-0.000 000 000 742 147 672 81 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 672 81(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 81(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 81| = 0.000 000 000 742 147 672 81


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 81.

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 81 × 2 = 0 + 0.000 000 001 484 295 345 62;
  • 2) 0.000 000 001 484 295 345 62 × 2 = 0 + 0.000 000 002 968 590 691 24;
  • 3) 0.000 000 002 968 590 691 24 × 2 = 0 + 0.000 000 005 937 181 382 48;
  • 4) 0.000 000 005 937 181 382 48 × 2 = 0 + 0.000 000 011 874 362 764 96;
  • 5) 0.000 000 011 874 362 764 96 × 2 = 0 + 0.000 000 023 748 725 529 92;
  • 6) 0.000 000 023 748 725 529 92 × 2 = 0 + 0.000 000 047 497 451 059 84;
  • 7) 0.000 000 047 497 451 059 84 × 2 = 0 + 0.000 000 094 994 902 119 68;
  • 8) 0.000 000 094 994 902 119 68 × 2 = 0 + 0.000 000 189 989 804 239 36;
  • 9) 0.000 000 189 989 804 239 36 × 2 = 0 + 0.000 000 379 979 608 478 72;
  • 10) 0.000 000 379 979 608 478 72 × 2 = 0 + 0.000 000 759 959 216 957 44;
  • 11) 0.000 000 759 959 216 957 44 × 2 = 0 + 0.000 001 519 918 433 914 88;
  • 12) 0.000 001 519 918 433 914 88 × 2 = 0 + 0.000 003 039 836 867 829 76;
  • 13) 0.000 003 039 836 867 829 76 × 2 = 0 + 0.000 006 079 673 735 659 52;
  • 14) 0.000 006 079 673 735 659 52 × 2 = 0 + 0.000 012 159 347 471 319 04;
  • 15) 0.000 012 159 347 471 319 04 × 2 = 0 + 0.000 024 318 694 942 638 08;
  • 16) 0.000 024 318 694 942 638 08 × 2 = 0 + 0.000 048 637 389 885 276 16;
  • 17) 0.000 048 637 389 885 276 16 × 2 = 0 + 0.000 097 274 779 770 552 32;
  • 18) 0.000 097 274 779 770 552 32 × 2 = 0 + 0.000 194 549 559 541 104 64;
  • 19) 0.000 194 549 559 541 104 64 × 2 = 0 + 0.000 389 099 119 082 209 28;
  • 20) 0.000 389 099 119 082 209 28 × 2 = 0 + 0.000 778 198 238 164 418 56;
  • 21) 0.000 778 198 238 164 418 56 × 2 = 0 + 0.001 556 396 476 328 837 12;
  • 22) 0.001 556 396 476 328 837 12 × 2 = 0 + 0.003 112 792 952 657 674 24;
  • 23) 0.003 112 792 952 657 674 24 × 2 = 0 + 0.006 225 585 905 315 348 48;
  • 24) 0.006 225 585 905 315 348 48 × 2 = 0 + 0.012 451 171 810 630 696 96;
  • 25) 0.012 451 171 810 630 696 96 × 2 = 0 + 0.024 902 343 621 261 393 92;
  • 26) 0.024 902 343 621 261 393 92 × 2 = 0 + 0.049 804 687 242 522 787 84;
  • 27) 0.049 804 687 242 522 787 84 × 2 = 0 + 0.099 609 374 485 045 575 68;
  • 28) 0.099 609 374 485 045 575 68 × 2 = 0 + 0.199 218 748 970 091 151 36;
  • 29) 0.199 218 748 970 091 151 36 × 2 = 0 + 0.398 437 497 940 182 302 72;
  • 30) 0.398 437 497 940 182 302 72 × 2 = 0 + 0.796 874 995 880 364 605 44;
  • 31) 0.796 874 995 880 364 605 44 × 2 = 1 + 0.593 749 991 760 729 210 88;
  • 32) 0.593 749 991 760 729 210 88 × 2 = 1 + 0.187 499 983 521 458 421 76;
  • 33) 0.187 499 983 521 458 421 76 × 2 = 0 + 0.374 999 967 042 916 843 52;
  • 34) 0.374 999 967 042 916 843 52 × 2 = 0 + 0.749 999 934 085 833 687 04;
  • 35) 0.749 999 934 085 833 687 04 × 2 = 1 + 0.499 999 868 171 667 374 08;
  • 36) 0.499 999 868 171 667 374 08 × 2 = 0 + 0.999 999 736 343 334 748 16;
  • 37) 0.999 999 736 343 334 748 16 × 2 = 1 + 0.999 999 472 686 669 496 32;
  • 38) 0.999 999 472 686 669 496 32 × 2 = 1 + 0.999 998 945 373 338 992 64;
  • 39) 0.999 998 945 373 338 992 64 × 2 = 1 + 0.999 997 890 746 677 985 28;
  • 40) 0.999 997 890 746 677 985 28 × 2 = 1 + 0.999 995 781 493 355 970 56;
  • 41) 0.999 995 781 493 355 970 56 × 2 = 1 + 0.999 991 562 986 711 941 12;
  • 42) 0.999 991 562 986 711 941 12 × 2 = 1 + 0.999 983 125 973 423 882 24;
  • 43) 0.999 983 125 973 423 882 24 × 2 = 1 + 0.999 966 251 946 847 764 48;
  • 44) 0.999 966 251 946 847 764 48 × 2 = 1 + 0.999 932 503 893 695 528 96;
  • 45) 0.999 932 503 893 695 528 96 × 2 = 1 + 0.999 865 007 787 391 057 92;
  • 46) 0.999 865 007 787 391 057 92 × 2 = 1 + 0.999 730 015 574 782 115 84;
  • 47) 0.999 730 015 574 782 115 84 × 2 = 1 + 0.999 460 031 149 564 231 68;
  • 48) 0.999 460 031 149 564 231 68 × 2 = 1 + 0.998 920 062 299 128 463 36;
  • 49) 0.998 920 062 299 128 463 36 × 2 = 1 + 0.997 840 124 598 256 926 72;
  • 50) 0.997 840 124 598 256 926 72 × 2 = 1 + 0.995 680 249 196 513 853 44;
  • 51) 0.995 680 249 196 513 853 44 × 2 = 1 + 0.991 360 498 393 027 706 88;
  • 52) 0.991 360 498 393 027 706 88 × 2 = 1 + 0.982 720 996 786 055 413 76;
  • 53) 0.982 720 996 786 055 413 76 × 2 = 1 + 0.965 441 993 572 110 827 52;
  • 54) 0.965 441 993 572 110 827 52 × 2 = 1 + 0.930 883 987 144 221 655 04;

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 81(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 81(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 81(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 81 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