-0.000 282 005 951 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 951 5(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
-0.000 282 005 951 5(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:

|-0.000 282 005 951 5| = 0.000 282 005 951 5


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: 0.000 282 005 951 5.

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) 0.000 282 005 951 5 × 2 = 0 + 0.000 564 011 903;
  • 2) 0.000 564 011 903 × 2 = 0 + 0.001 128 023 806;
  • 3) 0.001 128 023 806 × 2 = 0 + 0.002 256 047 612;
  • 4) 0.002 256 047 612 × 2 = 0 + 0.004 512 095 224;
  • 5) 0.004 512 095 224 × 2 = 0 + 0.009 024 190 448;
  • 6) 0.009 024 190 448 × 2 = 0 + 0.018 048 380 896;
  • 7) 0.018 048 380 896 × 2 = 0 + 0.036 096 761 792;
  • 8) 0.036 096 761 792 × 2 = 0 + 0.072 193 523 584;
  • 9) 0.072 193 523 584 × 2 = 0 + 0.144 387 047 168;
  • 10) 0.144 387 047 168 × 2 = 0 + 0.288 774 094 336;
  • 11) 0.288 774 094 336 × 2 = 0 + 0.577 548 188 672;
  • 12) 0.577 548 188 672 × 2 = 1 + 0.155 096 377 344;
  • 13) 0.155 096 377 344 × 2 = 0 + 0.310 192 754 688;
  • 14) 0.310 192 754 688 × 2 = 0 + 0.620 385 509 376;
  • 15) 0.620 385 509 376 × 2 = 1 + 0.240 771 018 752;
  • 16) 0.240 771 018 752 × 2 = 0 + 0.481 542 037 504;
  • 17) 0.481 542 037 504 × 2 = 0 + 0.963 084 075 008;
  • 18) 0.963 084 075 008 × 2 = 1 + 0.926 168 150 016;
  • 19) 0.926 168 150 016 × 2 = 1 + 0.852 336 300 032;
  • 20) 0.852 336 300 032 × 2 = 1 + 0.704 672 600 064;
  • 21) 0.704 672 600 064 × 2 = 1 + 0.409 345 200 128;
  • 22) 0.409 345 200 128 × 2 = 0 + 0.818 690 400 256;
  • 23) 0.818 690 400 256 × 2 = 1 + 0.637 380 800 512;
  • 24) 0.637 380 800 512 × 2 = 1 + 0.274 761 601 024;
  • 25) 0.274 761 601 024 × 2 = 0 + 0.549 523 202 048;
  • 26) 0.549 523 202 048 × 2 = 1 + 0.099 046 404 096;
  • 27) 0.099 046 404 096 × 2 = 0 + 0.198 092 808 192;
  • 28) 0.198 092 808 192 × 2 = 0 + 0.396 185 616 384;
  • 29) 0.396 185 616 384 × 2 = 0 + 0.792 371 232 768;
  • 30) 0.792 371 232 768 × 2 = 1 + 0.584 742 465 536;
  • 31) 0.584 742 465 536 × 2 = 1 + 0.169 484 931 072;
  • 32) 0.169 484 931 072 × 2 = 0 + 0.338 969 862 144;
  • 33) 0.338 969 862 144 × 2 = 0 + 0.677 939 724 288;
  • 34) 0.677 939 724 288 × 2 = 1 + 0.355 879 448 576;
  • 35) 0.355 879 448 576 × 2 = 0 + 0.711 758 897 152;
  • 36) 0.711 758 897 152 × 2 = 1 + 0.423 517 794 304;
  • 37) 0.423 517 794 304 × 2 = 0 + 0.847 035 588 608;
  • 38) 0.847 035 588 608 × 2 = 1 + 0.694 071 177 216;
  • 39) 0.694 071 177 216 × 2 = 1 + 0.388 142 354 432;
  • 40) 0.388 142 354 432 × 2 = 0 + 0.776 284 708 864;
  • 41) 0.776 284 708 864 × 2 = 1 + 0.552 569 417 728;
  • 42) 0.552 569 417 728 × 2 = 1 + 0.105 138 835 456;
  • 43) 0.105 138 835 456 × 2 = 0 + 0.210 277 670 912;
  • 44) 0.210 277 670 912 × 2 = 0 + 0.420 555 341 824;
  • 45) 0.420 555 341 824 × 2 = 0 + 0.841 110 683 648;
  • 46) 0.841 110 683 648 × 2 = 1 + 0.682 221 367 296;
  • 47) 0.682 221 367 296 × 2 = 1 + 0.364 442 734 592;
  • 48) 0.364 442 734 592 × 2 = 0 + 0.728 885 469 184;
  • 49) 0.728 885 469 184 × 2 = 1 + 0.457 770 938 368;
  • 50) 0.457 770 938 368 × 2 = 0 + 0.915 541 876 736;
  • 51) 0.915 541 876 736 × 2 = 1 + 0.831 083 753 472;
  • 52) 0.831 083 753 472 × 2 = 1 + 0.662 167 506 944;
  • 53) 0.662 167 506 944 × 2 = 1 + 0.324 335 013 888;
  • 54) 0.324 335 013 888 × 2 = 0 + 0.648 670 027 776;
  • 55) 0.648 670 027 776 × 2 = 1 + 0.297 340 055 552;
  • 56) 0.297 340 055 552 × 2 = 0 + 0.594 680 111 104;
  • 57) 0.594 680 111 104 × 2 = 1 + 0.189 360 222 208;
  • 58) 0.189 360 222 208 × 2 = 0 + 0.378 720 444 416;
  • 59) 0.378 720 444 416 × 2 = 0 + 0.757 440 888 832;
  • 60) 0.757 440 888 832 × 2 = 1 + 0.514 881 777 664;
  • 61) 0.514 881 777 664 × 2 = 1 + 0.029 763 555 328;
  • 62) 0.029 763 555 328 × 2 = 0 + 0.059 527 110 656;
  • 63) 0.059 527 110 656 × 2 = 0 + 0.119 054 221 312;
  • 64) 0.119 054 221 312 × 2 = 0 + 0.238 108 442 624;

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:


0.000 282 005 951 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 0110 1100 0110 1011 1010 1001 1000(2)

6. Positive number before normalization:

0.000 282 005 951 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 0110 1100 0110 1011 1010 1001 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:


0.000 282 005 951 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 0110 1100 0110 1011 1010 1001 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 0110 1100 0110 1011 1010 1001 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0101 0110 1100 0110 1011 1010 1001 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 0101 0110 1100 0110 1011 1010 1001 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 0101 0110 1100 0110 1011 1010 1001 1000 =

0010 0111 1011 0100 0110 0101 0110 1100 0110 1011 1010 1001 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 0101 0110 1100 0110 1011 1010 1001 1000


Decimal number -0.000 282 005 951 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0101 0110 1100 0110 1011 1010 1001 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