0.000 000 564 358 690 702 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 564 358 690 702(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
0.000 000 564 358 690 702(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: 0.000 000 564 358 690 702.

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) 0.000 000 564 358 690 702 × 2 = 0 + 0.000 001 128 717 381 404;
  • 2) 0.000 001 128 717 381 404 × 2 = 0 + 0.000 002 257 434 762 808;
  • 3) 0.000 002 257 434 762 808 × 2 = 0 + 0.000 004 514 869 525 616;
  • 4) 0.000 004 514 869 525 616 × 2 = 0 + 0.000 009 029 739 051 232;
  • 5) 0.000 009 029 739 051 232 × 2 = 0 + 0.000 018 059 478 102 464;
  • 6) 0.000 018 059 478 102 464 × 2 = 0 + 0.000 036 118 956 204 928;
  • 7) 0.000 036 118 956 204 928 × 2 = 0 + 0.000 072 237 912 409 856;
  • 8) 0.000 072 237 912 409 856 × 2 = 0 + 0.000 144 475 824 819 712;
  • 9) 0.000 144 475 824 819 712 × 2 = 0 + 0.000 288 951 649 639 424;
  • 10) 0.000 288 951 649 639 424 × 2 = 0 + 0.000 577 903 299 278 848;
  • 11) 0.000 577 903 299 278 848 × 2 = 0 + 0.001 155 806 598 557 696;
  • 12) 0.001 155 806 598 557 696 × 2 = 0 + 0.002 311 613 197 115 392;
  • 13) 0.002 311 613 197 115 392 × 2 = 0 + 0.004 623 226 394 230 784;
  • 14) 0.004 623 226 394 230 784 × 2 = 0 + 0.009 246 452 788 461 568;
  • 15) 0.009 246 452 788 461 568 × 2 = 0 + 0.018 492 905 576 923 136;
  • 16) 0.018 492 905 576 923 136 × 2 = 0 + 0.036 985 811 153 846 272;
  • 17) 0.036 985 811 153 846 272 × 2 = 0 + 0.073 971 622 307 692 544;
  • 18) 0.073 971 622 307 692 544 × 2 = 0 + 0.147 943 244 615 385 088;
  • 19) 0.147 943 244 615 385 088 × 2 = 0 + 0.295 886 489 230 770 176;
  • 20) 0.295 886 489 230 770 176 × 2 = 0 + 0.591 772 978 461 540 352;
  • 21) 0.591 772 978 461 540 352 × 2 = 1 + 0.183 545 956 923 080 704;
  • 22) 0.183 545 956 923 080 704 × 2 = 0 + 0.367 091 913 846 161 408;
  • 23) 0.367 091 913 846 161 408 × 2 = 0 + 0.734 183 827 692 322 816;
  • 24) 0.734 183 827 692 322 816 × 2 = 1 + 0.468 367 655 384 645 632;
  • 25) 0.468 367 655 384 645 632 × 2 = 0 + 0.936 735 310 769 291 264;
  • 26) 0.936 735 310 769 291 264 × 2 = 1 + 0.873 470 621 538 582 528;
  • 27) 0.873 470 621 538 582 528 × 2 = 1 + 0.746 941 243 077 165 056;
  • 28) 0.746 941 243 077 165 056 × 2 = 1 + 0.493 882 486 154 330 112;
  • 29) 0.493 882 486 154 330 112 × 2 = 0 + 0.987 764 972 308 660 224;
  • 30) 0.987 764 972 308 660 224 × 2 = 1 + 0.975 529 944 617 320 448;
  • 31) 0.975 529 944 617 320 448 × 2 = 1 + 0.951 059 889 234 640 896;
  • 32) 0.951 059 889 234 640 896 × 2 = 1 + 0.902 119 778 469 281 792;
  • 33) 0.902 119 778 469 281 792 × 2 = 1 + 0.804 239 556 938 563 584;
  • 34) 0.804 239 556 938 563 584 × 2 = 1 + 0.608 479 113 877 127 168;
  • 35) 0.608 479 113 877 127 168 × 2 = 1 + 0.216 958 227 754 254 336;
  • 36) 0.216 958 227 754 254 336 × 2 = 0 + 0.433 916 455 508 508 672;
  • 37) 0.433 916 455 508 508 672 × 2 = 0 + 0.867 832 911 017 017 344;
  • 38) 0.867 832 911 017 017 344 × 2 = 1 + 0.735 665 822 034 034 688;
  • 39) 0.735 665 822 034 034 688 × 2 = 1 + 0.471 331 644 068 069 376;
  • 40) 0.471 331 644 068 069 376 × 2 = 0 + 0.942 663 288 136 138 752;
  • 41) 0.942 663 288 136 138 752 × 2 = 1 + 0.885 326 576 272 277 504;
  • 42) 0.885 326 576 272 277 504 × 2 = 1 + 0.770 653 152 544 555 008;
  • 43) 0.770 653 152 544 555 008 × 2 = 1 + 0.541 306 305 089 110 016;
  • 44) 0.541 306 305 089 110 016 × 2 = 1 + 0.082 612 610 178 220 032;
  • 45) 0.082 612 610 178 220 032 × 2 = 0 + 0.165 225 220 356 440 064;
  • 46) 0.165 225 220 356 440 064 × 2 = 0 + 0.330 450 440 712 880 128;
  • 47) 0.330 450 440 712 880 128 × 2 = 0 + 0.660 900 881 425 760 256;
  • 48) 0.660 900 881 425 760 256 × 2 = 1 + 0.321 801 762 851 520 512;
  • 49) 0.321 801 762 851 520 512 × 2 = 0 + 0.643 603 525 703 041 024;
  • 50) 0.643 603 525 703 041 024 × 2 = 1 + 0.287 207 051 406 082 048;
  • 51) 0.287 207 051 406 082 048 × 2 = 0 + 0.574 414 102 812 164 096;
  • 52) 0.574 414 102 812 164 096 × 2 = 1 + 0.148 828 205 624 328 192;
  • 53) 0.148 828 205 624 328 192 × 2 = 0 + 0.297 656 411 248 656 384;
  • 54) 0.297 656 411 248 656 384 × 2 = 0 + 0.595 312 822 497 312 768;
  • 55) 0.595 312 822 497 312 768 × 2 = 1 + 0.190 625 644 994 625 536;
  • 56) 0.190 625 644 994 625 536 × 2 = 0 + 0.381 251 289 989 251 072;
  • 57) 0.381 251 289 989 251 072 × 2 = 0 + 0.762 502 579 978 502 144;
  • 58) 0.762 502 579 978 502 144 × 2 = 1 + 0.525 005 159 957 004 288;
  • 59) 0.525 005 159 957 004 288 × 2 = 1 + 0.050 010 319 914 008 576;
  • 60) 0.050 010 319 914 008 576 × 2 = 0 + 0.100 020 639 828 017 152;
  • 61) 0.100 020 639 828 017 152 × 2 = 0 + 0.200 041 279 656 034 304;
  • 62) 0.200 041 279 656 034 304 × 2 = 0 + 0.400 082 559 312 068 608;
  • 63) 0.400 082 559 312 068 608 × 2 = 0 + 0.800 165 118 624 137 216;
  • 64) 0.800 165 118 624 137 216 × 2 = 1 + 0.600 330 237 248 274 432;
  • 65) 0.600 330 237 248 274 432 × 2 = 1 + 0.200 660 474 496 548 864;
  • 66) 0.200 660 474 496 548 864 × 2 = 0 + 0.401 320 948 993 097 728;
  • 67) 0.401 320 948 993 097 728 × 2 = 0 + 0.802 641 897 986 195 456;
  • 68) 0.802 641 897 986 195 456 × 2 = 1 + 0.605 283 795 972 390 912;
  • 69) 0.605 283 795 972 390 912 × 2 = 1 + 0.210 567 591 944 781 824;
  • 70) 0.210 567 591 944 781 824 × 2 = 0 + 0.421 135 183 889 563 648;
  • 71) 0.421 135 183 889 563 648 × 2 = 0 + 0.842 270 367 779 127 296;
  • 72) 0.842 270 367 779 127 296 × 2 = 1 + 0.684 540 735 558 254 592;
  • 73) 0.684 540 735 558 254 592 × 2 = 1 + 0.369 081 471 116 509 184;

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:


0.000 000 564 358 690 702(10) =

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

5. Positive number before normalization:

0.000 000 564 358 690 702(10) =

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


0.000 000 564 358 690 702(10) =

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

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

1.0010 1110 1111 1100 1101 1110 0010 1010 0100 1100 0011 0011 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 1010 0100 1100 0011 0011 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 1010 0100 1100 0011 0011 0011 =

0010 1110 1111 1100 1101 1110 0010 1010 0100 1100 0011 0011 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 1010 0100 1100 0011 0011 0011


Decimal number 0.000 000 564 358 690 702 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1010 0100 1100 0011 0011 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