0.000 000 029 429 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 029 429(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
0.000 000 029 429(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: 0.000 000 029 429.

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) 0.000 000 029 429 × 2 = 0 + 0.000 000 058 858;
  • 2) 0.000 000 058 858 × 2 = 0 + 0.000 000 117 716;
  • 3) 0.000 000 117 716 × 2 = 0 + 0.000 000 235 432;
  • 4) 0.000 000 235 432 × 2 = 0 + 0.000 000 470 864;
  • 5) 0.000 000 470 864 × 2 = 0 + 0.000 000 941 728;
  • 6) 0.000 000 941 728 × 2 = 0 + 0.000 001 883 456;
  • 7) 0.000 001 883 456 × 2 = 0 + 0.000 003 766 912;
  • 8) 0.000 003 766 912 × 2 = 0 + 0.000 007 533 824;
  • 9) 0.000 007 533 824 × 2 = 0 + 0.000 015 067 648;
  • 10) 0.000 015 067 648 × 2 = 0 + 0.000 030 135 296;
  • 11) 0.000 030 135 296 × 2 = 0 + 0.000 060 270 592;
  • 12) 0.000 060 270 592 × 2 = 0 + 0.000 120 541 184;
  • 13) 0.000 120 541 184 × 2 = 0 + 0.000 241 082 368;
  • 14) 0.000 241 082 368 × 2 = 0 + 0.000 482 164 736;
  • 15) 0.000 482 164 736 × 2 = 0 + 0.000 964 329 472;
  • 16) 0.000 964 329 472 × 2 = 0 + 0.001 928 658 944;
  • 17) 0.001 928 658 944 × 2 = 0 + 0.003 857 317 888;
  • 18) 0.003 857 317 888 × 2 = 0 + 0.007 714 635 776;
  • 19) 0.007 714 635 776 × 2 = 0 + 0.015 429 271 552;
  • 20) 0.015 429 271 552 × 2 = 0 + 0.030 858 543 104;
  • 21) 0.030 858 543 104 × 2 = 0 + 0.061 717 086 208;
  • 22) 0.061 717 086 208 × 2 = 0 + 0.123 434 172 416;
  • 23) 0.123 434 172 416 × 2 = 0 + 0.246 868 344 832;
  • 24) 0.246 868 344 832 × 2 = 0 + 0.493 736 689 664;
  • 25) 0.493 736 689 664 × 2 = 0 + 0.987 473 379 328;
  • 26) 0.987 473 379 328 × 2 = 1 + 0.974 946 758 656;
  • 27) 0.974 946 758 656 × 2 = 1 + 0.949 893 517 312;
  • 28) 0.949 893 517 312 × 2 = 1 + 0.899 787 034 624;
  • 29) 0.899 787 034 624 × 2 = 1 + 0.799 574 069 248;
  • 30) 0.799 574 069 248 × 2 = 1 + 0.599 148 138 496;
  • 31) 0.599 148 138 496 × 2 = 1 + 0.198 296 276 992;
  • 32) 0.198 296 276 992 × 2 = 0 + 0.396 592 553 984;
  • 33) 0.396 592 553 984 × 2 = 0 + 0.793 185 107 968;
  • 34) 0.793 185 107 968 × 2 = 1 + 0.586 370 215 936;
  • 35) 0.586 370 215 936 × 2 = 1 + 0.172 740 431 872;
  • 36) 0.172 740 431 872 × 2 = 0 + 0.345 480 863 744;
  • 37) 0.345 480 863 744 × 2 = 0 + 0.690 961 727 488;
  • 38) 0.690 961 727 488 × 2 = 1 + 0.381 923 454 976;
  • 39) 0.381 923 454 976 × 2 = 0 + 0.763 846 909 952;
  • 40) 0.763 846 909 952 × 2 = 1 + 0.527 693 819 904;
  • 41) 0.527 693 819 904 × 2 = 1 + 0.055 387 639 808;
  • 42) 0.055 387 639 808 × 2 = 0 + 0.110 775 279 616;
  • 43) 0.110 775 279 616 × 2 = 0 + 0.221 550 559 232;
  • 44) 0.221 550 559 232 × 2 = 0 + 0.443 101 118 464;
  • 45) 0.443 101 118 464 × 2 = 0 + 0.886 202 236 928;
  • 46) 0.886 202 236 928 × 2 = 1 + 0.772 404 473 856;
  • 47) 0.772 404 473 856 × 2 = 1 + 0.544 808 947 712;
  • 48) 0.544 808 947 712 × 2 = 1 + 0.089 617 895 424;
  • 49) 0.089 617 895 424 × 2 = 0 + 0.179 235 790 848;
  • 50) 0.179 235 790 848 × 2 = 0 + 0.358 471 581 696;
  • 51) 0.358 471 581 696 × 2 = 0 + 0.716 943 163 392;
  • 52) 0.716 943 163 392 × 2 = 1 + 0.433 886 326 784;
  • 53) 0.433 886 326 784 × 2 = 0 + 0.867 772 653 568;
  • 54) 0.867 772 653 568 × 2 = 1 + 0.735 545 307 136;
  • 55) 0.735 545 307 136 × 2 = 1 + 0.471 090 614 272;
  • 56) 0.471 090 614 272 × 2 = 0 + 0.942 181 228 544;
  • 57) 0.942 181 228 544 × 2 = 1 + 0.884 362 457 088;
  • 58) 0.884 362 457 088 × 2 = 1 + 0.768 724 914 176;
  • 59) 0.768 724 914 176 × 2 = 1 + 0.537 449 828 352;
  • 60) 0.537 449 828 352 × 2 = 1 + 0.074 899 656 704;
  • 61) 0.074 899 656 704 × 2 = 0 + 0.149 799 313 408;
  • 62) 0.149 799 313 408 × 2 = 0 + 0.299 598 626 816;
  • 63) 0.299 598 626 816 × 2 = 0 + 0.599 197 253 632;
  • 64) 0.599 197 253 632 × 2 = 1 + 0.198 394 507 264;
  • 65) 0.198 394 507 264 × 2 = 0 + 0.396 789 014 528;
  • 66) 0.396 789 014 528 × 2 = 0 + 0.793 578 029 056;
  • 67) 0.793 578 029 056 × 2 = 1 + 0.587 156 058 112;
  • 68) 0.587 156 058 112 × 2 = 1 + 0.174 312 116 224;
  • 69) 0.174 312 116 224 × 2 = 0 + 0.348 624 232 448;
  • 70) 0.348 624 232 448 × 2 = 0 + 0.697 248 464 896;
  • 71) 0.697 248 464 896 × 2 = 1 + 0.394 496 929 792;
  • 72) 0.394 496 929 792 × 2 = 0 + 0.788 993 859 584;
  • 73) 0.788 993 859 584 × 2 = 1 + 0.577 987 719 168;
  • 74) 0.577 987 719 168 × 2 = 1 + 0.155 975 438 336;
  • 75) 0.155 975 438 336 × 2 = 0 + 0.311 950 876 672;
  • 76) 0.311 950 876 672 × 2 = 0 + 0.623 901 753 344;
  • 77) 0.623 901 753 344 × 2 = 1 + 0.247 803 506 688;
  • 78) 0.247 803 506 688 × 2 = 0 + 0.495 607 013 376;

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:


0.000 000 029 429(10) =

0.0000 0000 0000 0000 0000 0000 0111 1110 0110 0101 1000 0111 0001 0110 1111 0001 0011 0010 1100 10(2)

5. Positive number before normalization:

0.000 000 029 429(10) =

0.0000 0000 0000 0000 0000 0000 0111 1110 0110 0101 1000 0111 0001 0110 1111 0001 0011 0010 1100 10(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:


0.000 000 029 429(10) =

0.0000 0000 0000 0000 0000 0000 0111 1110 0110 0101 1000 0111 0001 0110 1111 0001 0011 0010 1100 10(2) =

0.0000 0000 0000 0000 0000 0000 0111 1110 0110 0101 1000 0111 0001 0110 1111 0001 0011 0010 1100 10(2) × 20 =

1.1111 1001 1001 0110 0001 1100 0101 1011 1100 0100 1100 1011 0010(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 1001 1001 0110 0001 1100 0101 1011 1100 0100 1100 1011 0010


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 1001 1001 0110 0001 1100 0101 1011 1100 0100 1100 1011 0010 =

1111 1001 1001 0110 0001 1100 0101 1011 1100 0100 1100 1011 0010


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 1001 1001 0110 0001 1100 0101 1011 1100 0100 1100 1011 0010


Decimal number 0.000 000 029 429 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0101 - 1111 1001 1001 0110 0001 1100 0101 1011 1100 0100 1100 1011 0010

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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