-0.000 000 000 742 147 672 59 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 672 59(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
-0.000 000 000 742 147 672 59(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:

|-0.000 000 000 742 147 672 59| = 0.000 000 000 742 147 672 59


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: 0.000 000 000 742 147 672 59.

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) 0.000 000 000 742 147 672 59 × 2 = 0 + 0.000 000 001 484 295 345 18;
  • 2) 0.000 000 001 484 295 345 18 × 2 = 0 + 0.000 000 002 968 590 690 36;
  • 3) 0.000 000 002 968 590 690 36 × 2 = 0 + 0.000 000 005 937 181 380 72;
  • 4) 0.000 000 005 937 181 380 72 × 2 = 0 + 0.000 000 011 874 362 761 44;
  • 5) 0.000 000 011 874 362 761 44 × 2 = 0 + 0.000 000 023 748 725 522 88;
  • 6) 0.000 000 023 748 725 522 88 × 2 = 0 + 0.000 000 047 497 451 045 76;
  • 7) 0.000 000 047 497 451 045 76 × 2 = 0 + 0.000 000 094 994 902 091 52;
  • 8) 0.000 000 094 994 902 091 52 × 2 = 0 + 0.000 000 189 989 804 183 04;
  • 9) 0.000 000 189 989 804 183 04 × 2 = 0 + 0.000 000 379 979 608 366 08;
  • 10) 0.000 000 379 979 608 366 08 × 2 = 0 + 0.000 000 759 959 216 732 16;
  • 11) 0.000 000 759 959 216 732 16 × 2 = 0 + 0.000 001 519 918 433 464 32;
  • 12) 0.000 001 519 918 433 464 32 × 2 = 0 + 0.000 003 039 836 866 928 64;
  • 13) 0.000 003 039 836 866 928 64 × 2 = 0 + 0.000 006 079 673 733 857 28;
  • 14) 0.000 006 079 673 733 857 28 × 2 = 0 + 0.000 012 159 347 467 714 56;
  • 15) 0.000 012 159 347 467 714 56 × 2 = 0 + 0.000 024 318 694 935 429 12;
  • 16) 0.000 024 318 694 935 429 12 × 2 = 0 + 0.000 048 637 389 870 858 24;
  • 17) 0.000 048 637 389 870 858 24 × 2 = 0 + 0.000 097 274 779 741 716 48;
  • 18) 0.000 097 274 779 741 716 48 × 2 = 0 + 0.000 194 549 559 483 432 96;
  • 19) 0.000 194 549 559 483 432 96 × 2 = 0 + 0.000 389 099 118 966 865 92;
  • 20) 0.000 389 099 118 966 865 92 × 2 = 0 + 0.000 778 198 237 933 731 84;
  • 21) 0.000 778 198 237 933 731 84 × 2 = 0 + 0.001 556 396 475 867 463 68;
  • 22) 0.001 556 396 475 867 463 68 × 2 = 0 + 0.003 112 792 951 734 927 36;
  • 23) 0.003 112 792 951 734 927 36 × 2 = 0 + 0.006 225 585 903 469 854 72;
  • 24) 0.006 225 585 903 469 854 72 × 2 = 0 + 0.012 451 171 806 939 709 44;
  • 25) 0.012 451 171 806 939 709 44 × 2 = 0 + 0.024 902 343 613 879 418 88;
  • 26) 0.024 902 343 613 879 418 88 × 2 = 0 + 0.049 804 687 227 758 837 76;
  • 27) 0.049 804 687 227 758 837 76 × 2 = 0 + 0.099 609 374 455 517 675 52;
  • 28) 0.099 609 374 455 517 675 52 × 2 = 0 + 0.199 218 748 911 035 351 04;
  • 29) 0.199 218 748 911 035 351 04 × 2 = 0 + 0.398 437 497 822 070 702 08;
  • 30) 0.398 437 497 822 070 702 08 × 2 = 0 + 0.796 874 995 644 141 404 16;
  • 31) 0.796 874 995 644 141 404 16 × 2 = 1 + 0.593 749 991 288 282 808 32;
  • 32) 0.593 749 991 288 282 808 32 × 2 = 1 + 0.187 499 982 576 565 616 64;
  • 33) 0.187 499 982 576 565 616 64 × 2 = 0 + 0.374 999 965 153 131 233 28;
  • 34) 0.374 999 965 153 131 233 28 × 2 = 0 + 0.749 999 930 306 262 466 56;
  • 35) 0.749 999 930 306 262 466 56 × 2 = 1 + 0.499 999 860 612 524 933 12;
  • 36) 0.499 999 860 612 524 933 12 × 2 = 0 + 0.999 999 721 225 049 866 24;
  • 37) 0.999 999 721 225 049 866 24 × 2 = 1 + 0.999 999 442 450 099 732 48;
  • 38) 0.999 999 442 450 099 732 48 × 2 = 1 + 0.999 998 884 900 199 464 96;
  • 39) 0.999 998 884 900 199 464 96 × 2 = 1 + 0.999 997 769 800 398 929 92;
  • 40) 0.999 997 769 800 398 929 92 × 2 = 1 + 0.999 995 539 600 797 859 84;
  • 41) 0.999 995 539 600 797 859 84 × 2 = 1 + 0.999 991 079 201 595 719 68;
  • 42) 0.999 991 079 201 595 719 68 × 2 = 1 + 0.999 982 158 403 191 439 36;
  • 43) 0.999 982 158 403 191 439 36 × 2 = 1 + 0.999 964 316 806 382 878 72;
  • 44) 0.999 964 316 806 382 878 72 × 2 = 1 + 0.999 928 633 612 765 757 44;
  • 45) 0.999 928 633 612 765 757 44 × 2 = 1 + 0.999 857 267 225 531 514 88;
  • 46) 0.999 857 267 225 531 514 88 × 2 = 1 + 0.999 714 534 451 063 029 76;
  • 47) 0.999 714 534 451 063 029 76 × 2 = 1 + 0.999 429 068 902 126 059 52;
  • 48) 0.999 429 068 902 126 059 52 × 2 = 1 + 0.998 858 137 804 252 119 04;
  • 49) 0.998 858 137 804 252 119 04 × 2 = 1 + 0.997 716 275 608 504 238 08;
  • 50) 0.997 716 275 608 504 238 08 × 2 = 1 + 0.995 432 551 217 008 476 16;
  • 51) 0.995 432 551 217 008 476 16 × 2 = 1 + 0.990 865 102 434 016 952 32;
  • 52) 0.990 865 102 434 016 952 32 × 2 = 1 + 0.981 730 204 868 033 904 64;
  • 53) 0.981 730 204 868 033 904 64 × 2 = 1 + 0.963 460 409 736 067 809 28;
  • 54) 0.963 460 409 736 067 809 28 × 2 = 1 + 0.926 920 819 472 135 618 56;

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:


0.000 000 000 742 147 672 59(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 672 59(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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:


0.000 000 000 742 147 672 59(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 111


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 1111 =

100 1011 1111 1111 1111 1111


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 1111


Decimal number -0.000 000 000 742 147 672 59 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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