-0.000 282 005 917 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 917(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
-0.000 282 005 917(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:

|-0.000 282 005 917| = 0.000 282 005 917


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: 0.000 282 005 917.

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) 0.000 282 005 917 × 2 = 0 + 0.000 564 011 834;
  • 2) 0.000 564 011 834 × 2 = 0 + 0.001 128 023 668;
  • 3) 0.001 128 023 668 × 2 = 0 + 0.002 256 047 336;
  • 4) 0.002 256 047 336 × 2 = 0 + 0.004 512 094 672;
  • 5) 0.004 512 094 672 × 2 = 0 + 0.009 024 189 344;
  • 6) 0.009 024 189 344 × 2 = 0 + 0.018 048 378 688;
  • 7) 0.018 048 378 688 × 2 = 0 + 0.036 096 757 376;
  • 8) 0.036 096 757 376 × 2 = 0 + 0.072 193 514 752;
  • 9) 0.072 193 514 752 × 2 = 0 + 0.144 387 029 504;
  • 10) 0.144 387 029 504 × 2 = 0 + 0.288 774 059 008;
  • 11) 0.288 774 059 008 × 2 = 0 + 0.577 548 118 016;
  • 12) 0.577 548 118 016 × 2 = 1 + 0.155 096 236 032;
  • 13) 0.155 096 236 032 × 2 = 0 + 0.310 192 472 064;
  • 14) 0.310 192 472 064 × 2 = 0 + 0.620 384 944 128;
  • 15) 0.620 384 944 128 × 2 = 1 + 0.240 769 888 256;
  • 16) 0.240 769 888 256 × 2 = 0 + 0.481 539 776 512;
  • 17) 0.481 539 776 512 × 2 = 0 + 0.963 079 553 024;
  • 18) 0.963 079 553 024 × 2 = 1 + 0.926 159 106 048;
  • 19) 0.926 159 106 048 × 2 = 1 + 0.852 318 212 096;
  • 20) 0.852 318 212 096 × 2 = 1 + 0.704 636 424 192;
  • 21) 0.704 636 424 192 × 2 = 1 + 0.409 272 848 384;
  • 22) 0.409 272 848 384 × 2 = 0 + 0.818 545 696 768;
  • 23) 0.818 545 696 768 × 2 = 1 + 0.637 091 393 536;
  • 24) 0.637 091 393 536 × 2 = 1 + 0.274 182 787 072;
  • 25) 0.274 182 787 072 × 2 = 0 + 0.548 365 574 144;
  • 26) 0.548 365 574 144 × 2 = 1 + 0.096 731 148 288;
  • 27) 0.096 731 148 288 × 2 = 0 + 0.193 462 296 576;
  • 28) 0.193 462 296 576 × 2 = 0 + 0.386 924 593 152;
  • 29) 0.386 924 593 152 × 2 = 0 + 0.773 849 186 304;
  • 30) 0.773 849 186 304 × 2 = 1 + 0.547 698 372 608;
  • 31) 0.547 698 372 608 × 2 = 1 + 0.095 396 745 216;
  • 32) 0.095 396 745 216 × 2 = 0 + 0.190 793 490 432;
  • 33) 0.190 793 490 432 × 2 = 0 + 0.381 586 980 864;
  • 34) 0.381 586 980 864 × 2 = 0 + 0.763 173 961 728;
  • 35) 0.763 173 961 728 × 2 = 1 + 0.526 347 923 456;
  • 36) 0.526 347 923 456 × 2 = 1 + 0.052 695 846 912;
  • 37) 0.052 695 846 912 × 2 = 0 + 0.105 391 693 824;
  • 38) 0.105 391 693 824 × 2 = 0 + 0.210 783 387 648;
  • 39) 0.210 783 387 648 × 2 = 0 + 0.421 566 775 296;
  • 40) 0.421 566 775 296 × 2 = 0 + 0.843 133 550 592;
  • 41) 0.843 133 550 592 × 2 = 1 + 0.686 267 101 184;
  • 42) 0.686 267 101 184 × 2 = 1 + 0.372 534 202 368;
  • 43) 0.372 534 202 368 × 2 = 0 + 0.745 068 404 736;
  • 44) 0.745 068 404 736 × 2 = 1 + 0.490 136 809 472;
  • 45) 0.490 136 809 472 × 2 = 0 + 0.980 273 618 944;
  • 46) 0.980 273 618 944 × 2 = 1 + 0.960 547 237 888;
  • 47) 0.960 547 237 888 × 2 = 1 + 0.921 094 475 776;
  • 48) 0.921 094 475 776 × 2 = 1 + 0.842 188 951 552;
  • 49) 0.842 188 951 552 × 2 = 1 + 0.684 377 903 104;
  • 50) 0.684 377 903 104 × 2 = 1 + 0.368 755 806 208;
  • 51) 0.368 755 806 208 × 2 = 0 + 0.737 511 612 416;
  • 52) 0.737 511 612 416 × 2 = 1 + 0.475 023 224 832;
  • 53) 0.475 023 224 832 × 2 = 0 + 0.950 046 449 664;
  • 54) 0.950 046 449 664 × 2 = 1 + 0.900 092 899 328;
  • 55) 0.900 092 899 328 × 2 = 1 + 0.800 185 798 656;
  • 56) 0.800 185 798 656 × 2 = 1 + 0.600 371 597 312;
  • 57) 0.600 371 597 312 × 2 = 1 + 0.200 743 194 624;
  • 58) 0.200 743 194 624 × 2 = 0 + 0.401 486 389 248;
  • 59) 0.401 486 389 248 × 2 = 0 + 0.802 972 778 496;
  • 60) 0.802 972 778 496 × 2 = 1 + 0.605 945 556 992;
  • 61) 0.605 945 556 992 × 2 = 1 + 0.211 891 113 984;
  • 62) 0.211 891 113 984 × 2 = 0 + 0.423 782 227 968;
  • 63) 0.423 782 227 968 × 2 = 0 + 0.847 564 455 936;
  • 64) 0.847 564 455 936 × 2 = 1 + 0.695 128 911 872;

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:


0.000 282 005 917(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 1101 0111 1101 0111 1001 1001(2)

6. Positive number before normalization:

0.000 282 005 917(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 1101 0111 1101 0111 1001 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:


0.000 282 005 917(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 1101 0111 1101 0111 1001 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 1101 0111 1101 0111 1001 1001(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0000 1101 0111 1101 0111 1001 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 1011 0100 0110 0011 0000 1101 0111 1101 0111 1001 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 1011 0100 0110 0011 0000 1101 0111 1101 0111 1001 1001 =

0010 0111 1011 0100 0110 0011 0000 1101 0111 1101 0111 1001 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 1011 0100 0110 0011 0000 1101 0111 1101 0111 1001 1001


Decimal number -0.000 282 005 917 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0000 1101 0111 1101 0111 1001 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