-0.000 281 986 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 986 7(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
-0.000 281 986 7(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:

|-0.000 281 986 7| = 0.000 281 986 7


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: 0.000 281 986 7.

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) 0.000 281 986 7 × 2 = 0 + 0.000 563 973 4;
  • 2) 0.000 563 973 4 × 2 = 0 + 0.001 127 946 8;
  • 3) 0.001 127 946 8 × 2 = 0 + 0.002 255 893 6;
  • 4) 0.002 255 893 6 × 2 = 0 + 0.004 511 787 2;
  • 5) 0.004 511 787 2 × 2 = 0 + 0.009 023 574 4;
  • 6) 0.009 023 574 4 × 2 = 0 + 0.018 047 148 8;
  • 7) 0.018 047 148 8 × 2 = 0 + 0.036 094 297 6;
  • 8) 0.036 094 297 6 × 2 = 0 + 0.072 188 595 2;
  • 9) 0.072 188 595 2 × 2 = 0 + 0.144 377 190 4;
  • 10) 0.144 377 190 4 × 2 = 0 + 0.288 754 380 8;
  • 11) 0.288 754 380 8 × 2 = 0 + 0.577 508 761 6;
  • 12) 0.577 508 761 6 × 2 = 1 + 0.155 017 523 2;
  • 13) 0.155 017 523 2 × 2 = 0 + 0.310 035 046 4;
  • 14) 0.310 035 046 4 × 2 = 0 + 0.620 070 092 8;
  • 15) 0.620 070 092 8 × 2 = 1 + 0.240 140 185 6;
  • 16) 0.240 140 185 6 × 2 = 0 + 0.480 280 371 2;
  • 17) 0.480 280 371 2 × 2 = 0 + 0.960 560 742 4;
  • 18) 0.960 560 742 4 × 2 = 1 + 0.921 121 484 8;
  • 19) 0.921 121 484 8 × 2 = 1 + 0.842 242 969 6;
  • 20) 0.842 242 969 6 × 2 = 1 + 0.684 485 939 2;
  • 21) 0.684 485 939 2 × 2 = 1 + 0.368 971 878 4;
  • 22) 0.368 971 878 4 × 2 = 0 + 0.737 943 756 8;
  • 23) 0.737 943 756 8 × 2 = 1 + 0.475 887 513 6;
  • 24) 0.475 887 513 6 × 2 = 0 + 0.951 775 027 2;
  • 25) 0.951 775 027 2 × 2 = 1 + 0.903 550 054 4;
  • 26) 0.903 550 054 4 × 2 = 1 + 0.807 100 108 8;
  • 27) 0.807 100 108 8 × 2 = 1 + 0.614 200 217 6;
  • 28) 0.614 200 217 6 × 2 = 1 + 0.228 400 435 2;
  • 29) 0.228 400 435 2 × 2 = 0 + 0.456 800 870 4;
  • 30) 0.456 800 870 4 × 2 = 0 + 0.913 601 740 8;
  • 31) 0.913 601 740 8 × 2 = 1 + 0.827 203 481 6;
  • 32) 0.827 203 481 6 × 2 = 1 + 0.654 406 963 2;
  • 33) 0.654 406 963 2 × 2 = 1 + 0.308 813 926 4;
  • 34) 0.308 813 926 4 × 2 = 0 + 0.617 627 852 8;
  • 35) 0.617 627 852 8 × 2 = 1 + 0.235 255 705 6;
  • 36) 0.235 255 705 6 × 2 = 0 + 0.470 511 411 2;
  • 37) 0.470 511 411 2 × 2 = 0 + 0.941 022 822 4;
  • 38) 0.941 022 822 4 × 2 = 1 + 0.882 045 644 8;
  • 39) 0.882 045 644 8 × 2 = 1 + 0.764 091 289 6;
  • 40) 0.764 091 289 6 × 2 = 1 + 0.528 182 579 2;
  • 41) 0.528 182 579 2 × 2 = 1 + 0.056 365 158 4;
  • 42) 0.056 365 158 4 × 2 = 0 + 0.112 730 316 8;
  • 43) 0.112 730 316 8 × 2 = 0 + 0.225 460 633 6;
  • 44) 0.225 460 633 6 × 2 = 0 + 0.450 921 267 2;
  • 45) 0.450 921 267 2 × 2 = 0 + 0.901 842 534 4;
  • 46) 0.901 842 534 4 × 2 = 1 + 0.803 685 068 8;
  • 47) 0.803 685 068 8 × 2 = 1 + 0.607 370 137 6;
  • 48) 0.607 370 137 6 × 2 = 1 + 0.214 740 275 2;
  • 49) 0.214 740 275 2 × 2 = 0 + 0.429 480 550 4;
  • 50) 0.429 480 550 4 × 2 = 0 + 0.858 961 100 8;
  • 51) 0.858 961 100 8 × 2 = 1 + 0.717 922 201 6;
  • 52) 0.717 922 201 6 × 2 = 1 + 0.435 844 403 2;
  • 53) 0.435 844 403 2 × 2 = 0 + 0.871 688 806 4;
  • 54) 0.871 688 806 4 × 2 = 1 + 0.743 377 612 8;
  • 55) 0.743 377 612 8 × 2 = 1 + 0.486 755 225 6;
  • 56) 0.486 755 225 6 × 2 = 0 + 0.973 510 451 2;
  • 57) 0.973 510 451 2 × 2 = 1 + 0.947 020 902 4;
  • 58) 0.947 020 902 4 × 2 = 1 + 0.894 041 804 8;
  • 59) 0.894 041 804 8 × 2 = 1 + 0.788 083 609 6;
  • 60) 0.788 083 609 6 × 2 = 1 + 0.576 167 219 2;
  • 61) 0.576 167 219 2 × 2 = 1 + 0.152 334 438 4;
  • 62) 0.152 334 438 4 × 2 = 0 + 0.304 668 876 8;
  • 63) 0.304 668 876 8 × 2 = 0 + 0.609 337 753 6;
  • 64) 0.609 337 753 6 × 2 = 1 + 0.218 675 507 2;

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:


0.000 281 986 7(10) =

0.0000 0000 0001 0010 0111 1010 1111 0011 1010 0111 1000 0111 0011 0110 1111 1001(2)

6. Positive number before normalization:

0.000 281 986 7(10) =

0.0000 0000 0001 0010 0111 1010 1111 0011 1010 0111 1000 0111 0011 0110 1111 1001(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:


0.000 281 986 7(10) =

0.0000 0000 0001 0010 0111 1010 1111 0011 1010 0111 1000 0111 0011 0110 1111 1001(2) =

0.0000 0000 0001 0010 0111 1010 1111 0011 1010 0111 1000 0111 0011 0110 1111 1001(2) × 20 =

1.0010 0111 1010 1111 0011 1010 0111 1000 0111 0011 0110 1111 1001(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 1010 1111 0011 1010 0111 1000 0111 0011 0110 1111 1001


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 1010 1111 0011 1010 0111 1000 0111 0011 0110 1111 1001 =

0010 0111 1010 1111 0011 1010 0111 1000 0111 0011 0110 1111 1001


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 1010 1111 0011 1010 0111 1000 0111 0011 0110 1111 1001


Decimal number -0.000 281 986 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1010 1111 0011 1010 0111 1000 0111 0011 0110 1111 1001

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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