-0.000 000 000 742 147 676 515 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 515(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
-0.000 000 000 742 147 676 515(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:

|-0.000 000 000 742 147 676 515| = 0.000 000 000 742 147 676 515


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: 0.000 000 000 742 147 676 515.

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) 0.000 000 000 742 147 676 515 × 2 = 0 + 0.000 000 001 484 295 353 03;
  • 2) 0.000 000 001 484 295 353 03 × 2 = 0 + 0.000 000 002 968 590 706 06;
  • 3) 0.000 000 002 968 590 706 06 × 2 = 0 + 0.000 000 005 937 181 412 12;
  • 4) 0.000 000 005 937 181 412 12 × 2 = 0 + 0.000 000 011 874 362 824 24;
  • 5) 0.000 000 011 874 362 824 24 × 2 = 0 + 0.000 000 023 748 725 648 48;
  • 6) 0.000 000 023 748 725 648 48 × 2 = 0 + 0.000 000 047 497 451 296 96;
  • 7) 0.000 000 047 497 451 296 96 × 2 = 0 + 0.000 000 094 994 902 593 92;
  • 8) 0.000 000 094 994 902 593 92 × 2 = 0 + 0.000 000 189 989 805 187 84;
  • 9) 0.000 000 189 989 805 187 84 × 2 = 0 + 0.000 000 379 979 610 375 68;
  • 10) 0.000 000 379 979 610 375 68 × 2 = 0 + 0.000 000 759 959 220 751 36;
  • 11) 0.000 000 759 959 220 751 36 × 2 = 0 + 0.000 001 519 918 441 502 72;
  • 12) 0.000 001 519 918 441 502 72 × 2 = 0 + 0.000 003 039 836 883 005 44;
  • 13) 0.000 003 039 836 883 005 44 × 2 = 0 + 0.000 006 079 673 766 010 88;
  • 14) 0.000 006 079 673 766 010 88 × 2 = 0 + 0.000 012 159 347 532 021 76;
  • 15) 0.000 012 159 347 532 021 76 × 2 = 0 + 0.000 024 318 695 064 043 52;
  • 16) 0.000 024 318 695 064 043 52 × 2 = 0 + 0.000 048 637 390 128 087 04;
  • 17) 0.000 048 637 390 128 087 04 × 2 = 0 + 0.000 097 274 780 256 174 08;
  • 18) 0.000 097 274 780 256 174 08 × 2 = 0 + 0.000 194 549 560 512 348 16;
  • 19) 0.000 194 549 560 512 348 16 × 2 = 0 + 0.000 389 099 121 024 696 32;
  • 20) 0.000 389 099 121 024 696 32 × 2 = 0 + 0.000 778 198 242 049 392 64;
  • 21) 0.000 778 198 242 049 392 64 × 2 = 0 + 0.001 556 396 484 098 785 28;
  • 22) 0.001 556 396 484 098 785 28 × 2 = 0 + 0.003 112 792 968 197 570 56;
  • 23) 0.003 112 792 968 197 570 56 × 2 = 0 + 0.006 225 585 936 395 141 12;
  • 24) 0.006 225 585 936 395 141 12 × 2 = 0 + 0.012 451 171 872 790 282 24;
  • 25) 0.012 451 171 872 790 282 24 × 2 = 0 + 0.024 902 343 745 580 564 48;
  • 26) 0.024 902 343 745 580 564 48 × 2 = 0 + 0.049 804 687 491 161 128 96;
  • 27) 0.049 804 687 491 161 128 96 × 2 = 0 + 0.099 609 374 982 322 257 92;
  • 28) 0.099 609 374 982 322 257 92 × 2 = 0 + 0.199 218 749 964 644 515 84;
  • 29) 0.199 218 749 964 644 515 84 × 2 = 0 + 0.398 437 499 929 289 031 68;
  • 30) 0.398 437 499 929 289 031 68 × 2 = 0 + 0.796 874 999 858 578 063 36;
  • 31) 0.796 874 999 858 578 063 36 × 2 = 1 + 0.593 749 999 717 156 126 72;
  • 32) 0.593 749 999 717 156 126 72 × 2 = 1 + 0.187 499 999 434 312 253 44;
  • 33) 0.187 499 999 434 312 253 44 × 2 = 0 + 0.374 999 998 868 624 506 88;
  • 34) 0.374 999 998 868 624 506 88 × 2 = 0 + 0.749 999 997 737 249 013 76;
  • 35) 0.749 999 997 737 249 013 76 × 2 = 1 + 0.499 999 995 474 498 027 52;
  • 36) 0.499 999 995 474 498 027 52 × 2 = 0 + 0.999 999 990 948 996 055 04;
  • 37) 0.999 999 990 948 996 055 04 × 2 = 1 + 0.999 999 981 897 992 110 08;
  • 38) 0.999 999 981 897 992 110 08 × 2 = 1 + 0.999 999 963 795 984 220 16;
  • 39) 0.999 999 963 795 984 220 16 × 2 = 1 + 0.999 999 927 591 968 440 32;
  • 40) 0.999 999 927 591 968 440 32 × 2 = 1 + 0.999 999 855 183 936 880 64;
  • 41) 0.999 999 855 183 936 880 64 × 2 = 1 + 0.999 999 710 367 873 761 28;
  • 42) 0.999 999 710 367 873 761 28 × 2 = 1 + 0.999 999 420 735 747 522 56;
  • 43) 0.999 999 420 735 747 522 56 × 2 = 1 + 0.999 998 841 471 495 045 12;
  • 44) 0.999 998 841 471 495 045 12 × 2 = 1 + 0.999 997 682 942 990 090 24;
  • 45) 0.999 997 682 942 990 090 24 × 2 = 1 + 0.999 995 365 885 980 180 48;
  • 46) 0.999 995 365 885 980 180 48 × 2 = 1 + 0.999 990 731 771 960 360 96;
  • 47) 0.999 990 731 771 960 360 96 × 2 = 1 + 0.999 981 463 543 920 721 92;
  • 48) 0.999 981 463 543 920 721 92 × 2 = 1 + 0.999 962 927 087 841 443 84;
  • 49) 0.999 962 927 087 841 443 84 × 2 = 1 + 0.999 925 854 175 682 887 68;
  • 50) 0.999 925 854 175 682 887 68 × 2 = 1 + 0.999 851 708 351 365 775 36;
  • 51) 0.999 851 708 351 365 775 36 × 2 = 1 + 0.999 703 416 702 731 550 72;
  • 52) 0.999 703 416 702 731 550 72 × 2 = 1 + 0.999 406 833 405 463 101 44;
  • 53) 0.999 406 833 405 463 101 44 × 2 = 1 + 0.998 813 666 810 926 202 88;
  • 54) 0.998 813 666 810 926 202 88 × 2 = 1 + 0.997 627 333 621 852 405 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:


0.000 000 000 742 147 676 515(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 515(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:


0.000 000 000 742 147 676 515(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 -0.000 000 000 742 147 676 515 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