0.000 000 564 358 690 631 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 564 358 690 631(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
0.000 000 564 358 690 631(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: 0.000 000 564 358 690 631.

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) 0.000 000 564 358 690 631 × 2 = 0 + 0.000 001 128 717 381 262;
  • 2) 0.000 001 128 717 381 262 × 2 = 0 + 0.000 002 257 434 762 524;
  • 3) 0.000 002 257 434 762 524 × 2 = 0 + 0.000 004 514 869 525 048;
  • 4) 0.000 004 514 869 525 048 × 2 = 0 + 0.000 009 029 739 050 096;
  • 5) 0.000 009 029 739 050 096 × 2 = 0 + 0.000 018 059 478 100 192;
  • 6) 0.000 018 059 478 100 192 × 2 = 0 + 0.000 036 118 956 200 384;
  • 7) 0.000 036 118 956 200 384 × 2 = 0 + 0.000 072 237 912 400 768;
  • 8) 0.000 072 237 912 400 768 × 2 = 0 + 0.000 144 475 824 801 536;
  • 9) 0.000 144 475 824 801 536 × 2 = 0 + 0.000 288 951 649 603 072;
  • 10) 0.000 288 951 649 603 072 × 2 = 0 + 0.000 577 903 299 206 144;
  • 11) 0.000 577 903 299 206 144 × 2 = 0 + 0.001 155 806 598 412 288;
  • 12) 0.001 155 806 598 412 288 × 2 = 0 + 0.002 311 613 196 824 576;
  • 13) 0.002 311 613 196 824 576 × 2 = 0 + 0.004 623 226 393 649 152;
  • 14) 0.004 623 226 393 649 152 × 2 = 0 + 0.009 246 452 787 298 304;
  • 15) 0.009 246 452 787 298 304 × 2 = 0 + 0.018 492 905 574 596 608;
  • 16) 0.018 492 905 574 596 608 × 2 = 0 + 0.036 985 811 149 193 216;
  • 17) 0.036 985 811 149 193 216 × 2 = 0 + 0.073 971 622 298 386 432;
  • 18) 0.073 971 622 298 386 432 × 2 = 0 + 0.147 943 244 596 772 864;
  • 19) 0.147 943 244 596 772 864 × 2 = 0 + 0.295 886 489 193 545 728;
  • 20) 0.295 886 489 193 545 728 × 2 = 0 + 0.591 772 978 387 091 456;
  • 21) 0.591 772 978 387 091 456 × 2 = 1 + 0.183 545 956 774 182 912;
  • 22) 0.183 545 956 774 182 912 × 2 = 0 + 0.367 091 913 548 365 824;
  • 23) 0.367 091 913 548 365 824 × 2 = 0 + 0.734 183 827 096 731 648;
  • 24) 0.734 183 827 096 731 648 × 2 = 1 + 0.468 367 654 193 463 296;
  • 25) 0.468 367 654 193 463 296 × 2 = 0 + 0.936 735 308 386 926 592;
  • 26) 0.936 735 308 386 926 592 × 2 = 1 + 0.873 470 616 773 853 184;
  • 27) 0.873 470 616 773 853 184 × 2 = 1 + 0.746 941 233 547 706 368;
  • 28) 0.746 941 233 547 706 368 × 2 = 1 + 0.493 882 467 095 412 736;
  • 29) 0.493 882 467 095 412 736 × 2 = 0 + 0.987 764 934 190 825 472;
  • 30) 0.987 764 934 190 825 472 × 2 = 1 + 0.975 529 868 381 650 944;
  • 31) 0.975 529 868 381 650 944 × 2 = 1 + 0.951 059 736 763 301 888;
  • 32) 0.951 059 736 763 301 888 × 2 = 1 + 0.902 119 473 526 603 776;
  • 33) 0.902 119 473 526 603 776 × 2 = 1 + 0.804 238 947 053 207 552;
  • 34) 0.804 238 947 053 207 552 × 2 = 1 + 0.608 477 894 106 415 104;
  • 35) 0.608 477 894 106 415 104 × 2 = 1 + 0.216 955 788 212 830 208;
  • 36) 0.216 955 788 212 830 208 × 2 = 0 + 0.433 911 576 425 660 416;
  • 37) 0.433 911 576 425 660 416 × 2 = 0 + 0.867 823 152 851 320 832;
  • 38) 0.867 823 152 851 320 832 × 2 = 1 + 0.735 646 305 702 641 664;
  • 39) 0.735 646 305 702 641 664 × 2 = 1 + 0.471 292 611 405 283 328;
  • 40) 0.471 292 611 405 283 328 × 2 = 0 + 0.942 585 222 810 566 656;
  • 41) 0.942 585 222 810 566 656 × 2 = 1 + 0.885 170 445 621 133 312;
  • 42) 0.885 170 445 621 133 312 × 2 = 1 + 0.770 340 891 242 266 624;
  • 43) 0.770 340 891 242 266 624 × 2 = 1 + 0.540 681 782 484 533 248;
  • 44) 0.540 681 782 484 533 248 × 2 = 1 + 0.081 363 564 969 066 496;
  • 45) 0.081 363 564 969 066 496 × 2 = 0 + 0.162 727 129 938 132 992;
  • 46) 0.162 727 129 938 132 992 × 2 = 0 + 0.325 454 259 876 265 984;
  • 47) 0.325 454 259 876 265 984 × 2 = 0 + 0.650 908 519 752 531 968;
  • 48) 0.650 908 519 752 531 968 × 2 = 1 + 0.301 817 039 505 063 936;
  • 49) 0.301 817 039 505 063 936 × 2 = 0 + 0.603 634 079 010 127 872;
  • 50) 0.603 634 079 010 127 872 × 2 = 1 + 0.207 268 158 020 255 744;
  • 51) 0.207 268 158 020 255 744 × 2 = 0 + 0.414 536 316 040 511 488;
  • 52) 0.414 536 316 040 511 488 × 2 = 0 + 0.829 072 632 081 022 976;
  • 53) 0.829 072 632 081 022 976 × 2 = 1 + 0.658 145 264 162 045 952;
  • 54) 0.658 145 264 162 045 952 × 2 = 1 + 0.316 290 528 324 091 904;
  • 55) 0.316 290 528 324 091 904 × 2 = 0 + 0.632 581 056 648 183 808;
  • 56) 0.632 581 056 648 183 808 × 2 = 1 + 0.265 162 113 296 367 616;
  • 57) 0.265 162 113 296 367 616 × 2 = 0 + 0.530 324 226 592 735 232;
  • 58) 0.530 324 226 592 735 232 × 2 = 1 + 0.060 648 453 185 470 464;
  • 59) 0.060 648 453 185 470 464 × 2 = 0 + 0.121 296 906 370 940 928;
  • 60) 0.121 296 906 370 940 928 × 2 = 0 + 0.242 593 812 741 881 856;
  • 61) 0.242 593 812 741 881 856 × 2 = 0 + 0.485 187 625 483 763 712;
  • 62) 0.485 187 625 483 763 712 × 2 = 0 + 0.970 375 250 967 527 424;
  • 63) 0.970 375 250 967 527 424 × 2 = 1 + 0.940 750 501 935 054 848;
  • 64) 0.940 750 501 935 054 848 × 2 = 1 + 0.881 501 003 870 109 696;
  • 65) 0.881 501 003 870 109 696 × 2 = 1 + 0.763 002 007 740 219 392;
  • 66) 0.763 002 007 740 219 392 × 2 = 1 + 0.526 004 015 480 438 784;
  • 67) 0.526 004 015 480 438 784 × 2 = 1 + 0.052 008 030 960 877 568;
  • 68) 0.052 008 030 960 877 568 × 2 = 0 + 0.104 016 061 921 755 136;
  • 69) 0.104 016 061 921 755 136 × 2 = 0 + 0.208 032 123 843 510 272;
  • 70) 0.208 032 123 843 510 272 × 2 = 0 + 0.416 064 247 687 020 544;
  • 71) 0.416 064 247 687 020 544 × 2 = 0 + 0.832 128 495 374 041 088;
  • 72) 0.832 128 495 374 041 088 × 2 = 1 + 0.664 256 990 748 082 176;
  • 73) 0.664 256 990 748 082 176 × 2 = 1 + 0.328 513 981 496 164 352;

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:


0.000 000 564 358 690 631(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0100 1101 0100 0011 1110 0001 1(2)

5. Positive number before normalization:

0.000 000 564 358 690 631(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0100 1101 0100 0011 1110 0001 1(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:


0.000 000 564 358 690 631(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0100 1101 0100 0011 1110 0001 1(2) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0100 1101 0100 0011 1110 0001 1(2) × 20 =

1.0010 1110 1111 1100 1101 1110 0010 1001 1010 1000 0111 1100 0011(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 1001 1010 1000 0111 1100 0011


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 1001 1010 1000 0111 1100 0011 =

0010 1110 1111 1100 1101 1110 0010 1001 1010 1000 0111 1100 0011


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 1001 1010 1000 0111 1100 0011


Decimal number 0.000 000 564 358 690 631 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1001 1010 1000 0111 1100 0011

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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