-0.000 000 000 742 147 676 19 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 19(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 19(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 19| = 0.000 000 000 742 147 676 19


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 19.

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 19 × 2 = 0 + 0.000 000 001 484 295 352 38;
  • 2) 0.000 000 001 484 295 352 38 × 2 = 0 + 0.000 000 002 968 590 704 76;
  • 3) 0.000 000 002 968 590 704 76 × 2 = 0 + 0.000 000 005 937 181 409 52;
  • 4) 0.000 000 005 937 181 409 52 × 2 = 0 + 0.000 000 011 874 362 819 04;
  • 5) 0.000 000 011 874 362 819 04 × 2 = 0 + 0.000 000 023 748 725 638 08;
  • 6) 0.000 000 023 748 725 638 08 × 2 = 0 + 0.000 000 047 497 451 276 16;
  • 7) 0.000 000 047 497 451 276 16 × 2 = 0 + 0.000 000 094 994 902 552 32;
  • 8) 0.000 000 094 994 902 552 32 × 2 = 0 + 0.000 000 189 989 805 104 64;
  • 9) 0.000 000 189 989 805 104 64 × 2 = 0 + 0.000 000 379 979 610 209 28;
  • 10) 0.000 000 379 979 610 209 28 × 2 = 0 + 0.000 000 759 959 220 418 56;
  • 11) 0.000 000 759 959 220 418 56 × 2 = 0 + 0.000 001 519 918 440 837 12;
  • 12) 0.000 001 519 918 440 837 12 × 2 = 0 + 0.000 003 039 836 881 674 24;
  • 13) 0.000 003 039 836 881 674 24 × 2 = 0 + 0.000 006 079 673 763 348 48;
  • 14) 0.000 006 079 673 763 348 48 × 2 = 0 + 0.000 012 159 347 526 696 96;
  • 15) 0.000 012 159 347 526 696 96 × 2 = 0 + 0.000 024 318 695 053 393 92;
  • 16) 0.000 024 318 695 053 393 92 × 2 = 0 + 0.000 048 637 390 106 787 84;
  • 17) 0.000 048 637 390 106 787 84 × 2 = 0 + 0.000 097 274 780 213 575 68;
  • 18) 0.000 097 274 780 213 575 68 × 2 = 0 + 0.000 194 549 560 427 151 36;
  • 19) 0.000 194 549 560 427 151 36 × 2 = 0 + 0.000 389 099 120 854 302 72;
  • 20) 0.000 389 099 120 854 302 72 × 2 = 0 + 0.000 778 198 241 708 605 44;
  • 21) 0.000 778 198 241 708 605 44 × 2 = 0 + 0.001 556 396 483 417 210 88;
  • 22) 0.001 556 396 483 417 210 88 × 2 = 0 + 0.003 112 792 966 834 421 76;
  • 23) 0.003 112 792 966 834 421 76 × 2 = 0 + 0.006 225 585 933 668 843 52;
  • 24) 0.006 225 585 933 668 843 52 × 2 = 0 + 0.012 451 171 867 337 687 04;
  • 25) 0.012 451 171 867 337 687 04 × 2 = 0 + 0.024 902 343 734 675 374 08;
  • 26) 0.024 902 343 734 675 374 08 × 2 = 0 + 0.049 804 687 469 350 748 16;
  • 27) 0.049 804 687 469 350 748 16 × 2 = 0 + 0.099 609 374 938 701 496 32;
  • 28) 0.099 609 374 938 701 496 32 × 2 = 0 + 0.199 218 749 877 402 992 64;
  • 29) 0.199 218 749 877 402 992 64 × 2 = 0 + 0.398 437 499 754 805 985 28;
  • 30) 0.398 437 499 754 805 985 28 × 2 = 0 + 0.796 874 999 509 611 970 56;
  • 31) 0.796 874 999 509 611 970 56 × 2 = 1 + 0.593 749 999 019 223 941 12;
  • 32) 0.593 749 999 019 223 941 12 × 2 = 1 + 0.187 499 998 038 447 882 24;
  • 33) 0.187 499 998 038 447 882 24 × 2 = 0 + 0.374 999 996 076 895 764 48;
  • 34) 0.374 999 996 076 895 764 48 × 2 = 0 + 0.749 999 992 153 791 528 96;
  • 35) 0.749 999 992 153 791 528 96 × 2 = 1 + 0.499 999 984 307 583 057 92;
  • 36) 0.499 999 984 307 583 057 92 × 2 = 0 + 0.999 999 968 615 166 115 84;
  • 37) 0.999 999 968 615 166 115 84 × 2 = 1 + 0.999 999 937 230 332 231 68;
  • 38) 0.999 999 937 230 332 231 68 × 2 = 1 + 0.999 999 874 460 664 463 36;
  • 39) 0.999 999 874 460 664 463 36 × 2 = 1 + 0.999 999 748 921 328 926 72;
  • 40) 0.999 999 748 921 328 926 72 × 2 = 1 + 0.999 999 497 842 657 853 44;
  • 41) 0.999 999 497 842 657 853 44 × 2 = 1 + 0.999 998 995 685 315 706 88;
  • 42) 0.999 998 995 685 315 706 88 × 2 = 1 + 0.999 997 991 370 631 413 76;
  • 43) 0.999 997 991 370 631 413 76 × 2 = 1 + 0.999 995 982 741 262 827 52;
  • 44) 0.999 995 982 741 262 827 52 × 2 = 1 + 0.999 991 965 482 525 655 04;
  • 45) 0.999 991 965 482 525 655 04 × 2 = 1 + 0.999 983 930 965 051 310 08;
  • 46) 0.999 983 930 965 051 310 08 × 2 = 1 + 0.999 967 861 930 102 620 16;
  • 47) 0.999 967 861 930 102 620 16 × 2 = 1 + 0.999 935 723 860 205 240 32;
  • 48) 0.999 935 723 860 205 240 32 × 2 = 1 + 0.999 871 447 720 410 480 64;
  • 49) 0.999 871 447 720 410 480 64 × 2 = 1 + 0.999 742 895 440 820 961 28;
  • 50) 0.999 742 895 440 820 961 28 × 2 = 1 + 0.999 485 790 881 641 922 56;
  • 51) 0.999 485 790 881 641 922 56 × 2 = 1 + 0.998 971 581 763 283 845 12;
  • 52) 0.998 971 581 763 283 845 12 × 2 = 1 + 0.997 943 163 526 567 690 24;
  • 53) 0.997 943 163 526 567 690 24 × 2 = 1 + 0.995 886 327 053 135 380 48;
  • 54) 0.995 886 327 053 135 380 48 × 2 = 1 + 0.991 772 654 106 270 760 96;

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 19(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 19(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 19(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 19 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