-0.000 282 005 910 18 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 910 18(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
-0.000 282 005 910 18(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:

|-0.000 282 005 910 18| = 0.000 282 005 910 18


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: 0.000 282 005 910 18.

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) 0.000 282 005 910 18 × 2 = 0 + 0.000 564 011 820 36;
  • 2) 0.000 564 011 820 36 × 2 = 0 + 0.001 128 023 640 72;
  • 3) 0.001 128 023 640 72 × 2 = 0 + 0.002 256 047 281 44;
  • 4) 0.002 256 047 281 44 × 2 = 0 + 0.004 512 094 562 88;
  • 5) 0.004 512 094 562 88 × 2 = 0 + 0.009 024 189 125 76;
  • 6) 0.009 024 189 125 76 × 2 = 0 + 0.018 048 378 251 52;
  • 7) 0.018 048 378 251 52 × 2 = 0 + 0.036 096 756 503 04;
  • 8) 0.036 096 756 503 04 × 2 = 0 + 0.072 193 513 006 08;
  • 9) 0.072 193 513 006 08 × 2 = 0 + 0.144 387 026 012 16;
  • 10) 0.144 387 026 012 16 × 2 = 0 + 0.288 774 052 024 32;
  • 11) 0.288 774 052 024 32 × 2 = 0 + 0.577 548 104 048 64;
  • 12) 0.577 548 104 048 64 × 2 = 1 + 0.155 096 208 097 28;
  • 13) 0.155 096 208 097 28 × 2 = 0 + 0.310 192 416 194 56;
  • 14) 0.310 192 416 194 56 × 2 = 0 + 0.620 384 832 389 12;
  • 15) 0.620 384 832 389 12 × 2 = 1 + 0.240 769 664 778 24;
  • 16) 0.240 769 664 778 24 × 2 = 0 + 0.481 539 329 556 48;
  • 17) 0.481 539 329 556 48 × 2 = 0 + 0.963 078 659 112 96;
  • 18) 0.963 078 659 112 96 × 2 = 1 + 0.926 157 318 225 92;
  • 19) 0.926 157 318 225 92 × 2 = 1 + 0.852 314 636 451 84;
  • 20) 0.852 314 636 451 84 × 2 = 1 + 0.704 629 272 903 68;
  • 21) 0.704 629 272 903 68 × 2 = 1 + 0.409 258 545 807 36;
  • 22) 0.409 258 545 807 36 × 2 = 0 + 0.818 517 091 614 72;
  • 23) 0.818 517 091 614 72 × 2 = 1 + 0.637 034 183 229 44;
  • 24) 0.637 034 183 229 44 × 2 = 1 + 0.274 068 366 458 88;
  • 25) 0.274 068 366 458 88 × 2 = 0 + 0.548 136 732 917 76;
  • 26) 0.548 136 732 917 76 × 2 = 1 + 0.096 273 465 835 52;
  • 27) 0.096 273 465 835 52 × 2 = 0 + 0.192 546 931 671 04;
  • 28) 0.192 546 931 671 04 × 2 = 0 + 0.385 093 863 342 08;
  • 29) 0.385 093 863 342 08 × 2 = 0 + 0.770 187 726 684 16;
  • 30) 0.770 187 726 684 16 × 2 = 1 + 0.540 375 453 368 32;
  • 31) 0.540 375 453 368 32 × 2 = 1 + 0.080 750 906 736 64;
  • 32) 0.080 750 906 736 64 × 2 = 0 + 0.161 501 813 473 28;
  • 33) 0.161 501 813 473 28 × 2 = 0 + 0.323 003 626 946 56;
  • 34) 0.323 003 626 946 56 × 2 = 0 + 0.646 007 253 893 12;
  • 35) 0.646 007 253 893 12 × 2 = 1 + 0.292 014 507 786 24;
  • 36) 0.292 014 507 786 24 × 2 = 0 + 0.584 029 015 572 48;
  • 37) 0.584 029 015 572 48 × 2 = 1 + 0.168 058 031 144 96;
  • 38) 0.168 058 031 144 96 × 2 = 0 + 0.336 116 062 289 92;
  • 39) 0.336 116 062 289 92 × 2 = 0 + 0.672 232 124 579 84;
  • 40) 0.672 232 124 579 84 × 2 = 1 + 0.344 464 249 159 68;
  • 41) 0.344 464 249 159 68 × 2 = 0 + 0.688 928 498 319 36;
  • 42) 0.688 928 498 319 36 × 2 = 1 + 0.377 856 996 638 72;
  • 43) 0.377 856 996 638 72 × 2 = 0 + 0.755 713 993 277 44;
  • 44) 0.755 713 993 277 44 × 2 = 1 + 0.511 427 986 554 88;
  • 45) 0.511 427 986 554 88 × 2 = 1 + 0.022 855 973 109 76;
  • 46) 0.022 855 973 109 76 × 2 = 0 + 0.045 711 946 219 52;
  • 47) 0.045 711 946 219 52 × 2 = 0 + 0.091 423 892 439 04;
  • 48) 0.091 423 892 439 04 × 2 = 0 + 0.182 847 784 878 08;
  • 49) 0.182 847 784 878 08 × 2 = 0 + 0.365 695 569 756 16;
  • 50) 0.365 695 569 756 16 × 2 = 0 + 0.731 391 139 512 32;
  • 51) 0.731 391 139 512 32 × 2 = 1 + 0.462 782 279 024 64;
  • 52) 0.462 782 279 024 64 × 2 = 0 + 0.925 564 558 049 28;
  • 53) 0.925 564 558 049 28 × 2 = 1 + 0.851 129 116 098 56;
  • 54) 0.851 129 116 098 56 × 2 = 1 + 0.702 258 232 197 12;
  • 55) 0.702 258 232 197 12 × 2 = 1 + 0.404 516 464 394 24;
  • 56) 0.404 516 464 394 24 × 2 = 0 + 0.809 032 928 788 48;
  • 57) 0.809 032 928 788 48 × 2 = 1 + 0.618 065 857 576 96;
  • 58) 0.618 065 857 576 96 × 2 = 1 + 0.236 131 715 153 92;
  • 59) 0.236 131 715 153 92 × 2 = 0 + 0.472 263 430 307 84;
  • 60) 0.472 263 430 307 84 × 2 = 0 + 0.944 526 860 615 68;
  • 61) 0.944 526 860 615 68 × 2 = 1 + 0.889 053 721 231 36;
  • 62) 0.889 053 721 231 36 × 2 = 1 + 0.778 107 442 462 72;
  • 63) 0.778 107 442 462 72 × 2 = 1 + 0.556 214 884 925 44;
  • 64) 0.556 214 884 925 44 × 2 = 1 + 0.112 429 769 850 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:


0.000 282 005 910 18(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 0101 1000 0010 1110 1100 1111(2)

6. Positive number before normalization:

0.000 282 005 910 18(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 0101 1000 0010 1110 1100 1111(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:


0.000 282 005 910 18(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 0101 1000 0010 1110 1100 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 0101 1000 0010 1110 1100 1111(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1001 0101 1000 0010 1110 1100 1111(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 1001 0101 1000 0010 1110 1100 1111


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 1001 0101 1000 0010 1110 1100 1111 =

0010 0111 1011 0100 0110 0010 1001 0101 1000 0010 1110 1100 1111


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 1001 0101 1000 0010 1110 1100 1111


Decimal number -0.000 282 005 910 18 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1001 0101 1000 0010 1110 1100 1111

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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