-0.000 282 003 32 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 003 32(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 282 003 32(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 282 003 32| = 0.000 282 003 32


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: 0.000 282 003 32.

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) 0.000 282 003 32 × 2 = 0 + 0.000 564 006 64;
  • 2) 0.000 564 006 64 × 2 = 0 + 0.001 128 013 28;
  • 3) 0.001 128 013 28 × 2 = 0 + 0.002 256 026 56;
  • 4) 0.002 256 026 56 × 2 = 0 + 0.004 512 053 12;
  • 5) 0.004 512 053 12 × 2 = 0 + 0.009 024 106 24;
  • 6) 0.009 024 106 24 × 2 = 0 + 0.018 048 212 48;
  • 7) 0.018 048 212 48 × 2 = 0 + 0.036 096 424 96;
  • 8) 0.036 096 424 96 × 2 = 0 + 0.072 192 849 92;
  • 9) 0.072 192 849 92 × 2 = 0 + 0.144 385 699 84;
  • 10) 0.144 385 699 84 × 2 = 0 + 0.288 771 399 68;
  • 11) 0.288 771 399 68 × 2 = 0 + 0.577 542 799 36;
  • 12) 0.577 542 799 36 × 2 = 1 + 0.155 085 598 72;
  • 13) 0.155 085 598 72 × 2 = 0 + 0.310 171 197 44;
  • 14) 0.310 171 197 44 × 2 = 0 + 0.620 342 394 88;
  • 15) 0.620 342 394 88 × 2 = 1 + 0.240 684 789 76;
  • 16) 0.240 684 789 76 × 2 = 0 + 0.481 369 579 52;
  • 17) 0.481 369 579 52 × 2 = 0 + 0.962 739 159 04;
  • 18) 0.962 739 159 04 × 2 = 1 + 0.925 478 318 08;
  • 19) 0.925 478 318 08 × 2 = 1 + 0.850 956 636 16;
  • 20) 0.850 956 636 16 × 2 = 1 + 0.701 913 272 32;
  • 21) 0.701 913 272 32 × 2 = 1 + 0.403 826 544 64;
  • 22) 0.403 826 544 64 × 2 = 0 + 0.807 653 089 28;
  • 23) 0.807 653 089 28 × 2 = 1 + 0.615 306 178 56;
  • 24) 0.615 306 178 56 × 2 = 1 + 0.230 612 357 12;
  • 25) 0.230 612 357 12 × 2 = 0 + 0.461 224 714 24;
  • 26) 0.461 224 714 24 × 2 = 0 + 0.922 449 428 48;
  • 27) 0.922 449 428 48 × 2 = 1 + 0.844 898 856 96;
  • 28) 0.844 898 856 96 × 2 = 1 + 0.689 797 713 92;
  • 29) 0.689 797 713 92 × 2 = 1 + 0.379 595 427 84;
  • 30) 0.379 595 427 84 × 2 = 0 + 0.759 190 855 68;
  • 31) 0.759 190 855 68 × 2 = 1 + 0.518 381 711 36;
  • 32) 0.518 381 711 36 × 2 = 1 + 0.036 763 422 72;
  • 33) 0.036 763 422 72 × 2 = 0 + 0.073 526 845 44;
  • 34) 0.073 526 845 44 × 2 = 0 + 0.147 053 690 88;
  • 35) 0.147 053 690 88 × 2 = 0 + 0.294 107 381 76;
  • 36) 0.294 107 381 76 × 2 = 0 + 0.588 214 763 52;
  • 37) 0.588 214 763 52 × 2 = 1 + 0.176 429 527 04;
  • 38) 0.176 429 527 04 × 2 = 0 + 0.352 859 054 08;
  • 39) 0.352 859 054 08 × 2 = 0 + 0.705 718 108 16;
  • 40) 0.705 718 108 16 × 2 = 1 + 0.411 436 216 32;
  • 41) 0.411 436 216 32 × 2 = 0 + 0.822 872 432 64;
  • 42) 0.822 872 432 64 × 2 = 1 + 0.645 744 865 28;
  • 43) 0.645 744 865 28 × 2 = 1 + 0.291 489 730 56;
  • 44) 0.291 489 730 56 × 2 = 0 + 0.582 979 461 12;
  • 45) 0.582 979 461 12 × 2 = 1 + 0.165 958 922 24;
  • 46) 0.165 958 922 24 × 2 = 0 + 0.331 917 844 48;
  • 47) 0.331 917 844 48 × 2 = 0 + 0.663 835 688 96;
  • 48) 0.663 835 688 96 × 2 = 1 + 0.327 671 377 92;
  • 49) 0.327 671 377 92 × 2 = 0 + 0.655 342 755 84;
  • 50) 0.655 342 755 84 × 2 = 1 + 0.310 685 511 68;
  • 51) 0.310 685 511 68 × 2 = 0 + 0.621 371 023 36;
  • 52) 0.621 371 023 36 × 2 = 1 + 0.242 742 046 72;
  • 53) 0.242 742 046 72 × 2 = 0 + 0.485 484 093 44;
  • 54) 0.485 484 093 44 × 2 = 0 + 0.970 968 186 88;
  • 55) 0.970 968 186 88 × 2 = 1 + 0.941 936 373 76;
  • 56) 0.941 936 373 76 × 2 = 1 + 0.883 872 747 52;
  • 57) 0.883 872 747 52 × 2 = 1 + 0.767 745 495 04;
  • 58) 0.767 745 495 04 × 2 = 1 + 0.535 490 990 08;
  • 59) 0.535 490 990 08 × 2 = 1 + 0.070 981 980 16;
  • 60) 0.070 981 980 16 × 2 = 0 + 0.141 963 960 32;
  • 61) 0.141 963 960 32 × 2 = 0 + 0.283 927 920 64;
  • 62) 0.283 927 920 64 × 2 = 0 + 0.567 855 841 28;
  • 63) 0.567 855 841 28 × 2 = 1 + 0.135 711 682 56;
  • 64) 0.135 711 682 56 × 2 = 0 + 0.271 423 365 12;

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:


0.000 282 003 32(10) =

0.0000 0000 0001 0010 0111 1011 0011 1011 0000 1001 0110 1001 0101 0011 1110 0010(2)

6. Positive number before normalization:

0.000 282 003 32(10) =

0.0000 0000 0001 0010 0111 1011 0011 1011 0000 1001 0110 1001 0101 0011 1110 0010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 003 32(10) =

0.0000 0000 0001 0010 0111 1011 0011 1011 0000 1001 0110 1001 0101 0011 1110 0010(2) =

0.0000 0000 0001 0010 0111 1011 0011 1011 0000 1001 0110 1001 0101 0011 1110 0010(2) × 20 =

1.0010 0111 1011 0011 1011 0000 1001 0110 1001 0101 0011 1110 0010(2) × 2-12


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -12

Mantissa (not normalized):
1.0010 0111 1011 0011 1011 0000 1001 0110 1001 0101 0011 1110 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

1011(10) =

011 1111 0011(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 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0111 1011 0011 1011 0000 1001 0110 1001 0101 0011 1110 0010 =

0010 0111 1011 0011 1011 0000 1001 0110 1001 0101 0011 1110 0010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0011

Mantissa (52 bits) =
0010 0111 1011 0011 1011 0000 1001 0110 1001 0101 0011 1110 0010


Decimal number -0.000 282 003 32 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1011 0000 1001 0110 1001 0101 0011 1110 0010

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 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:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100