-0.000 282 005 914 513 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 513(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
-0.000 282 005 914 513(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:

|-0.000 282 005 914 513| = 0.000 282 005 914 513


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: 0.000 282 005 914 513.

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) 0.000 282 005 914 513 × 2 = 0 + 0.000 564 011 829 026;
  • 2) 0.000 564 011 829 026 × 2 = 0 + 0.001 128 023 658 052;
  • 3) 0.001 128 023 658 052 × 2 = 0 + 0.002 256 047 316 104;
  • 4) 0.002 256 047 316 104 × 2 = 0 + 0.004 512 094 632 208;
  • 5) 0.004 512 094 632 208 × 2 = 0 + 0.009 024 189 264 416;
  • 6) 0.009 024 189 264 416 × 2 = 0 + 0.018 048 378 528 832;
  • 7) 0.018 048 378 528 832 × 2 = 0 + 0.036 096 757 057 664;
  • 8) 0.036 096 757 057 664 × 2 = 0 + 0.072 193 514 115 328;
  • 9) 0.072 193 514 115 328 × 2 = 0 + 0.144 387 028 230 656;
  • 10) 0.144 387 028 230 656 × 2 = 0 + 0.288 774 056 461 312;
  • 11) 0.288 774 056 461 312 × 2 = 0 + 0.577 548 112 922 624;
  • 12) 0.577 548 112 922 624 × 2 = 1 + 0.155 096 225 845 248;
  • 13) 0.155 096 225 845 248 × 2 = 0 + 0.310 192 451 690 496;
  • 14) 0.310 192 451 690 496 × 2 = 0 + 0.620 384 903 380 992;
  • 15) 0.620 384 903 380 992 × 2 = 1 + 0.240 769 806 761 984;
  • 16) 0.240 769 806 761 984 × 2 = 0 + 0.481 539 613 523 968;
  • 17) 0.481 539 613 523 968 × 2 = 0 + 0.963 079 227 047 936;
  • 18) 0.963 079 227 047 936 × 2 = 1 + 0.926 158 454 095 872;
  • 19) 0.926 158 454 095 872 × 2 = 1 + 0.852 316 908 191 744;
  • 20) 0.852 316 908 191 744 × 2 = 1 + 0.704 633 816 383 488;
  • 21) 0.704 633 816 383 488 × 2 = 1 + 0.409 267 632 766 976;
  • 22) 0.409 267 632 766 976 × 2 = 0 + 0.818 535 265 533 952;
  • 23) 0.818 535 265 533 952 × 2 = 1 + 0.637 070 531 067 904;
  • 24) 0.637 070 531 067 904 × 2 = 1 + 0.274 141 062 135 808;
  • 25) 0.274 141 062 135 808 × 2 = 0 + 0.548 282 124 271 616;
  • 26) 0.548 282 124 271 616 × 2 = 1 + 0.096 564 248 543 232;
  • 27) 0.096 564 248 543 232 × 2 = 0 + 0.193 128 497 086 464;
  • 28) 0.193 128 497 086 464 × 2 = 0 + 0.386 256 994 172 928;
  • 29) 0.386 256 994 172 928 × 2 = 0 + 0.772 513 988 345 856;
  • 30) 0.772 513 988 345 856 × 2 = 1 + 0.545 027 976 691 712;
  • 31) 0.545 027 976 691 712 × 2 = 1 + 0.090 055 953 383 424;
  • 32) 0.090 055 953 383 424 × 2 = 0 + 0.180 111 906 766 848;
  • 33) 0.180 111 906 766 848 × 2 = 0 + 0.360 223 813 533 696;
  • 34) 0.360 223 813 533 696 × 2 = 0 + 0.720 447 627 067 392;
  • 35) 0.720 447 627 067 392 × 2 = 1 + 0.440 895 254 134 784;
  • 36) 0.440 895 254 134 784 × 2 = 0 + 0.881 790 508 269 568;
  • 37) 0.881 790 508 269 568 × 2 = 1 + 0.763 581 016 539 136;
  • 38) 0.763 581 016 539 136 × 2 = 1 + 0.527 162 033 078 272;
  • 39) 0.527 162 033 078 272 × 2 = 1 + 0.054 324 066 156 544;
  • 40) 0.054 324 066 156 544 × 2 = 0 + 0.108 648 132 313 088;
  • 41) 0.108 648 132 313 088 × 2 = 0 + 0.217 296 264 626 176;
  • 42) 0.217 296 264 626 176 × 2 = 0 + 0.434 592 529 252 352;
  • 43) 0.434 592 529 252 352 × 2 = 0 + 0.869 185 058 504 704;
  • 44) 0.869 185 058 504 704 × 2 = 1 + 0.738 370 117 009 408;
  • 45) 0.738 370 117 009 408 × 2 = 1 + 0.476 740 234 018 816;
  • 46) 0.476 740 234 018 816 × 2 = 0 + 0.953 480 468 037 632;
  • 47) 0.953 480 468 037 632 × 2 = 1 + 0.906 960 936 075 264;
  • 48) 0.906 960 936 075 264 × 2 = 1 + 0.813 921 872 150 528;
  • 49) 0.813 921 872 150 528 × 2 = 1 + 0.627 843 744 301 056;
  • 50) 0.627 843 744 301 056 × 2 = 1 + 0.255 687 488 602 112;
  • 51) 0.255 687 488 602 112 × 2 = 0 + 0.511 374 977 204 224;
  • 52) 0.511 374 977 204 224 × 2 = 1 + 0.022 749 954 408 448;
  • 53) 0.022 749 954 408 448 × 2 = 0 + 0.045 499 908 816 896;
  • 54) 0.045 499 908 816 896 × 2 = 0 + 0.090 999 817 633 792;
  • 55) 0.090 999 817 633 792 × 2 = 0 + 0.181 999 635 267 584;
  • 56) 0.181 999 635 267 584 × 2 = 0 + 0.363 999 270 535 168;
  • 57) 0.363 999 270 535 168 × 2 = 0 + 0.727 998 541 070 336;
  • 58) 0.727 998 541 070 336 × 2 = 1 + 0.455 997 082 140 672;
  • 59) 0.455 997 082 140 672 × 2 = 0 + 0.911 994 164 281 344;
  • 60) 0.911 994 164 281 344 × 2 = 1 + 0.823 988 328 562 688;
  • 61) 0.823 988 328 562 688 × 2 = 1 + 0.647 976 657 125 376;
  • 62) 0.647 976 657 125 376 × 2 = 1 + 0.295 953 314 250 752;
  • 63) 0.295 953 314 250 752 × 2 = 0 + 0.591 906 628 501 504;
  • 64) 0.591 906 628 501 504 × 2 = 1 + 0.183 813 257 003 008;

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:


0.000 282 005 914 513(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1011 1101 0000 0101 1101(2)

6. Positive number before normalization:

0.000 282 005 914 513(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1011 1101 0000 0101 1101(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:


0.000 282 005 914 513(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1011 1101 0000 0101 1101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1011 1101 0000 0101 1101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0001 1011 1101 0000 0101 1101(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 1110 0001 1011 1101 0000 0101 1101


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 1110 0001 1011 1101 0000 0101 1101 =

0010 0111 1011 0100 0110 0010 1110 0001 1011 1101 0000 0101 1101


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 1110 0001 1011 1101 0000 0101 1101


Decimal number -0.000 282 005 914 513 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0001 1011 1101 0000 0101 1101

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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