-0.000 000 000 742 147 39 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 39(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
-0.000 000 000 742 147 39(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:

|-0.000 000 000 742 147 39| = 0.000 000 000 742 147 39


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: 0.000 000 000 742 147 39.

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) 0.000 000 000 742 147 39 × 2 = 0 + 0.000 000 001 484 294 78;
  • 2) 0.000 000 001 484 294 78 × 2 = 0 + 0.000 000 002 968 589 56;
  • 3) 0.000 000 002 968 589 56 × 2 = 0 + 0.000 000 005 937 179 12;
  • 4) 0.000 000 005 937 179 12 × 2 = 0 + 0.000 000 011 874 358 24;
  • 5) 0.000 000 011 874 358 24 × 2 = 0 + 0.000 000 023 748 716 48;
  • 6) 0.000 000 023 748 716 48 × 2 = 0 + 0.000 000 047 497 432 96;
  • 7) 0.000 000 047 497 432 96 × 2 = 0 + 0.000 000 094 994 865 92;
  • 8) 0.000 000 094 994 865 92 × 2 = 0 + 0.000 000 189 989 731 84;
  • 9) 0.000 000 189 989 731 84 × 2 = 0 + 0.000 000 379 979 463 68;
  • 10) 0.000 000 379 979 463 68 × 2 = 0 + 0.000 000 759 958 927 36;
  • 11) 0.000 000 759 958 927 36 × 2 = 0 + 0.000 001 519 917 854 72;
  • 12) 0.000 001 519 917 854 72 × 2 = 0 + 0.000 003 039 835 709 44;
  • 13) 0.000 003 039 835 709 44 × 2 = 0 + 0.000 006 079 671 418 88;
  • 14) 0.000 006 079 671 418 88 × 2 = 0 + 0.000 012 159 342 837 76;
  • 15) 0.000 012 159 342 837 76 × 2 = 0 + 0.000 024 318 685 675 52;
  • 16) 0.000 024 318 685 675 52 × 2 = 0 + 0.000 048 637 371 351 04;
  • 17) 0.000 048 637 371 351 04 × 2 = 0 + 0.000 097 274 742 702 08;
  • 18) 0.000 097 274 742 702 08 × 2 = 0 + 0.000 194 549 485 404 16;
  • 19) 0.000 194 549 485 404 16 × 2 = 0 + 0.000 389 098 970 808 32;
  • 20) 0.000 389 098 970 808 32 × 2 = 0 + 0.000 778 197 941 616 64;
  • 21) 0.000 778 197 941 616 64 × 2 = 0 + 0.001 556 395 883 233 28;
  • 22) 0.001 556 395 883 233 28 × 2 = 0 + 0.003 112 791 766 466 56;
  • 23) 0.003 112 791 766 466 56 × 2 = 0 + 0.006 225 583 532 933 12;
  • 24) 0.006 225 583 532 933 12 × 2 = 0 + 0.012 451 167 065 866 24;
  • 25) 0.012 451 167 065 866 24 × 2 = 0 + 0.024 902 334 131 732 48;
  • 26) 0.024 902 334 131 732 48 × 2 = 0 + 0.049 804 668 263 464 96;
  • 27) 0.049 804 668 263 464 96 × 2 = 0 + 0.099 609 336 526 929 92;
  • 28) 0.099 609 336 526 929 92 × 2 = 0 + 0.199 218 673 053 859 84;
  • 29) 0.199 218 673 053 859 84 × 2 = 0 + 0.398 437 346 107 719 68;
  • 30) 0.398 437 346 107 719 68 × 2 = 0 + 0.796 874 692 215 439 36;
  • 31) 0.796 874 692 215 439 36 × 2 = 1 + 0.593 749 384 430 878 72;
  • 32) 0.593 749 384 430 878 72 × 2 = 1 + 0.187 498 768 861 757 44;
  • 33) 0.187 498 768 861 757 44 × 2 = 0 + 0.374 997 537 723 514 88;
  • 34) 0.374 997 537 723 514 88 × 2 = 0 + 0.749 995 075 447 029 76;
  • 35) 0.749 995 075 447 029 76 × 2 = 1 + 0.499 990 150 894 059 52;
  • 36) 0.499 990 150 894 059 52 × 2 = 0 + 0.999 980 301 788 119 04;
  • 37) 0.999 980 301 788 119 04 × 2 = 1 + 0.999 960 603 576 238 08;
  • 38) 0.999 960 603 576 238 08 × 2 = 1 + 0.999 921 207 152 476 16;
  • 39) 0.999 921 207 152 476 16 × 2 = 1 + 0.999 842 414 304 952 32;
  • 40) 0.999 842 414 304 952 32 × 2 = 1 + 0.999 684 828 609 904 64;
  • 41) 0.999 684 828 609 904 64 × 2 = 1 + 0.999 369 657 219 809 28;
  • 42) 0.999 369 657 219 809 28 × 2 = 1 + 0.998 739 314 439 618 56;
  • 43) 0.998 739 314 439 618 56 × 2 = 1 + 0.997 478 628 879 237 12;
  • 44) 0.997 478 628 879 237 12 × 2 = 1 + 0.994 957 257 758 474 24;
  • 45) 0.994 957 257 758 474 24 × 2 = 1 + 0.989 914 515 516 948 48;
  • 46) 0.989 914 515 516 948 48 × 2 = 1 + 0.979 829 031 033 896 96;
  • 47) 0.979 829 031 033 896 96 × 2 = 1 + 0.959 658 062 067 793 92;
  • 48) 0.959 658 062 067 793 92 × 2 = 1 + 0.919 316 124 135 587 84;
  • 49) 0.919 316 124 135 587 84 × 2 = 1 + 0.838 632 248 271 175 68;
  • 50) 0.838 632 248 271 175 68 × 2 = 1 + 0.677 264 496 542 351 36;
  • 51) 0.677 264 496 542 351 36 × 2 = 1 + 0.354 528 993 084 702 72;
  • 52) 0.354 528 993 084 702 72 × 2 = 0 + 0.709 057 986 169 405 44;
  • 53) 0.709 057 986 169 405 44 × 2 = 1 + 0.418 115 972 338 810 88;
  • 54) 0.418 115 972 338 810 88 × 2 = 0 + 0.836 231 944 677 621 76;

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:


0.000 000 000 742 147 39(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 39(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(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:


0.000 000 000 742 147 39(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2) × 20 =

1.1001 0111 1111 1111 1111 010(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 0111 1111 1111 1111 010


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 1011 1111 1111 1111 1010 =

100 1011 1111 1111 1111 1010


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 1011 1111 1111 1111 1010


Decimal number -0.000 000 000 742 147 39 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1010

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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