-0.000 000 000 742 147 675 87 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 87(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 87(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 87| = 0.000 000 000 742 147 675 87


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 87.

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 87 × 2 = 0 + 0.000 000 001 484 295 351 74;
  • 2) 0.000 000 001 484 295 351 74 × 2 = 0 + 0.000 000 002 968 590 703 48;
  • 3) 0.000 000 002 968 590 703 48 × 2 = 0 + 0.000 000 005 937 181 406 96;
  • 4) 0.000 000 005 937 181 406 96 × 2 = 0 + 0.000 000 011 874 362 813 92;
  • 5) 0.000 000 011 874 362 813 92 × 2 = 0 + 0.000 000 023 748 725 627 84;
  • 6) 0.000 000 023 748 725 627 84 × 2 = 0 + 0.000 000 047 497 451 255 68;
  • 7) 0.000 000 047 497 451 255 68 × 2 = 0 + 0.000 000 094 994 902 511 36;
  • 8) 0.000 000 094 994 902 511 36 × 2 = 0 + 0.000 000 189 989 805 022 72;
  • 9) 0.000 000 189 989 805 022 72 × 2 = 0 + 0.000 000 379 979 610 045 44;
  • 10) 0.000 000 379 979 610 045 44 × 2 = 0 + 0.000 000 759 959 220 090 88;
  • 11) 0.000 000 759 959 220 090 88 × 2 = 0 + 0.000 001 519 918 440 181 76;
  • 12) 0.000 001 519 918 440 181 76 × 2 = 0 + 0.000 003 039 836 880 363 52;
  • 13) 0.000 003 039 836 880 363 52 × 2 = 0 + 0.000 006 079 673 760 727 04;
  • 14) 0.000 006 079 673 760 727 04 × 2 = 0 + 0.000 012 159 347 521 454 08;
  • 15) 0.000 012 159 347 521 454 08 × 2 = 0 + 0.000 024 318 695 042 908 16;
  • 16) 0.000 024 318 695 042 908 16 × 2 = 0 + 0.000 048 637 390 085 816 32;
  • 17) 0.000 048 637 390 085 816 32 × 2 = 0 + 0.000 097 274 780 171 632 64;
  • 18) 0.000 097 274 780 171 632 64 × 2 = 0 + 0.000 194 549 560 343 265 28;
  • 19) 0.000 194 549 560 343 265 28 × 2 = 0 + 0.000 389 099 120 686 530 56;
  • 20) 0.000 389 099 120 686 530 56 × 2 = 0 + 0.000 778 198 241 373 061 12;
  • 21) 0.000 778 198 241 373 061 12 × 2 = 0 + 0.001 556 396 482 746 122 24;
  • 22) 0.001 556 396 482 746 122 24 × 2 = 0 + 0.003 112 792 965 492 244 48;
  • 23) 0.003 112 792 965 492 244 48 × 2 = 0 + 0.006 225 585 930 984 488 96;
  • 24) 0.006 225 585 930 984 488 96 × 2 = 0 + 0.012 451 171 861 968 977 92;
  • 25) 0.012 451 171 861 968 977 92 × 2 = 0 + 0.024 902 343 723 937 955 84;
  • 26) 0.024 902 343 723 937 955 84 × 2 = 0 + 0.049 804 687 447 875 911 68;
  • 27) 0.049 804 687 447 875 911 68 × 2 = 0 + 0.099 609 374 895 751 823 36;
  • 28) 0.099 609 374 895 751 823 36 × 2 = 0 + 0.199 218 749 791 503 646 72;
  • 29) 0.199 218 749 791 503 646 72 × 2 = 0 + 0.398 437 499 583 007 293 44;
  • 30) 0.398 437 499 583 007 293 44 × 2 = 0 + 0.796 874 999 166 014 586 88;
  • 31) 0.796 874 999 166 014 586 88 × 2 = 1 + 0.593 749 998 332 029 173 76;
  • 32) 0.593 749 998 332 029 173 76 × 2 = 1 + 0.187 499 996 664 058 347 52;
  • 33) 0.187 499 996 664 058 347 52 × 2 = 0 + 0.374 999 993 328 116 695 04;
  • 34) 0.374 999 993 328 116 695 04 × 2 = 0 + 0.749 999 986 656 233 390 08;
  • 35) 0.749 999 986 656 233 390 08 × 2 = 1 + 0.499 999 973 312 466 780 16;
  • 36) 0.499 999 973 312 466 780 16 × 2 = 0 + 0.999 999 946 624 933 560 32;
  • 37) 0.999 999 946 624 933 560 32 × 2 = 1 + 0.999 999 893 249 867 120 64;
  • 38) 0.999 999 893 249 867 120 64 × 2 = 1 + 0.999 999 786 499 734 241 28;
  • 39) 0.999 999 786 499 734 241 28 × 2 = 1 + 0.999 999 572 999 468 482 56;
  • 40) 0.999 999 572 999 468 482 56 × 2 = 1 + 0.999 999 145 998 936 965 12;
  • 41) 0.999 999 145 998 936 965 12 × 2 = 1 + 0.999 998 291 997 873 930 24;
  • 42) 0.999 998 291 997 873 930 24 × 2 = 1 + 0.999 996 583 995 747 860 48;
  • 43) 0.999 996 583 995 747 860 48 × 2 = 1 + 0.999 993 167 991 495 720 96;
  • 44) 0.999 993 167 991 495 720 96 × 2 = 1 + 0.999 986 335 982 991 441 92;
  • 45) 0.999 986 335 982 991 441 92 × 2 = 1 + 0.999 972 671 965 982 883 84;
  • 46) 0.999 972 671 965 982 883 84 × 2 = 1 + 0.999 945 343 931 965 767 68;
  • 47) 0.999 945 343 931 965 767 68 × 2 = 1 + 0.999 890 687 863 931 535 36;
  • 48) 0.999 890 687 863 931 535 36 × 2 = 1 + 0.999 781 375 727 863 070 72;
  • 49) 0.999 781 375 727 863 070 72 × 2 = 1 + 0.999 562 751 455 726 141 44;
  • 50) 0.999 562 751 455 726 141 44 × 2 = 1 + 0.999 125 502 911 452 282 88;
  • 51) 0.999 125 502 911 452 282 88 × 2 = 1 + 0.998 251 005 822 904 565 76;
  • 52) 0.998 251 005 822 904 565 76 × 2 = 1 + 0.996 502 011 645 809 131 52;
  • 53) 0.996 502 011 645 809 131 52 × 2 = 1 + 0.993 004 023 291 618 263 04;
  • 54) 0.993 004 023 291 618 263 04 × 2 = 1 + 0.986 008 046 583 236 526 08;

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 87(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 87(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 87(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 87 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