The equation of the given representation is:

$0.M X 2^{Exponent-Base}$

**The 0 in 0.M is not implicit, but explicit**

Now, given number -7.5

$(-7.5)_{10} = (-111.1)_{2} = -0.1111 * 2^{3} \rightarrow\left[1\right]$

Now, base is given as 16 with excess 64, that is

$16^{x-64} = 2^{4(x-64)} = 2^{4x-256} \rightarrow\left[2\right]$

Here, x is the exponent. Now from equation 1 and 2, we get

$4x-256=3$

$x\approx 65$

Now, equation 1 becomes:

$-0.1111 * 2^{65-64}$

This gives S=1, M = 011110000000000000000000 and E = 1000001

$(1\; 01111\underbrace{000\dots0}_{\text{19 zeroes}} \; 1000001)_2 = (BC000041)_{16}$

## 4 Answers

Here, mantissa is represented in normalized representation and exponent in $\text{excess-}64$ (subtract $64$ to get actual value).

a. We have to represent $-(7.5)_{10} = -(111.1)_2$.

Now we are using base $16$ for exponent. So, mantissa will be $.01111$ and this makes exponent as $1 (4$ bit positions and no hiding first $1$ as in $\text{IEEE}\; 754$ as this is not mentioned in question) which in $\text{excess-}64$ will be $64 + 1 = 65.$ Number being negative sign bit is $1.$ So, we get

$(1\; 01111\underbrace{000\dots0}_{\text{19 zeroes}} \; 1000001)_2 = (\text{BC}000041)_{16}$

b. Largest value will be with largest possible mantissa, largest possible exponent and positive sign bit. So, this will be all $1$'s except sign bit which will be

$0.\underbrace{111\dots 1}_{\text{24 ones}} \times 16^{127-64} = \left( 1 - 2^{24}\right) \times 16^{63} = \left(1- 2^{-24}\right) \times 16^{63}$

$($Again we did not add implicit $1$ as in $\text{IEEE}\; 754)$

$2^x = 10^{y} \implies y = \log 2^x = x \log 2$

So, $(1-2^{-24}) \times 16^{63} \\= \left(1 - 10^{-24 \log 2}\right) \times 10^{63 \log 16}\\ \approx \left(1 - 10^{-7}\right) \times 10^{76}\\ = 10^{76}$

Not directly relevant here, but a useful read: https://jeapostrophe.github.io/courses/2015/fall/305/notes/dist/reading/help-floating-point.pdf

### 21 Comments

@ G.K.T

Because in base 16, 4 bits are used to represent each Byte.

Each byte is shown as a pair of nibbles (*4 bits*) .

@Arjun sir and bikram sir

In ques https://gateoverflow.in/2729/gate1996_1-25

we have used first bit after decimal as 1 because it was normalized form but not IEEE.And same is the case here.So why in current question we are able to store fraction as .0111 as normalized form with first bit after decimal as 0

So I have read all the comments and understood the Part A answer. But when this answer is converted to decimal, we don't get -7.5. Mantissa is .01111 and unbiased exponent is 1. Now it is given in the question that the floating point number is normalized. Hence the number is 1.01111 * 16^1 (with sign bit as 1). But this number is not equal to -7.5. Now 7.5 in binary is 111.1 . If we want to keep this number normalized, we must have a 1 before the binary point. We have 1.111 * 2^2 = 1.111 * 16^(0.5). Biased exponent is 64.5 which I think cannot be expressed. What am I doing wrong? Please help @Arjun sir @Bikram sir @Akash Kanase sir

for largest value, M=all 1's but as it is mentioned in the question it is fraction and binary point is at the left end of the fraction bits.

so, fraction field value=0.111111111111111111111111 (0.all 1's) = (1-2^24)

exponent = all 1s (1111111) =2^7-1 =127

exponent value =16^(127-bias) =16^(127-64)= 16^63

sign= 0 (+ve number)

value = (1-2^24) * 16^63

represent in power of 10

16^63 =10^x

take log both side

63 log16=x log 10 (base of log is 10)

63*1.2=x

x=75.6

so value will be (1-2^24)*10^75.6

PS: correct me if i am going wrong somewhere

a)

-7.5 Due to minus sign we will get sing bit as 1.

7.5 = 111.1 * 16^0

As Base is 16 & We need to normalize we need to at least normalize to 16^1. (This answer maybe different if we choose not to normalize, Though I think we can not use either implicit or explicit normalization correctly here.

It will be 0.0111*16^1.

So now our biased exponent is 1+64 = 65.

Mantissa is (0.) 0111000...

When we save it we get hex as BC000041

2. Maximum value possible is $(1- 2^{-24}) \times 16^{(127-64)}$.

WHen converted into base 10 we get $7.237006 \times 10^{75}$.

### 5 Comments

@Sambhrant Maurya yes 0 is a valid exponent but mantissa is represented as M in **0.M **, that's why we do normalization.

Normalized form is $(-1)^S (0.M)\times 16^{E-64}$

First express (-7.5) into binary

How to write (-7.5) in binary

## first write 7.5 into binary = (111.1)

Do not worry about negative sign here sign bit will contain 1 because above number is negative.

### And base 16 is given hence we will convert into power of 16(we know 2^4=16)

So we can write 111.1*2^0 = 0.01111*2^4

### 0.01111*2^4 = 0.01111*16^1( here 16 is a base and 1 is exponent)

Excess 64 is given hence 1+64=65

### 65 in binary= 1000001(we will represent 65 in exponent field)

Now we have to represent into floating point

### Sign bit = 1

### Mantissa = 011110...0(19 zero consecutive)

### Exponent = 1000001

## ( 1 011110....0 1000001 ) in floating point

convert it into hexadecimal and make a pair of 4 digit from right side. ( for example 0001=1 in hexadecimal )

## then you will get (BC000041) in hexadecimal

-7.5= - 111.1 =- 1.111 * 2^2

sign bit =1

biased exponent = 65 as (65 - 63 = 2) = 0010001

fraction = 111000000000000000000000 (three 1s followed by twenty one 0s)

normalized representation = 1 111000000000000000000000 0010001

hexadecimal representation = F0000011

b) largest value :

sign bit = 0 (positive)

exponent = 1111110

fraction = 111111111111111111111111 ( twenty four 1s )

normalized number = 0 111111111111111111111111 1111110

= 2147483646

= 2.147483646 * 10^9

CORRECT ME IF I AM WRONG

### 1 comment

BTW..65 binary is 1000001 which brings ur answer to F0000041

if we use 64 biasing i am getting F0000042

can some confirm which biasing to use??

for part b)

the mantissa will look like

1.1...24 1's...1 which is 2.16777215

exponent will be (126-64 or 63) (say 126-64 = 62)

so..=2.16777215 X 10^62

correct me also if wrong :)