Question

Convert the following number into 32bit IEEE 754 floating point representation.

0.000101

Answer 1

32 bit IEEE 754 floating point representation

1 bit Sign (S)

8 bits Exponent (E)

23 bits Mantissa(M)

Representation: number = (-1)^S * (1.M) * ​​​​​​2^E-127

Givennumber:0.000101

(i)Consideringthegivennumberisindecimal

• 1) 0.000 101 × 2 = 0 + 0.000 202;
• 2) 0.000 202 × 2 = 0 + 0.000 404;
• 3) 0.000 404 × 2 = 0 + 0.000 808;
• 4) 0.000 808 × 2 = 0 + 0.001 616;
• 5) 0.001 616 × 2 = 0 + 0.003 232;
• 6) 0.003 232 × 2 = 0 + 0.006 464;
• 7) 0.006 464 × 2 = 0 + 0.012 928;
• 8) 0.012 928 × 2 = 0 + 0.025 856;
• 9) 0.025 856 × 2 = 0 + 0.051 712;
• 10) 0.051 712 × 2 = 0 + 0.103 424;
• 11) 0.103 424 × 2 = 0 + 0.206 848;
• 12) 0.206 848 × 2 = 0 + 0.413 696;
• 13) 0.413 696 × 2 = 0 + 0.827 392;
• 14) 0.827 392 × 2 = 1 + 0.654 784;
• 15) 0.654 784 × 2 = 1 + 0.309 568;
• 16) 0.309 568 × 2 = 0 + 0.619 136;
• 17) 0.619 136 × 2 = 1 + 0.238 272;
• 18) 0.238 272 × 2 = 0 + 0.476 544;
• 19) 0.476 544 × 2 = 0 + 0.953 088;
• 20) 0.953 088 × 2 = 1 + 0.906 176;
• 21) 0.906 176 × 2 = 1 + 0.812 352;
• 22) 0.812 352 × 2 = 1 + 0.624 704;
• 23) 0.624 704 × 2 = 1 + 0.249 408;
• 24) 0.249 408 × 2 = 0 + 0.498 816;
• 25) 0.498 816 × 2 = 0 + 0.997 632;
• 26) 0.997 632 × 2 = 1 + 0.995 264;
• 27) 0.995 264 × 2 = 1 + 0.990 528;
• 28) 0.990 528 × 2 = 1 + 0.981 056;
• 29) 0.981 056 × 2 = 1 + 0.962 112;
• 30) 0.962 112 × 2 = 1 + 0.924 224;
• 31) 0.924 224 × 2 = 1 + 0.848 448;
• 32) 0.848 448 × 2 = 1 + 0.696 896;
• 33) 0.696 896 × 2 = 1 + 0.393 792;
• 34) 0.393 792 × 2 = 0 + 0.787 584;
• 35) 0.787 584 × 2 = 1 + 0.575 168;
• 36) 0.575 168 × 2 = 1 + 0.150 336;
• 37) 0.150 336 × 2 = 0 + 0.300 672;

and so on..

So,

0.000101₍₁₀₎ =

0.0000000000000110100111100111111110110₍₂₎

= 1.10100111100111111110110₍₂₎ × 2^-14

Now let's adjust the expont

(-14 + 127)₍₁₀₎ = 113₍₁₀₎

• 113 ÷ 2 = 56 + 1;
• 56 ÷ 2 = 28 + 0;
• 28 ÷ 2 = 14 + 0;
• 14 ÷ 2 = 7 + 0;
• 7 ÷ 2 = 3 + 1;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;

So, (113)₁₀ = 01110001₍₂₎

Sign bit will be 0 since given number is positive

Therefore, (0.000101)₁₀ is represented in 32 bit IEEE floating point representation as follows

0

01110001

10100111100111111110110

(ii) Considering the given number is in binary

(0.000101)₂ = (1.01000000000000000000000)₂ * 2^-4

Now let's adjust the expont
(-4+127)₁₀ = (123)₁₀

• 123 ÷ 2 = 61 + 1;
• 61 ÷ 2 = 30 + 1;
• 30 ÷ 2 = 15 + 0;
• 15 ÷ 2 = 7 + 1;
• 7 ÷ 2 = 3 + 1;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;

(123)₁₀ = (01111011)₂

Sign bit will be 0, since given number is positive.

Therefore (0.000101)₂ is represented in 32 bit IEEE floating point representation as follows

0

01111011

01000000000000000000000