-0.000 000 000 744 47 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 744 47(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 744 47(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 744 47| = 0.000 000 000 744 47


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 744 47.

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) 0.000 000 000 744 47 × 2 = 0 + 0.000 000 001 488 94;
  • 2) 0.000 000 001 488 94 × 2 = 0 + 0.000 000 002 977 88;
  • 3) 0.000 000 002 977 88 × 2 = 0 + 0.000 000 005 955 76;
  • 4) 0.000 000 005 955 76 × 2 = 0 + 0.000 000 011 911 52;
  • 5) 0.000 000 011 911 52 × 2 = 0 + 0.000 000 023 823 04;
  • 6) 0.000 000 023 823 04 × 2 = 0 + 0.000 000 047 646 08;
  • 7) 0.000 000 047 646 08 × 2 = 0 + 0.000 000 095 292 16;
  • 8) 0.000 000 095 292 16 × 2 = 0 + 0.000 000 190 584 32;
  • 9) 0.000 000 190 584 32 × 2 = 0 + 0.000 000 381 168 64;
  • 10) 0.000 000 381 168 64 × 2 = 0 + 0.000 000 762 337 28;
  • 11) 0.000 000 762 337 28 × 2 = 0 + 0.000 001 524 674 56;
  • 12) 0.000 001 524 674 56 × 2 = 0 + 0.000 003 049 349 12;
  • 13) 0.000 003 049 349 12 × 2 = 0 + 0.000 006 098 698 24;
  • 14) 0.000 006 098 698 24 × 2 = 0 + 0.000 012 197 396 48;
  • 15) 0.000 012 197 396 48 × 2 = 0 + 0.000 024 394 792 96;
  • 16) 0.000 024 394 792 96 × 2 = 0 + 0.000 048 789 585 92;
  • 17) 0.000 048 789 585 92 × 2 = 0 + 0.000 097 579 171 84;
  • 18) 0.000 097 579 171 84 × 2 = 0 + 0.000 195 158 343 68;
  • 19) 0.000 195 158 343 68 × 2 = 0 + 0.000 390 316 687 36;
  • 20) 0.000 390 316 687 36 × 2 = 0 + 0.000 780 633 374 72;
  • 21) 0.000 780 633 374 72 × 2 = 0 + 0.001 561 266 749 44;
  • 22) 0.001 561 266 749 44 × 2 = 0 + 0.003 122 533 498 88;
  • 23) 0.003 122 533 498 88 × 2 = 0 + 0.006 245 066 997 76;
  • 24) 0.006 245 066 997 76 × 2 = 0 + 0.012 490 133 995 52;
  • 25) 0.012 490 133 995 52 × 2 = 0 + 0.024 980 267 991 04;
  • 26) 0.024 980 267 991 04 × 2 = 0 + 0.049 960 535 982 08;
  • 27) 0.049 960 535 982 08 × 2 = 0 + 0.099 921 071 964 16;
  • 28) 0.099 921 071 964 16 × 2 = 0 + 0.199 842 143 928 32;
  • 29) 0.199 842 143 928 32 × 2 = 0 + 0.399 684 287 856 64;
  • 30) 0.399 684 287 856 64 × 2 = 0 + 0.799 368 575 713 28;
  • 31) 0.799 368 575 713 28 × 2 = 1 + 0.598 737 151 426 56;
  • 32) 0.598 737 151 426 56 × 2 = 1 + 0.197 474 302 853 12;
  • 33) 0.197 474 302 853 12 × 2 = 0 + 0.394 948 605 706 24;
  • 34) 0.394 948 605 706 24 × 2 = 0 + 0.789 897 211 412 48;
  • 35) 0.789 897 211 412 48 × 2 = 1 + 0.579 794 422 824 96;
  • 36) 0.579 794 422 824 96 × 2 = 1 + 0.159 588 845 649 92;
  • 37) 0.159 588 845 649 92 × 2 = 0 + 0.319 177 691 299 84;
  • 38) 0.319 177 691 299 84 × 2 = 0 + 0.638 355 382 599 68;
  • 39) 0.638 355 382 599 68 × 2 = 1 + 0.276 710 765 199 36;
  • 40) 0.276 710 765 199 36 × 2 = 0 + 0.553 421 530 398 72;
  • 41) 0.553 421 530 398 72 × 2 = 1 + 0.106 843 060 797 44;
  • 42) 0.106 843 060 797 44 × 2 = 0 + 0.213 686 121 594 88;
  • 43) 0.213 686 121 594 88 × 2 = 0 + 0.427 372 243 189 76;
  • 44) 0.427 372 243 189 76 × 2 = 0 + 0.854 744 486 379 52;
  • 45) 0.854 744 486 379 52 × 2 = 1 + 0.709 488 972 759 04;
  • 46) 0.709 488 972 759 04 × 2 = 1 + 0.418 977 945 518 08;
  • 47) 0.418 977 945 518 08 × 2 = 0 + 0.837 955 891 036 16;
  • 48) 0.837 955 891 036 16 × 2 = 1 + 0.675 911 782 072 32;
  • 49) 0.675 911 782 072 32 × 2 = 1 + 0.351 823 564 144 64;
  • 50) 0.351 823 564 144 64 × 2 = 0 + 0.703 647 128 289 28;
  • 51) 0.703 647 128 289 28 × 2 = 1 + 0.407 294 256 578 56;
  • 52) 0.407 294 256 578 56 × 2 = 0 + 0.814 588 513 157 12;
  • 53) 0.814 588 513 157 12 × 2 = 1 + 0.629 177 026 314 24;
  • 54) 0.629 177 026 314 24 × 2 = 1 + 0.258 354 052 628 48;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


5. 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:


0.000 000 000 744 47(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1000 1101 1010 11(2)

6. Positive number before normalization:

0.000 000 000 744 47(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1000 1101 1010 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 744 47(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1000 1101 1010 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0010 1000 1101 1010 11(2) × 20 =

1.1001 1001 0100 0110 1101 011(2) × 2-31


8. 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 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 1001 0100 0110 1101 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1100 1010 0011 0110 1011 =

100 1100 1010 0011 0110 1011


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

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1100 1010 0011 0110 1011


Decimal number -0.000 000 000 744 47 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 1010 0011 0110 1011

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

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

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111