0.000 000 564 358 690 935 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 564 358 690 935(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
0.000 000 564 358 690 935(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 564 358 690 935.

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) 0.000 000 564 358 690 935 × 2 = 0 + 0.000 001 128 717 381 87;
  • 2) 0.000 001 128 717 381 87 × 2 = 0 + 0.000 002 257 434 763 74;
  • 3) 0.000 002 257 434 763 74 × 2 = 0 + 0.000 004 514 869 527 48;
  • 4) 0.000 004 514 869 527 48 × 2 = 0 + 0.000 009 029 739 054 96;
  • 5) 0.000 009 029 739 054 96 × 2 = 0 + 0.000 018 059 478 109 92;
  • 6) 0.000 018 059 478 109 92 × 2 = 0 + 0.000 036 118 956 219 84;
  • 7) 0.000 036 118 956 219 84 × 2 = 0 + 0.000 072 237 912 439 68;
  • 8) 0.000 072 237 912 439 68 × 2 = 0 + 0.000 144 475 824 879 36;
  • 9) 0.000 144 475 824 879 36 × 2 = 0 + 0.000 288 951 649 758 72;
  • 10) 0.000 288 951 649 758 72 × 2 = 0 + 0.000 577 903 299 517 44;
  • 11) 0.000 577 903 299 517 44 × 2 = 0 + 0.001 155 806 599 034 88;
  • 12) 0.001 155 806 599 034 88 × 2 = 0 + 0.002 311 613 198 069 76;
  • 13) 0.002 311 613 198 069 76 × 2 = 0 + 0.004 623 226 396 139 52;
  • 14) 0.004 623 226 396 139 52 × 2 = 0 + 0.009 246 452 792 279 04;
  • 15) 0.009 246 452 792 279 04 × 2 = 0 + 0.018 492 905 584 558 08;
  • 16) 0.018 492 905 584 558 08 × 2 = 0 + 0.036 985 811 169 116 16;
  • 17) 0.036 985 811 169 116 16 × 2 = 0 + 0.073 971 622 338 232 32;
  • 18) 0.073 971 622 338 232 32 × 2 = 0 + 0.147 943 244 676 464 64;
  • 19) 0.147 943 244 676 464 64 × 2 = 0 + 0.295 886 489 352 929 28;
  • 20) 0.295 886 489 352 929 28 × 2 = 0 + 0.591 772 978 705 858 56;
  • 21) 0.591 772 978 705 858 56 × 2 = 1 + 0.183 545 957 411 717 12;
  • 22) 0.183 545 957 411 717 12 × 2 = 0 + 0.367 091 914 823 434 24;
  • 23) 0.367 091 914 823 434 24 × 2 = 0 + 0.734 183 829 646 868 48;
  • 24) 0.734 183 829 646 868 48 × 2 = 1 + 0.468 367 659 293 736 96;
  • 25) 0.468 367 659 293 736 96 × 2 = 0 + 0.936 735 318 587 473 92;
  • 26) 0.936 735 318 587 473 92 × 2 = 1 + 0.873 470 637 174 947 84;
  • 27) 0.873 470 637 174 947 84 × 2 = 1 + 0.746 941 274 349 895 68;
  • 28) 0.746 941 274 349 895 68 × 2 = 1 + 0.493 882 548 699 791 36;
  • 29) 0.493 882 548 699 791 36 × 2 = 0 + 0.987 765 097 399 582 72;
  • 30) 0.987 765 097 399 582 72 × 2 = 1 + 0.975 530 194 799 165 44;
  • 31) 0.975 530 194 799 165 44 × 2 = 1 + 0.951 060 389 598 330 88;
  • 32) 0.951 060 389 598 330 88 × 2 = 1 + 0.902 120 779 196 661 76;
  • 33) 0.902 120 779 196 661 76 × 2 = 1 + 0.804 241 558 393 323 52;
  • 34) 0.804 241 558 393 323 52 × 2 = 1 + 0.608 483 116 786 647 04;
  • 35) 0.608 483 116 786 647 04 × 2 = 1 + 0.216 966 233 573 294 08;
  • 36) 0.216 966 233 573 294 08 × 2 = 0 + 0.433 932 467 146 588 16;
  • 37) 0.433 932 467 146 588 16 × 2 = 0 + 0.867 864 934 293 176 32;
  • 38) 0.867 864 934 293 176 32 × 2 = 1 + 0.735 729 868 586 352 64;
  • 39) 0.735 729 868 586 352 64 × 2 = 1 + 0.471 459 737 172 705 28;
  • 40) 0.471 459 737 172 705 28 × 2 = 0 + 0.942 919 474 345 410 56;
  • 41) 0.942 919 474 345 410 56 × 2 = 1 + 0.885 838 948 690 821 12;
  • 42) 0.885 838 948 690 821 12 × 2 = 1 + 0.771 677 897 381 642 24;
  • 43) 0.771 677 897 381 642 24 × 2 = 1 + 0.543 355 794 763 284 48;
  • 44) 0.543 355 794 763 284 48 × 2 = 1 + 0.086 711 589 526 568 96;
  • 45) 0.086 711 589 526 568 96 × 2 = 0 + 0.173 423 179 053 137 92;
  • 46) 0.173 423 179 053 137 92 × 2 = 0 + 0.346 846 358 106 275 84;
  • 47) 0.346 846 358 106 275 84 × 2 = 0 + 0.693 692 716 212 551 68;
  • 48) 0.693 692 716 212 551 68 × 2 = 1 + 0.387 385 432 425 103 36;
  • 49) 0.387 385 432 425 103 36 × 2 = 0 + 0.774 770 864 850 206 72;
  • 50) 0.774 770 864 850 206 72 × 2 = 1 + 0.549 541 729 700 413 44;
  • 51) 0.549 541 729 700 413 44 × 2 = 1 + 0.099 083 459 400 826 88;
  • 52) 0.099 083 459 400 826 88 × 2 = 0 + 0.198 166 918 801 653 76;
  • 53) 0.198 166 918 801 653 76 × 2 = 0 + 0.396 333 837 603 307 52;
  • 54) 0.396 333 837 603 307 52 × 2 = 0 + 0.792 667 675 206 615 04;
  • 55) 0.792 667 675 206 615 04 × 2 = 1 + 0.585 335 350 413 230 08;
  • 56) 0.585 335 350 413 230 08 × 2 = 1 + 0.170 670 700 826 460 16;
  • 57) 0.170 670 700 826 460 16 × 2 = 0 + 0.341 341 401 652 920 32;
  • 58) 0.341 341 401 652 920 32 × 2 = 0 + 0.682 682 803 305 840 64;
  • 59) 0.682 682 803 305 840 64 × 2 = 1 + 0.365 365 606 611 681 28;
  • 60) 0.365 365 606 611 681 28 × 2 = 0 + 0.730 731 213 223 362 56;
  • 61) 0.730 731 213 223 362 56 × 2 = 1 + 0.461 462 426 446 725 12;
  • 62) 0.461 462 426 446 725 12 × 2 = 0 + 0.922 924 852 893 450 24;
  • 63) 0.922 924 852 893 450 24 × 2 = 1 + 0.845 849 705 786 900 48;
  • 64) 0.845 849 705 786 900 48 × 2 = 1 + 0.691 699 411 573 800 96;
  • 65) 0.691 699 411 573 800 96 × 2 = 1 + 0.383 398 823 147 601 92;
  • 66) 0.383 398 823 147 601 92 × 2 = 0 + 0.766 797 646 295 203 84;
  • 67) 0.766 797 646 295 203 84 × 2 = 1 + 0.533 595 292 590 407 68;
  • 68) 0.533 595 292 590 407 68 × 2 = 1 + 0.067 190 585 180 815 36;
  • 69) 0.067 190 585 180 815 36 × 2 = 0 + 0.134 381 170 361 630 72;
  • 70) 0.134 381 170 361 630 72 × 2 = 0 + 0.268 762 340 723 261 44;
  • 71) 0.268 762 340 723 261 44 × 2 = 0 + 0.537 524 681 446 522 88;
  • 72) 0.537 524 681 446 522 88 × 2 = 1 + 0.075 049 362 893 045 76;
  • 73) 0.075 049 362 893 045 76 × 2 = 0 + 0.150 098 725 786 091 52;

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).


4. 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:


0.000 000 564 358 690 935(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0110 0011 0010 1011 1011 0001 0(2)

5. Positive number before normalization:

0.000 000 564 358 690 935(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0110 0011 0010 1011 1011 0001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 564 358 690 935(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0110 0011 0010 1011 1011 0001 0(2) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0110 0011 0010 1011 1011 0001 0(2) × 20 =

1.0010 1110 1111 1100 1101 1110 0010 1100 0110 0101 0111 0110 0010(2) × 2-21


7. 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 0 (a positive number)

Exponent (unadjusted): -21

Mantissa (not normalized):
1.0010 1110 1111 1100 1101 1110 0010 1100 0110 0101 0111 0110 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-21 + 2(11-1) - 1 =

(-21 + 1 023)(10) =

1 002(10)


9. 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;
  • 1 002 ÷ 2 = 501 + 0;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1002(10) =

011 1110 1010(2)


11. 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 1110 1111 1100 1101 1110 0010 1100 0110 0101 0111 0110 0010 =

0010 1110 1111 1100 1101 1110 0010 1100 0110 0101 0111 0110 0010


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1110 1010

Mantissa (52 bits) =
0010 1110 1111 1100 1101 1110 0010 1100 0110 0101 0111 0110 0010


Decimal number 0.000 000 564 358 690 935 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1100 0110 0101 0111 0110 0010

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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