-0.000 282 005 913 68 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 68(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
-0.000 282 005 913 68(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:

|-0.000 282 005 913 68| = 0.000 282 005 913 68


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: 0.000 282 005 913 68.

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) 0.000 282 005 913 68 × 2 = 0 + 0.000 564 011 827 36;
  • 2) 0.000 564 011 827 36 × 2 = 0 + 0.001 128 023 654 72;
  • 3) 0.001 128 023 654 72 × 2 = 0 + 0.002 256 047 309 44;
  • 4) 0.002 256 047 309 44 × 2 = 0 + 0.004 512 094 618 88;
  • 5) 0.004 512 094 618 88 × 2 = 0 + 0.009 024 189 237 76;
  • 6) 0.009 024 189 237 76 × 2 = 0 + 0.018 048 378 475 52;
  • 7) 0.018 048 378 475 52 × 2 = 0 + 0.036 096 756 951 04;
  • 8) 0.036 096 756 951 04 × 2 = 0 + 0.072 193 513 902 08;
  • 9) 0.072 193 513 902 08 × 2 = 0 + 0.144 387 027 804 16;
  • 10) 0.144 387 027 804 16 × 2 = 0 + 0.288 774 055 608 32;
  • 11) 0.288 774 055 608 32 × 2 = 0 + 0.577 548 111 216 64;
  • 12) 0.577 548 111 216 64 × 2 = 1 + 0.155 096 222 433 28;
  • 13) 0.155 096 222 433 28 × 2 = 0 + 0.310 192 444 866 56;
  • 14) 0.310 192 444 866 56 × 2 = 0 + 0.620 384 889 733 12;
  • 15) 0.620 384 889 733 12 × 2 = 1 + 0.240 769 779 466 24;
  • 16) 0.240 769 779 466 24 × 2 = 0 + 0.481 539 558 932 48;
  • 17) 0.481 539 558 932 48 × 2 = 0 + 0.963 079 117 864 96;
  • 18) 0.963 079 117 864 96 × 2 = 1 + 0.926 158 235 729 92;
  • 19) 0.926 158 235 729 92 × 2 = 1 + 0.852 316 471 459 84;
  • 20) 0.852 316 471 459 84 × 2 = 1 + 0.704 632 942 919 68;
  • 21) 0.704 632 942 919 68 × 2 = 1 + 0.409 265 885 839 36;
  • 22) 0.409 265 885 839 36 × 2 = 0 + 0.818 531 771 678 72;
  • 23) 0.818 531 771 678 72 × 2 = 1 + 0.637 063 543 357 44;
  • 24) 0.637 063 543 357 44 × 2 = 1 + 0.274 127 086 714 88;
  • 25) 0.274 127 086 714 88 × 2 = 0 + 0.548 254 173 429 76;
  • 26) 0.548 254 173 429 76 × 2 = 1 + 0.096 508 346 859 52;
  • 27) 0.096 508 346 859 52 × 2 = 0 + 0.193 016 693 719 04;
  • 28) 0.193 016 693 719 04 × 2 = 0 + 0.386 033 387 438 08;
  • 29) 0.386 033 387 438 08 × 2 = 0 + 0.772 066 774 876 16;
  • 30) 0.772 066 774 876 16 × 2 = 1 + 0.544 133 549 752 32;
  • 31) 0.544 133 549 752 32 × 2 = 1 + 0.088 267 099 504 64;
  • 32) 0.088 267 099 504 64 × 2 = 0 + 0.176 534 199 009 28;
  • 33) 0.176 534 199 009 28 × 2 = 0 + 0.353 068 398 018 56;
  • 34) 0.353 068 398 018 56 × 2 = 0 + 0.706 136 796 037 12;
  • 35) 0.706 136 796 037 12 × 2 = 1 + 0.412 273 592 074 24;
  • 36) 0.412 273 592 074 24 × 2 = 0 + 0.824 547 184 148 48;
  • 37) 0.824 547 184 148 48 × 2 = 1 + 0.649 094 368 296 96;
  • 38) 0.649 094 368 296 96 × 2 = 1 + 0.298 188 736 593 92;
  • 39) 0.298 188 736 593 92 × 2 = 0 + 0.596 377 473 187 84;
  • 40) 0.596 377 473 187 84 × 2 = 1 + 0.192 754 946 375 68;
  • 41) 0.192 754 946 375 68 × 2 = 0 + 0.385 509 892 751 36;
  • 42) 0.385 509 892 751 36 × 2 = 0 + 0.771 019 785 502 72;
  • 43) 0.771 019 785 502 72 × 2 = 1 + 0.542 039 571 005 44;
  • 44) 0.542 039 571 005 44 × 2 = 1 + 0.084 079 142 010 88;
  • 45) 0.084 079 142 010 88 × 2 = 0 + 0.168 158 284 021 76;
  • 46) 0.168 158 284 021 76 × 2 = 0 + 0.336 316 568 043 52;
  • 47) 0.336 316 568 043 52 × 2 = 0 + 0.672 633 136 087 04;
  • 48) 0.672 633 136 087 04 × 2 = 1 + 0.345 266 272 174 08;
  • 49) 0.345 266 272 174 08 × 2 = 0 + 0.690 532 544 348 16;
  • 50) 0.690 532 544 348 16 × 2 = 1 + 0.381 065 088 696 32;
  • 51) 0.381 065 088 696 32 × 2 = 0 + 0.762 130 177 392 64;
  • 52) 0.762 130 177 392 64 × 2 = 1 + 0.524 260 354 785 28;
  • 53) 0.524 260 354 785 28 × 2 = 1 + 0.048 520 709 570 56;
  • 54) 0.048 520 709 570 56 × 2 = 0 + 0.097 041 419 141 12;
  • 55) 0.097 041 419 141 12 × 2 = 0 + 0.194 082 838 282 24;
  • 56) 0.194 082 838 282 24 × 2 = 0 + 0.388 165 676 564 48;
  • 57) 0.388 165 676 564 48 × 2 = 0 + 0.776 331 353 128 96;
  • 58) 0.776 331 353 128 96 × 2 = 1 + 0.552 662 706 257 92;
  • 59) 0.552 662 706 257 92 × 2 = 1 + 0.105 325 412 515 84;
  • 60) 0.105 325 412 515 84 × 2 = 0 + 0.210 650 825 031 68;
  • 61) 0.210 650 825 031 68 × 2 = 0 + 0.421 301 650 063 36;
  • 62) 0.421 301 650 063 36 × 2 = 0 + 0.842 603 300 126 72;
  • 63) 0.842 603 300 126 72 × 2 = 1 + 0.685 206 600 253 44;
  • 64) 0.685 206 600 253 44 × 2 = 1 + 0.370 413 200 506 88;

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:


0.000 282 005 913 68(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0011 0001 0101 1000 0110 0011(2)

6. Positive number before normalization:

0.000 282 005 913 68(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0011 0001 0101 1000 0110 0011(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:


0.000 282 005 913 68(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0011 0001 0101 1000 0110 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0011 0001 0101 1000 0110 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 0011 0001 0101 1000 0110 0011(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 0010 1101 0011 0001 0101 1000 0110 0011


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 0010 1101 0011 0001 0101 1000 0110 0011 =

0010 0111 1011 0100 0110 0010 1101 0011 0001 0101 1000 0110 0011


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 0010 1101 0011 0001 0101 1000 0110 0011


Decimal number -0.000 282 005 913 68 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 0011 0001 0101 1000 0110 0011

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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