#Number System

Question

Sign bit (1 bit)

Exponent (5 bit)

Mantissa (9 bit) (normalized form)

Numbers in the exponent are in two's complement form. Smallest positive number this machine can represent.

i.) 0 11111 100000000

ii.) 0 00000 100000000

iii.) 0 00000 000000001

iv.) 0 10000 100000000

(or any other option you should provide me because I think this options are not correct  )

Answer 1

Option iii is right.

Comments

According to you if you are taking all exponent as zero then it become Denormalized number

Exponent all zero

Mantissa non zero

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No biasing is mentioned and the exponent is in 2's complement form then the smallest exponent that can be represented in 5 bits is (-2^5-1) =(10000)₂.

Mantissa is in normalized form. Normalization can be of 2 types 1.M and 0.1M (where 1 is explicit i.e. it has to be written in the mantissa). So for eg 2.5 is to be represented in 0.1M form then we do like ->  2.5=10.100000.. = 0.10100.....0x 2². Mantissa is [1010....0].

To add to this, another normalized form 0.1M where 1 is implicit also exists. In that case the mantissa will look like [010...0] because here the 1st '1' after point is implicit.

I think option D. [0 10000 10000..0]. As mantissa has explicit '1' (according to my assumption) so the first '1' will always stay no matter what and to be the smallest no. the other bits have to be 0.

But not sure. These things are really confusing :P

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it is not given no is stored in ieee format so answer will be 4

Answer 2

Here exponent is in 2's complement form so smallest exponent possible with 5 bits is -2^{^4} that is -16(10000 in 2's complement).

and mantissa is normalized that means 1.M format so we can mantissa bits to 0.

so number will look like 0  10000 000000000 .