0.000 000 029 364 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 029 364(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
0.000 000 029 364(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: 0.000 000 029 364.

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) 0.000 000 029 364 × 2 = 0 + 0.000 000 058 728;
  • 2) 0.000 000 058 728 × 2 = 0 + 0.000 000 117 456;
  • 3) 0.000 000 117 456 × 2 = 0 + 0.000 000 234 912;
  • 4) 0.000 000 234 912 × 2 = 0 + 0.000 000 469 824;
  • 5) 0.000 000 469 824 × 2 = 0 + 0.000 000 939 648;
  • 6) 0.000 000 939 648 × 2 = 0 + 0.000 001 879 296;
  • 7) 0.000 001 879 296 × 2 = 0 + 0.000 003 758 592;
  • 8) 0.000 003 758 592 × 2 = 0 + 0.000 007 517 184;
  • 9) 0.000 007 517 184 × 2 = 0 + 0.000 015 034 368;
  • 10) 0.000 015 034 368 × 2 = 0 + 0.000 030 068 736;
  • 11) 0.000 030 068 736 × 2 = 0 + 0.000 060 137 472;
  • 12) 0.000 060 137 472 × 2 = 0 + 0.000 120 274 944;
  • 13) 0.000 120 274 944 × 2 = 0 + 0.000 240 549 888;
  • 14) 0.000 240 549 888 × 2 = 0 + 0.000 481 099 776;
  • 15) 0.000 481 099 776 × 2 = 0 + 0.000 962 199 552;
  • 16) 0.000 962 199 552 × 2 = 0 + 0.001 924 399 104;
  • 17) 0.001 924 399 104 × 2 = 0 + 0.003 848 798 208;
  • 18) 0.003 848 798 208 × 2 = 0 + 0.007 697 596 416;
  • 19) 0.007 697 596 416 × 2 = 0 + 0.015 395 192 832;
  • 20) 0.015 395 192 832 × 2 = 0 + 0.030 790 385 664;
  • 21) 0.030 790 385 664 × 2 = 0 + 0.061 580 771 328;
  • 22) 0.061 580 771 328 × 2 = 0 + 0.123 161 542 656;
  • 23) 0.123 161 542 656 × 2 = 0 + 0.246 323 085 312;
  • 24) 0.246 323 085 312 × 2 = 0 + 0.492 646 170 624;
  • 25) 0.492 646 170 624 × 2 = 0 + 0.985 292 341 248;
  • 26) 0.985 292 341 248 × 2 = 1 + 0.970 584 682 496;
  • 27) 0.970 584 682 496 × 2 = 1 + 0.941 169 364 992;
  • 28) 0.941 169 364 992 × 2 = 1 + 0.882 338 729 984;
  • 29) 0.882 338 729 984 × 2 = 1 + 0.764 677 459 968;
  • 30) 0.764 677 459 968 × 2 = 1 + 0.529 354 919 936;
  • 31) 0.529 354 919 936 × 2 = 1 + 0.058 709 839 872;
  • 32) 0.058 709 839 872 × 2 = 0 + 0.117 419 679 744;
  • 33) 0.117 419 679 744 × 2 = 0 + 0.234 839 359 488;
  • 34) 0.234 839 359 488 × 2 = 0 + 0.469 678 718 976;
  • 35) 0.469 678 718 976 × 2 = 0 + 0.939 357 437 952;
  • 36) 0.939 357 437 952 × 2 = 1 + 0.878 714 875 904;
  • 37) 0.878 714 875 904 × 2 = 1 + 0.757 429 751 808;
  • 38) 0.757 429 751 808 × 2 = 1 + 0.514 859 503 616;
  • 39) 0.514 859 503 616 × 2 = 1 + 0.029 719 007 232;
  • 40) 0.029 719 007 232 × 2 = 0 + 0.059 438 014 464;
  • 41) 0.059 438 014 464 × 2 = 0 + 0.118 876 028 928;
  • 42) 0.118 876 028 928 × 2 = 0 + 0.237 752 057 856;
  • 43) 0.237 752 057 856 × 2 = 0 + 0.475 504 115 712;
  • 44) 0.475 504 115 712 × 2 = 0 + 0.951 008 231 424;
  • 45) 0.951 008 231 424 × 2 = 1 + 0.902 016 462 848;
  • 46) 0.902 016 462 848 × 2 = 1 + 0.804 032 925 696;
  • 47) 0.804 032 925 696 × 2 = 1 + 0.608 065 851 392;
  • 48) 0.608 065 851 392 × 2 = 1 + 0.216 131 702 784;
  • 49) 0.216 131 702 784 × 2 = 0 + 0.432 263 405 568;
  • 50) 0.432 263 405 568 × 2 = 0 + 0.864 526 811 136;
  • 51) 0.864 526 811 136 × 2 = 1 + 0.729 053 622 272;
  • 52) 0.729 053 622 272 × 2 = 1 + 0.458 107 244 544;
  • 53) 0.458 107 244 544 × 2 = 0 + 0.916 214 489 088;
  • 54) 0.916 214 489 088 × 2 = 1 + 0.832 428 978 176;
  • 55) 0.832 428 978 176 × 2 = 1 + 0.664 857 956 352;
  • 56) 0.664 857 956 352 × 2 = 1 + 0.329 715 912 704;
  • 57) 0.329 715 912 704 × 2 = 0 + 0.659 431 825 408;
  • 58) 0.659 431 825 408 × 2 = 1 + 0.318 863 650 816;
  • 59) 0.318 863 650 816 × 2 = 0 + 0.637 727 301 632;
  • 60) 0.637 727 301 632 × 2 = 1 + 0.275 454 603 264;
  • 61) 0.275 454 603 264 × 2 = 0 + 0.550 909 206 528;
  • 62) 0.550 909 206 528 × 2 = 1 + 0.101 818 413 056;
  • 63) 0.101 818 413 056 × 2 = 0 + 0.203 636 826 112;
  • 64) 0.203 636 826 112 × 2 = 0 + 0.407 273 652 224;
  • 65) 0.407 273 652 224 × 2 = 0 + 0.814 547 304 448;
  • 66) 0.814 547 304 448 × 2 = 1 + 0.629 094 608 896;
  • 67) 0.629 094 608 896 × 2 = 1 + 0.258 189 217 792;
  • 68) 0.258 189 217 792 × 2 = 0 + 0.516 378 435 584;
  • 69) 0.516 378 435 584 × 2 = 1 + 0.032 756 871 168;
  • 70) 0.032 756 871 168 × 2 = 0 + 0.065 513 742 336;
  • 71) 0.065 513 742 336 × 2 = 0 + 0.131 027 484 672;
  • 72) 0.131 027 484 672 × 2 = 0 + 0.262 054 969 344;
  • 73) 0.262 054 969 344 × 2 = 0 + 0.524 109 938 688;
  • 74) 0.524 109 938 688 × 2 = 1 + 0.048 219 877 376;
  • 75) 0.048 219 877 376 × 2 = 0 + 0.096 439 754 752;
  • 76) 0.096 439 754 752 × 2 = 0 + 0.192 879 509 504;
  • 77) 0.192 879 509 504 × 2 = 0 + 0.385 759 019 008;
  • 78) 0.385 759 019 008 × 2 = 0 + 0.771 518 038 016;

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:


0.000 000 029 364(10) =

0.0000 0000 0000 0000 0000 0000 0111 1110 0001 1110 0000 1111 0011 0111 0101 0100 0110 1000 0100 00(2)

5. Positive number before normalization:

0.000 000 029 364(10) =

0.0000 0000 0000 0000 0000 0000 0111 1110 0001 1110 0000 1111 0011 0111 0101 0100 0110 1000 0100 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 029 364(10) =

0.0000 0000 0000 0000 0000 0000 0111 1110 0001 1110 0000 1111 0011 0111 0101 0100 0110 1000 0100 00(2) =

0.0000 0000 0000 0000 0000 0000 0111 1110 0001 1110 0000 1111 0011 0111 0101 0100 0110 1000 0100 00(2) × 20 =

1.1111 1000 0111 1000 0011 1100 1101 1101 0101 0001 1010 0001 0000(2) × 2-26


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): -26

Mantissa (not normalized):
1.1111 1000 0111 1000 0011 1100 1101 1101 0101 0001 1010 0001 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(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;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 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) =

997(10) =

011 1110 0101(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. 1111 1000 0111 1000 0011 1100 1101 1101 0101 0001 1010 0001 0000 =

1111 1000 0111 1000 0011 1100 1101 1101 0101 0001 1010 0001 0000


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 0101

Mantissa (52 bits) =
1111 1000 0111 1000 0011 1100 1101 1101 0101 0001 1010 0001 0000


Decimal number 0.000 000 029 364 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0101 - 1111 1000 0111 1000 0011 1100 1101 1101 0101 0001 1010 0001 0000

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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