0.000 000 564 358 691 012 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 564 358 691 012(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
0.000 000 564 358 691 012(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: 0.000 000 564 358 691 012.

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) 0.000 000 564 358 691 012 × 2 = 0 + 0.000 001 128 717 382 024;
  • 2) 0.000 001 128 717 382 024 × 2 = 0 + 0.000 002 257 434 764 048;
  • 3) 0.000 002 257 434 764 048 × 2 = 0 + 0.000 004 514 869 528 096;
  • 4) 0.000 004 514 869 528 096 × 2 = 0 + 0.000 009 029 739 056 192;
  • 5) 0.000 009 029 739 056 192 × 2 = 0 + 0.000 018 059 478 112 384;
  • 6) 0.000 018 059 478 112 384 × 2 = 0 + 0.000 036 118 956 224 768;
  • 7) 0.000 036 118 956 224 768 × 2 = 0 + 0.000 072 237 912 449 536;
  • 8) 0.000 072 237 912 449 536 × 2 = 0 + 0.000 144 475 824 899 072;
  • 9) 0.000 144 475 824 899 072 × 2 = 0 + 0.000 288 951 649 798 144;
  • 10) 0.000 288 951 649 798 144 × 2 = 0 + 0.000 577 903 299 596 288;
  • 11) 0.000 577 903 299 596 288 × 2 = 0 + 0.001 155 806 599 192 576;
  • 12) 0.001 155 806 599 192 576 × 2 = 0 + 0.002 311 613 198 385 152;
  • 13) 0.002 311 613 198 385 152 × 2 = 0 + 0.004 623 226 396 770 304;
  • 14) 0.004 623 226 396 770 304 × 2 = 0 + 0.009 246 452 793 540 608;
  • 15) 0.009 246 452 793 540 608 × 2 = 0 + 0.018 492 905 587 081 216;
  • 16) 0.018 492 905 587 081 216 × 2 = 0 + 0.036 985 811 174 162 432;
  • 17) 0.036 985 811 174 162 432 × 2 = 0 + 0.073 971 622 348 324 864;
  • 18) 0.073 971 622 348 324 864 × 2 = 0 + 0.147 943 244 696 649 728;
  • 19) 0.147 943 244 696 649 728 × 2 = 0 + 0.295 886 489 393 299 456;
  • 20) 0.295 886 489 393 299 456 × 2 = 0 + 0.591 772 978 786 598 912;
  • 21) 0.591 772 978 786 598 912 × 2 = 1 + 0.183 545 957 573 197 824;
  • 22) 0.183 545 957 573 197 824 × 2 = 0 + 0.367 091 915 146 395 648;
  • 23) 0.367 091 915 146 395 648 × 2 = 0 + 0.734 183 830 292 791 296;
  • 24) 0.734 183 830 292 791 296 × 2 = 1 + 0.468 367 660 585 582 592;
  • 25) 0.468 367 660 585 582 592 × 2 = 0 + 0.936 735 321 171 165 184;
  • 26) 0.936 735 321 171 165 184 × 2 = 1 + 0.873 470 642 342 330 368;
  • 27) 0.873 470 642 342 330 368 × 2 = 1 + 0.746 941 284 684 660 736;
  • 28) 0.746 941 284 684 660 736 × 2 = 1 + 0.493 882 569 369 321 472;
  • 29) 0.493 882 569 369 321 472 × 2 = 0 + 0.987 765 138 738 642 944;
  • 30) 0.987 765 138 738 642 944 × 2 = 1 + 0.975 530 277 477 285 888;
  • 31) 0.975 530 277 477 285 888 × 2 = 1 + 0.951 060 554 954 571 776;
  • 32) 0.951 060 554 954 571 776 × 2 = 1 + 0.902 121 109 909 143 552;
  • 33) 0.902 121 109 909 143 552 × 2 = 1 + 0.804 242 219 818 287 104;
  • 34) 0.804 242 219 818 287 104 × 2 = 1 + 0.608 484 439 636 574 208;
  • 35) 0.608 484 439 636 574 208 × 2 = 1 + 0.216 968 879 273 148 416;
  • 36) 0.216 968 879 273 148 416 × 2 = 0 + 0.433 937 758 546 296 832;
  • 37) 0.433 937 758 546 296 832 × 2 = 0 + 0.867 875 517 092 593 664;
  • 38) 0.867 875 517 092 593 664 × 2 = 1 + 0.735 751 034 185 187 328;
  • 39) 0.735 751 034 185 187 328 × 2 = 1 + 0.471 502 068 370 374 656;
  • 40) 0.471 502 068 370 374 656 × 2 = 0 + 0.943 004 136 740 749 312;
  • 41) 0.943 004 136 740 749 312 × 2 = 1 + 0.886 008 273 481 498 624;
  • 42) 0.886 008 273 481 498 624 × 2 = 1 + 0.772 016 546 962 997 248;
  • 43) 0.772 016 546 962 997 248 × 2 = 1 + 0.544 033 093 925 994 496;
  • 44) 0.544 033 093 925 994 496 × 2 = 1 + 0.088 066 187 851 988 992;
  • 45) 0.088 066 187 851 988 992 × 2 = 0 + 0.176 132 375 703 977 984;
  • 46) 0.176 132 375 703 977 984 × 2 = 0 + 0.352 264 751 407 955 968;
  • 47) 0.352 264 751 407 955 968 × 2 = 0 + 0.704 529 502 815 911 936;
  • 48) 0.704 529 502 815 911 936 × 2 = 1 + 0.409 059 005 631 823 872;
  • 49) 0.409 059 005 631 823 872 × 2 = 0 + 0.818 118 011 263 647 744;
  • 50) 0.818 118 011 263 647 744 × 2 = 1 + 0.636 236 022 527 295 488;
  • 51) 0.636 236 022 527 295 488 × 2 = 1 + 0.272 472 045 054 590 976;
  • 52) 0.272 472 045 054 590 976 × 2 = 0 + 0.544 944 090 109 181 952;
  • 53) 0.544 944 090 109 181 952 × 2 = 1 + 0.089 888 180 218 363 904;
  • 54) 0.089 888 180 218 363 904 × 2 = 0 + 0.179 776 360 436 727 808;
  • 55) 0.179 776 360 436 727 808 × 2 = 0 + 0.359 552 720 873 455 616;
  • 56) 0.359 552 720 873 455 616 × 2 = 0 + 0.719 105 441 746 911 232;
  • 57) 0.719 105 441 746 911 232 × 2 = 1 + 0.438 210 883 493 822 464;
  • 58) 0.438 210 883 493 822 464 × 2 = 0 + 0.876 421 766 987 644 928;
  • 59) 0.876 421 766 987 644 928 × 2 = 1 + 0.752 843 533 975 289 856;
  • 60) 0.752 843 533 975 289 856 × 2 = 1 + 0.505 687 067 950 579 712;
  • 61) 0.505 687 067 950 579 712 × 2 = 1 + 0.011 374 135 901 159 424;
  • 62) 0.011 374 135 901 159 424 × 2 = 0 + 0.022 748 271 802 318 848;
  • 63) 0.022 748 271 802 318 848 × 2 = 0 + 0.045 496 543 604 637 696;
  • 64) 0.045 496 543 604 637 696 × 2 = 0 + 0.090 993 087 209 275 392;
  • 65) 0.090 993 087 209 275 392 × 2 = 0 + 0.181 986 174 418 550 784;
  • 66) 0.181 986 174 418 550 784 × 2 = 0 + 0.363 972 348 837 101 568;
  • 67) 0.363 972 348 837 101 568 × 2 = 0 + 0.727 944 697 674 203 136;
  • 68) 0.727 944 697 674 203 136 × 2 = 1 + 0.455 889 395 348 406 272;
  • 69) 0.455 889 395 348 406 272 × 2 = 0 + 0.911 778 790 696 812 544;
  • 70) 0.911 778 790 696 812 544 × 2 = 1 + 0.823 557 581 393 625 088;
  • 71) 0.823 557 581 393 625 088 × 2 = 1 + 0.647 115 162 787 250 176;
  • 72) 0.647 115 162 787 250 176 × 2 = 1 + 0.294 230 325 574 500 352;
  • 73) 0.294 230 325 574 500 352 × 2 = 0 + 0.588 460 651 149 000 704;

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:


0.000 000 564 358 691 012(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0110 1000 1011 1000 0001 0111 0(2)

5. Positive number before normalization:

0.000 000 564 358 691 012(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0110 1000 1011 1000 0001 0111 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:


0.000 000 564 358 691 012(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0110 1000 1011 1000 0001 0111 0(2) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0110 1000 1011 1000 0001 0111 0(2) × 20 =

1.0010 1110 1111 1100 1101 1110 0010 1101 0001 0111 0000 0010 1110(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 1101 0001 0111 0000 0010 1110


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 1101 0001 0111 0000 0010 1110 =

0010 1110 1111 1100 1101 1110 0010 1101 0001 0111 0000 0010 1110


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 1101 0001 0111 0000 0010 1110


Decimal number 0.000 000 564 358 691 012 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1101 0001 0111 0000 0010 1110

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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