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

Convert decimal 0.000 000 564 358 690 754(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
0.000 000 564 358 690 754(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: 0.000 000 564 358 690 754.

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) 0.000 000 564 358 690 754 × 2 = 0 + 0.000 001 128 717 381 508;
  • 2) 0.000 001 128 717 381 508 × 2 = 0 + 0.000 002 257 434 763 016;
  • 3) 0.000 002 257 434 763 016 × 2 = 0 + 0.000 004 514 869 526 032;
  • 4) 0.000 004 514 869 526 032 × 2 = 0 + 0.000 009 029 739 052 064;
  • 5) 0.000 009 029 739 052 064 × 2 = 0 + 0.000 018 059 478 104 128;
  • 6) 0.000 018 059 478 104 128 × 2 = 0 + 0.000 036 118 956 208 256;
  • 7) 0.000 036 118 956 208 256 × 2 = 0 + 0.000 072 237 912 416 512;
  • 8) 0.000 072 237 912 416 512 × 2 = 0 + 0.000 144 475 824 833 024;
  • 9) 0.000 144 475 824 833 024 × 2 = 0 + 0.000 288 951 649 666 048;
  • 10) 0.000 288 951 649 666 048 × 2 = 0 + 0.000 577 903 299 332 096;
  • 11) 0.000 577 903 299 332 096 × 2 = 0 + 0.001 155 806 598 664 192;
  • 12) 0.001 155 806 598 664 192 × 2 = 0 + 0.002 311 613 197 328 384;
  • 13) 0.002 311 613 197 328 384 × 2 = 0 + 0.004 623 226 394 656 768;
  • 14) 0.004 623 226 394 656 768 × 2 = 0 + 0.009 246 452 789 313 536;
  • 15) 0.009 246 452 789 313 536 × 2 = 0 + 0.018 492 905 578 627 072;
  • 16) 0.018 492 905 578 627 072 × 2 = 0 + 0.036 985 811 157 254 144;
  • 17) 0.036 985 811 157 254 144 × 2 = 0 + 0.073 971 622 314 508 288;
  • 18) 0.073 971 622 314 508 288 × 2 = 0 + 0.147 943 244 629 016 576;
  • 19) 0.147 943 244 629 016 576 × 2 = 0 + 0.295 886 489 258 033 152;
  • 20) 0.295 886 489 258 033 152 × 2 = 0 + 0.591 772 978 516 066 304;
  • 21) 0.591 772 978 516 066 304 × 2 = 1 + 0.183 545 957 032 132 608;
  • 22) 0.183 545 957 032 132 608 × 2 = 0 + 0.367 091 914 064 265 216;
  • 23) 0.367 091 914 064 265 216 × 2 = 0 + 0.734 183 828 128 530 432;
  • 24) 0.734 183 828 128 530 432 × 2 = 1 + 0.468 367 656 257 060 864;
  • 25) 0.468 367 656 257 060 864 × 2 = 0 + 0.936 735 312 514 121 728;
  • 26) 0.936 735 312 514 121 728 × 2 = 1 + 0.873 470 625 028 243 456;
  • 27) 0.873 470 625 028 243 456 × 2 = 1 + 0.746 941 250 056 486 912;
  • 28) 0.746 941 250 056 486 912 × 2 = 1 + 0.493 882 500 112 973 824;
  • 29) 0.493 882 500 112 973 824 × 2 = 0 + 0.987 765 000 225 947 648;
  • 30) 0.987 765 000 225 947 648 × 2 = 1 + 0.975 530 000 451 895 296;
  • 31) 0.975 530 000 451 895 296 × 2 = 1 + 0.951 060 000 903 790 592;
  • 32) 0.951 060 000 903 790 592 × 2 = 1 + 0.902 120 001 807 581 184;
  • 33) 0.902 120 001 807 581 184 × 2 = 1 + 0.804 240 003 615 162 368;
  • 34) 0.804 240 003 615 162 368 × 2 = 1 + 0.608 480 007 230 324 736;
  • 35) 0.608 480 007 230 324 736 × 2 = 1 + 0.216 960 014 460 649 472;
  • 36) 0.216 960 014 460 649 472 × 2 = 0 + 0.433 920 028 921 298 944;
  • 37) 0.433 920 028 921 298 944 × 2 = 0 + 0.867 840 057 842 597 888;
  • 38) 0.867 840 057 842 597 888 × 2 = 1 + 0.735 680 115 685 195 776;
  • 39) 0.735 680 115 685 195 776 × 2 = 1 + 0.471 360 231 370 391 552;
  • 40) 0.471 360 231 370 391 552 × 2 = 0 + 0.942 720 462 740 783 104;
  • 41) 0.942 720 462 740 783 104 × 2 = 1 + 0.885 440 925 481 566 208;
  • 42) 0.885 440 925 481 566 208 × 2 = 1 + 0.770 881 850 963 132 416;
  • 43) 0.770 881 850 963 132 416 × 2 = 1 + 0.541 763 701 926 264 832;
  • 44) 0.541 763 701 926 264 832 × 2 = 1 + 0.083 527 403 852 529 664;
  • 45) 0.083 527 403 852 529 664 × 2 = 0 + 0.167 054 807 705 059 328;
  • 46) 0.167 054 807 705 059 328 × 2 = 0 + 0.334 109 615 410 118 656;
  • 47) 0.334 109 615 410 118 656 × 2 = 0 + 0.668 219 230 820 237 312;
  • 48) 0.668 219 230 820 237 312 × 2 = 1 + 0.336 438 461 640 474 624;
  • 49) 0.336 438 461 640 474 624 × 2 = 0 + 0.672 876 923 280 949 248;
  • 50) 0.672 876 923 280 949 248 × 2 = 1 + 0.345 753 846 561 898 496;
  • 51) 0.345 753 846 561 898 496 × 2 = 0 + 0.691 507 693 123 796 992;
  • 52) 0.691 507 693 123 796 992 × 2 = 1 + 0.383 015 386 247 593 984;
  • 53) 0.383 015 386 247 593 984 × 2 = 0 + 0.766 030 772 495 187 968;
  • 54) 0.766 030 772 495 187 968 × 2 = 1 + 0.532 061 544 990 375 936;
  • 55) 0.532 061 544 990 375 936 × 2 = 1 + 0.064 123 089 980 751 872;
  • 56) 0.064 123 089 980 751 872 × 2 = 0 + 0.128 246 179 961 503 744;
  • 57) 0.128 246 179 961 503 744 × 2 = 0 + 0.256 492 359 923 007 488;
  • 58) 0.256 492 359 923 007 488 × 2 = 0 + 0.512 984 719 846 014 976;
  • 59) 0.512 984 719 846 014 976 × 2 = 1 + 0.025 969 439 692 029 952;
  • 60) 0.025 969 439 692 029 952 × 2 = 0 + 0.051 938 879 384 059 904;
  • 61) 0.051 938 879 384 059 904 × 2 = 0 + 0.103 877 758 768 119 808;
  • 62) 0.103 877 758 768 119 808 × 2 = 0 + 0.207 755 517 536 239 616;
  • 63) 0.207 755 517 536 239 616 × 2 = 0 + 0.415 511 035 072 479 232;
  • 64) 0.415 511 035 072 479 232 × 2 = 0 + 0.831 022 070 144 958 464;
  • 65) 0.831 022 070 144 958 464 × 2 = 1 + 0.662 044 140 289 916 928;
  • 66) 0.662 044 140 289 916 928 × 2 = 1 + 0.324 088 280 579 833 856;
  • 67) 0.324 088 280 579 833 856 × 2 = 0 + 0.648 176 561 159 667 712;
  • 68) 0.648 176 561 159 667 712 × 2 = 1 + 0.296 353 122 319 335 424;
  • 69) 0.296 353 122 319 335 424 × 2 = 0 + 0.592 706 244 638 670 848;
  • 70) 0.592 706 244 638 670 848 × 2 = 1 + 0.185 412 489 277 341 696;
  • 71) 0.185 412 489 277 341 696 × 2 = 0 + 0.370 824 978 554 683 392;
  • 72) 0.370 824 978 554 683 392 × 2 = 0 + 0.741 649 957 109 366 784;
  • 73) 0.741 649 957 109 366 784 × 2 = 1 + 0.483 299 914 218 733 568;

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:


0.000 000 564 358 690 754(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0110 0010 0000 1101 0100 1(2)

5. Positive number before normalization:

0.000 000 564 358 690 754(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0110 0010 0000 1101 0100 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:


0.000 000 564 358 690 754(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0110 0010 0000 1101 0100 1(2) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0110 0010 0000 1101 0100 1(2) × 20 =

1.0010 1110 1111 1100 1101 1110 0010 1010 1100 0100 0001 1010 1001(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 1010 1100 0100 0001 1010 1001


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 1010 1100 0100 0001 1010 1001 =

0010 1110 1111 1100 1101 1110 0010 1010 1100 0100 0001 1010 1001


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 1010 1100 0100 0001 1010 1001


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

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1010 1100 0100 0001 1010 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