32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 16.000 122 100 114 822 387 2 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 16.000 122 100 114 822 387 2(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 16.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


16(10) =


1 0000(2)


3. Convert to binary (base 2) the fractional part: 0.000 122 100 114 822 387 2.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 122 100 114 822 387 2 × 2 = 0 + 0.000 244 200 229 644 774 4;
  • 2) 0.000 244 200 229 644 774 4 × 2 = 0 + 0.000 488 400 459 289 548 8;
  • 3) 0.000 488 400 459 289 548 8 × 2 = 0 + 0.000 976 800 918 579 097 6;
  • 4) 0.000 976 800 918 579 097 6 × 2 = 0 + 0.001 953 601 837 158 195 2;
  • 5) 0.001 953 601 837 158 195 2 × 2 = 0 + 0.003 907 203 674 316 390 4;
  • 6) 0.003 907 203 674 316 390 4 × 2 = 0 + 0.007 814 407 348 632 780 8;
  • 7) 0.007 814 407 348 632 780 8 × 2 = 0 + 0.015 628 814 697 265 561 6;
  • 8) 0.015 628 814 697 265 561 6 × 2 = 0 + 0.031 257 629 394 531 123 2;
  • 9) 0.031 257 629 394 531 123 2 × 2 = 0 + 0.062 515 258 789 062 246 4;
  • 10) 0.062 515 258 789 062 246 4 × 2 = 0 + 0.125 030 517 578 124 492 8;
  • 11) 0.125 030 517 578 124 492 8 × 2 = 0 + 0.250 061 035 156 248 985 6;
  • 12) 0.250 061 035 156 248 985 6 × 2 = 0 + 0.500 122 070 312 497 971 2;
  • 13) 0.500 122 070 312 497 971 2 × 2 = 1 + 0.000 244 140 624 995 942 4;
  • 14) 0.000 244 140 624 995 942 4 × 2 = 0 + 0.000 488 281 249 991 884 8;
  • 15) 0.000 488 281 249 991 884 8 × 2 = 0 + 0.000 976 562 499 983 769 6;
  • 16) 0.000 976 562 499 983 769 6 × 2 = 0 + 0.001 953 124 999 967 539 2;
  • 17) 0.001 953 124 999 967 539 2 × 2 = 0 + 0.003 906 249 999 935 078 4;
  • 18) 0.003 906 249 999 935 078 4 × 2 = 0 + 0.007 812 499 999 870 156 8;
  • 19) 0.007 812 499 999 870 156 8 × 2 = 0 + 0.015 624 999 999 740 313 6;
  • 20) 0.015 624 999 999 740 313 6 × 2 = 0 + 0.031 249 999 999 480 627 2;
  • 21) 0.031 249 999 999 480 627 2 × 2 = 0 + 0.062 499 999 998 961 254 4;
  • 22) 0.062 499 999 998 961 254 4 × 2 = 0 + 0.124 999 999 997 922 508 8;
  • 23) 0.124 999 999 997 922 508 8 × 2 = 0 + 0.249 999 999 995 845 017 6;
  • 24) 0.249 999 999 995 845 017 6 × 2 = 0 + 0.499 999 999 991 690 035 2;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 122 100 114 822 387 2(10) =


0.0000 0000 0000 1000 0000 0000(2)


5. Positive number before normalization:

16.000 122 100 114 822 387 2(10) =


1 0000.0000 0000 0000 1000 0000 0000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


16.000 122 100 114 822 387 2(10) =


1 0000.0000 0000 0000 1000 0000 0000(2) =


1 0000.0000 0000 0000 1000 0000 0000(2) × 20 =


1.0000 0000 0000 0000 1000 0000 0000(2) × 24


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)


Exponent (unadjusted): 4


Mantissa (not normalized):
1.0000 0000 0000 0000 1000 0000 0000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


4 + 2(8-1) - 1 =


(4 + 127)(10) =


131(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 131 ÷ 2 = 65 + 1;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


131(10) =


1000 0011(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.


b) Adjust its length to 23 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 000 0000 0000 0000 0100 0000 0 0000 =


000 0000 0000 0000 0100 0000


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (8 bits) =
1000 0011


Mantissa (23 bits) =
000 0000 0000 0000 0100 0000


The base ten decimal number 16.000 122 100 114 822 387 2 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0011 - 000 0000 0000 0000 0100 0000

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