-0.000 000 000 742 147 675 82 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 82(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
-0.000 000 000 742 147 675 82(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:

|-0.000 000 000 742 147 675 82| = 0.000 000 000 742 147 675 82


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: 0.000 000 000 742 147 675 82.

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) 0.000 000 000 742 147 675 82 × 2 = 0 + 0.000 000 001 484 295 351 64;
  • 2) 0.000 000 001 484 295 351 64 × 2 = 0 + 0.000 000 002 968 590 703 28;
  • 3) 0.000 000 002 968 590 703 28 × 2 = 0 + 0.000 000 005 937 181 406 56;
  • 4) 0.000 000 005 937 181 406 56 × 2 = 0 + 0.000 000 011 874 362 813 12;
  • 5) 0.000 000 011 874 362 813 12 × 2 = 0 + 0.000 000 023 748 725 626 24;
  • 6) 0.000 000 023 748 725 626 24 × 2 = 0 + 0.000 000 047 497 451 252 48;
  • 7) 0.000 000 047 497 451 252 48 × 2 = 0 + 0.000 000 094 994 902 504 96;
  • 8) 0.000 000 094 994 902 504 96 × 2 = 0 + 0.000 000 189 989 805 009 92;
  • 9) 0.000 000 189 989 805 009 92 × 2 = 0 + 0.000 000 379 979 610 019 84;
  • 10) 0.000 000 379 979 610 019 84 × 2 = 0 + 0.000 000 759 959 220 039 68;
  • 11) 0.000 000 759 959 220 039 68 × 2 = 0 + 0.000 001 519 918 440 079 36;
  • 12) 0.000 001 519 918 440 079 36 × 2 = 0 + 0.000 003 039 836 880 158 72;
  • 13) 0.000 003 039 836 880 158 72 × 2 = 0 + 0.000 006 079 673 760 317 44;
  • 14) 0.000 006 079 673 760 317 44 × 2 = 0 + 0.000 012 159 347 520 634 88;
  • 15) 0.000 012 159 347 520 634 88 × 2 = 0 + 0.000 024 318 695 041 269 76;
  • 16) 0.000 024 318 695 041 269 76 × 2 = 0 + 0.000 048 637 390 082 539 52;
  • 17) 0.000 048 637 390 082 539 52 × 2 = 0 + 0.000 097 274 780 165 079 04;
  • 18) 0.000 097 274 780 165 079 04 × 2 = 0 + 0.000 194 549 560 330 158 08;
  • 19) 0.000 194 549 560 330 158 08 × 2 = 0 + 0.000 389 099 120 660 316 16;
  • 20) 0.000 389 099 120 660 316 16 × 2 = 0 + 0.000 778 198 241 320 632 32;
  • 21) 0.000 778 198 241 320 632 32 × 2 = 0 + 0.001 556 396 482 641 264 64;
  • 22) 0.001 556 396 482 641 264 64 × 2 = 0 + 0.003 112 792 965 282 529 28;
  • 23) 0.003 112 792 965 282 529 28 × 2 = 0 + 0.006 225 585 930 565 058 56;
  • 24) 0.006 225 585 930 565 058 56 × 2 = 0 + 0.012 451 171 861 130 117 12;
  • 25) 0.012 451 171 861 130 117 12 × 2 = 0 + 0.024 902 343 722 260 234 24;
  • 26) 0.024 902 343 722 260 234 24 × 2 = 0 + 0.049 804 687 444 520 468 48;
  • 27) 0.049 804 687 444 520 468 48 × 2 = 0 + 0.099 609 374 889 040 936 96;
  • 28) 0.099 609 374 889 040 936 96 × 2 = 0 + 0.199 218 749 778 081 873 92;
  • 29) 0.199 218 749 778 081 873 92 × 2 = 0 + 0.398 437 499 556 163 747 84;
  • 30) 0.398 437 499 556 163 747 84 × 2 = 0 + 0.796 874 999 112 327 495 68;
  • 31) 0.796 874 999 112 327 495 68 × 2 = 1 + 0.593 749 998 224 654 991 36;
  • 32) 0.593 749 998 224 654 991 36 × 2 = 1 + 0.187 499 996 449 309 982 72;
  • 33) 0.187 499 996 449 309 982 72 × 2 = 0 + 0.374 999 992 898 619 965 44;
  • 34) 0.374 999 992 898 619 965 44 × 2 = 0 + 0.749 999 985 797 239 930 88;
  • 35) 0.749 999 985 797 239 930 88 × 2 = 1 + 0.499 999 971 594 479 861 76;
  • 36) 0.499 999 971 594 479 861 76 × 2 = 0 + 0.999 999 943 188 959 723 52;
  • 37) 0.999 999 943 188 959 723 52 × 2 = 1 + 0.999 999 886 377 919 447 04;
  • 38) 0.999 999 886 377 919 447 04 × 2 = 1 + 0.999 999 772 755 838 894 08;
  • 39) 0.999 999 772 755 838 894 08 × 2 = 1 + 0.999 999 545 511 677 788 16;
  • 40) 0.999 999 545 511 677 788 16 × 2 = 1 + 0.999 999 091 023 355 576 32;
  • 41) 0.999 999 091 023 355 576 32 × 2 = 1 + 0.999 998 182 046 711 152 64;
  • 42) 0.999 998 182 046 711 152 64 × 2 = 1 + 0.999 996 364 093 422 305 28;
  • 43) 0.999 996 364 093 422 305 28 × 2 = 1 + 0.999 992 728 186 844 610 56;
  • 44) 0.999 992 728 186 844 610 56 × 2 = 1 + 0.999 985 456 373 689 221 12;
  • 45) 0.999 985 456 373 689 221 12 × 2 = 1 + 0.999 970 912 747 378 442 24;
  • 46) 0.999 970 912 747 378 442 24 × 2 = 1 + 0.999 941 825 494 756 884 48;
  • 47) 0.999 941 825 494 756 884 48 × 2 = 1 + 0.999 883 650 989 513 768 96;
  • 48) 0.999 883 650 989 513 768 96 × 2 = 1 + 0.999 767 301 979 027 537 92;
  • 49) 0.999 767 301 979 027 537 92 × 2 = 1 + 0.999 534 603 958 055 075 84;
  • 50) 0.999 534 603 958 055 075 84 × 2 = 1 + 0.999 069 207 916 110 151 68;
  • 51) 0.999 069 207 916 110 151 68 × 2 = 1 + 0.998 138 415 832 220 303 36;
  • 52) 0.998 138 415 832 220 303 36 × 2 = 1 + 0.996 276 831 664 440 606 72;
  • 53) 0.996 276 831 664 440 606 72 × 2 = 1 + 0.992 553 663 328 881 213 44;
  • 54) 0.992 553 663 328 881 213 44 × 2 = 1 + 0.985 107 326 657 762 426 88;

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:


0.000 000 000 742 147 675 82(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 82(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:


0.000 000 000 742 147 675 82(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 -0.000 000 000 742 147 675 82 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