-0.000 282 005 895 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 895(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
-0.000 282 005 895(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:

|-0.000 282 005 895| = 0.000 282 005 895


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: 0.000 282 005 895.

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) 0.000 282 005 895 × 2 = 0 + 0.000 564 011 79;
  • 2) 0.000 564 011 79 × 2 = 0 + 0.001 128 023 58;
  • 3) 0.001 128 023 58 × 2 = 0 + 0.002 256 047 16;
  • 4) 0.002 256 047 16 × 2 = 0 + 0.004 512 094 32;
  • 5) 0.004 512 094 32 × 2 = 0 + 0.009 024 188 64;
  • 6) 0.009 024 188 64 × 2 = 0 + 0.018 048 377 28;
  • 7) 0.018 048 377 28 × 2 = 0 + 0.036 096 754 56;
  • 8) 0.036 096 754 56 × 2 = 0 + 0.072 193 509 12;
  • 9) 0.072 193 509 12 × 2 = 0 + 0.144 387 018 24;
  • 10) 0.144 387 018 24 × 2 = 0 + 0.288 774 036 48;
  • 11) 0.288 774 036 48 × 2 = 0 + 0.577 548 072 96;
  • 12) 0.577 548 072 96 × 2 = 1 + 0.155 096 145 92;
  • 13) 0.155 096 145 92 × 2 = 0 + 0.310 192 291 84;
  • 14) 0.310 192 291 84 × 2 = 0 + 0.620 384 583 68;
  • 15) 0.620 384 583 68 × 2 = 1 + 0.240 769 167 36;
  • 16) 0.240 769 167 36 × 2 = 0 + 0.481 538 334 72;
  • 17) 0.481 538 334 72 × 2 = 0 + 0.963 076 669 44;
  • 18) 0.963 076 669 44 × 2 = 1 + 0.926 153 338 88;
  • 19) 0.926 153 338 88 × 2 = 1 + 0.852 306 677 76;
  • 20) 0.852 306 677 76 × 2 = 1 + 0.704 613 355 52;
  • 21) 0.704 613 355 52 × 2 = 1 + 0.409 226 711 04;
  • 22) 0.409 226 711 04 × 2 = 0 + 0.818 453 422 08;
  • 23) 0.818 453 422 08 × 2 = 1 + 0.636 906 844 16;
  • 24) 0.636 906 844 16 × 2 = 1 + 0.273 813 688 32;
  • 25) 0.273 813 688 32 × 2 = 0 + 0.547 627 376 64;
  • 26) 0.547 627 376 64 × 2 = 1 + 0.095 254 753 28;
  • 27) 0.095 254 753 28 × 2 = 0 + 0.190 509 506 56;
  • 28) 0.190 509 506 56 × 2 = 0 + 0.381 019 013 12;
  • 29) 0.381 019 013 12 × 2 = 0 + 0.762 038 026 24;
  • 30) 0.762 038 026 24 × 2 = 1 + 0.524 076 052 48;
  • 31) 0.524 076 052 48 × 2 = 1 + 0.048 152 104 96;
  • 32) 0.048 152 104 96 × 2 = 0 + 0.096 304 209 92;
  • 33) 0.096 304 209 92 × 2 = 0 + 0.192 608 419 84;
  • 34) 0.192 608 419 84 × 2 = 0 + 0.385 216 839 68;
  • 35) 0.385 216 839 68 × 2 = 0 + 0.770 433 679 36;
  • 36) 0.770 433 679 36 × 2 = 1 + 0.540 867 358 72;
  • 37) 0.540 867 358 72 × 2 = 1 + 0.081 734 717 44;
  • 38) 0.081 734 717 44 × 2 = 0 + 0.163 469 434 88;
  • 39) 0.163 469 434 88 × 2 = 0 + 0.326 938 869 76;
  • 40) 0.326 938 869 76 × 2 = 0 + 0.653 877 739 52;
  • 41) 0.653 877 739 52 × 2 = 1 + 0.307 755 479 04;
  • 42) 0.307 755 479 04 × 2 = 0 + 0.615 510 958 08;
  • 43) 0.615 510 958 08 × 2 = 1 + 0.231 021 916 16;
  • 44) 0.231 021 916 16 × 2 = 0 + 0.462 043 832 32;
  • 45) 0.462 043 832 32 × 2 = 0 + 0.924 087 664 64;
  • 46) 0.924 087 664 64 × 2 = 1 + 0.848 175 329 28;
  • 47) 0.848 175 329 28 × 2 = 1 + 0.696 350 658 56;
  • 48) 0.696 350 658 56 × 2 = 1 + 0.392 701 317 12;
  • 49) 0.392 701 317 12 × 2 = 0 + 0.785 402 634 24;
  • 50) 0.785 402 634 24 × 2 = 1 + 0.570 805 268 48;
  • 51) 0.570 805 268 48 × 2 = 1 + 0.141 610 536 96;
  • 52) 0.141 610 536 96 × 2 = 0 + 0.283 221 073 92;
  • 53) 0.283 221 073 92 × 2 = 0 + 0.566 442 147 84;
  • 54) 0.566 442 147 84 × 2 = 1 + 0.132 884 295 68;
  • 55) 0.132 884 295 68 × 2 = 0 + 0.265 768 591 36;
  • 56) 0.265 768 591 36 × 2 = 0 + 0.531 537 182 72;
  • 57) 0.531 537 182 72 × 2 = 1 + 0.063 074 365 44;
  • 58) 0.063 074 365 44 × 2 = 0 + 0.126 148 730 88;
  • 59) 0.126 148 730 88 × 2 = 0 + 0.252 297 461 76;
  • 60) 0.252 297 461 76 × 2 = 0 + 0.504 594 923 52;
  • 61) 0.504 594 923 52 × 2 = 1 + 0.009 189 847 04;
  • 62) 0.009 189 847 04 × 2 = 0 + 0.018 379 694 08;
  • 63) 0.018 379 694 08 × 2 = 0 + 0.036 759 388 16;
  • 64) 0.036 759 388 16 × 2 = 0 + 0.073 518 776 32;

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:


0.000 282 005 895(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1000 1010 0111 0110 0100 1000 1000(2)

6. Positive number before normalization:

0.000 282 005 895(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1000 1010 0111 0110 0100 1000 1000(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:


0.000 282 005 895(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1000 1010 0111 0110 0100 1000 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1000 1010 0111 0110 0100 1000 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0001 1000 1010 0111 0110 0100 1000 1000(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 0100 0110 0001 1000 1010 0111 0110 0100 1000 1000


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 0100 0110 0001 1000 1010 0111 0110 0100 1000 1000 =

0010 0111 1011 0100 0110 0001 1000 1010 0111 0110 0100 1000 1000


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 0100 0110 0001 1000 1010 0111 0110 0100 1000 1000


Decimal number -0.000 282 005 895 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 1000 1010 0111 0110 0100 1000 1000

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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