0.000 000 000 000 028 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 028(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
0.000 000 000 000 028(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: 0.000 000 000 000 028.

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) 0.000 000 000 000 028 × 2 = 0 + 0.000 000 000 000 056;
  • 2) 0.000 000 000 000 056 × 2 = 0 + 0.000 000 000 000 112;
  • 3) 0.000 000 000 000 112 × 2 = 0 + 0.000 000 000 000 224;
  • 4) 0.000 000 000 000 224 × 2 = 0 + 0.000 000 000 000 448;
  • 5) 0.000 000 000 000 448 × 2 = 0 + 0.000 000 000 000 896;
  • 6) 0.000 000 000 000 896 × 2 = 0 + 0.000 000 000 001 792;
  • 7) 0.000 000 000 001 792 × 2 = 0 + 0.000 000 000 003 584;
  • 8) 0.000 000 000 003 584 × 2 = 0 + 0.000 000 000 007 168;
  • 9) 0.000 000 000 007 168 × 2 = 0 + 0.000 000 000 014 336;
  • 10) 0.000 000 000 014 336 × 2 = 0 + 0.000 000 000 028 672;
  • 11) 0.000 000 000 028 672 × 2 = 0 + 0.000 000 000 057 344;
  • 12) 0.000 000 000 057 344 × 2 = 0 + 0.000 000 000 114 688;
  • 13) 0.000 000 000 114 688 × 2 = 0 + 0.000 000 000 229 376;
  • 14) 0.000 000 000 229 376 × 2 = 0 + 0.000 000 000 458 752;
  • 15) 0.000 000 000 458 752 × 2 = 0 + 0.000 000 000 917 504;
  • 16) 0.000 000 000 917 504 × 2 = 0 + 0.000 000 001 835 008;
  • 17) 0.000 000 001 835 008 × 2 = 0 + 0.000 000 003 670 016;
  • 18) 0.000 000 003 670 016 × 2 = 0 + 0.000 000 007 340 032;
  • 19) 0.000 000 007 340 032 × 2 = 0 + 0.000 000 014 680 064;
  • 20) 0.000 000 014 680 064 × 2 = 0 + 0.000 000 029 360 128;
  • 21) 0.000 000 029 360 128 × 2 = 0 + 0.000 000 058 720 256;
  • 22) 0.000 000 058 720 256 × 2 = 0 + 0.000 000 117 440 512;
  • 23) 0.000 000 117 440 512 × 2 = 0 + 0.000 000 234 881 024;
  • 24) 0.000 000 234 881 024 × 2 = 0 + 0.000 000 469 762 048;
  • 25) 0.000 000 469 762 048 × 2 = 0 + 0.000 000 939 524 096;
  • 26) 0.000 000 939 524 096 × 2 = 0 + 0.000 001 879 048 192;
  • 27) 0.000 001 879 048 192 × 2 = 0 + 0.000 003 758 096 384;
  • 28) 0.000 003 758 096 384 × 2 = 0 + 0.000 007 516 192 768;
  • 29) 0.000 007 516 192 768 × 2 = 0 + 0.000 015 032 385 536;
  • 30) 0.000 015 032 385 536 × 2 = 0 + 0.000 030 064 771 072;
  • 31) 0.000 030 064 771 072 × 2 = 0 + 0.000 060 129 542 144;
  • 32) 0.000 060 129 542 144 × 2 = 0 + 0.000 120 259 084 288;
  • 33) 0.000 120 259 084 288 × 2 = 0 + 0.000 240 518 168 576;
  • 34) 0.000 240 518 168 576 × 2 = 0 + 0.000 481 036 337 152;
  • 35) 0.000 481 036 337 152 × 2 = 0 + 0.000 962 072 674 304;
  • 36) 0.000 962 072 674 304 × 2 = 0 + 0.001 924 145 348 608;
  • 37) 0.001 924 145 348 608 × 2 = 0 + 0.003 848 290 697 216;
  • 38) 0.003 848 290 697 216 × 2 = 0 + 0.007 696 581 394 432;
  • 39) 0.007 696 581 394 432 × 2 = 0 + 0.015 393 162 788 864;
  • 40) 0.015 393 162 788 864 × 2 = 0 + 0.030 786 325 577 728;
  • 41) 0.030 786 325 577 728 × 2 = 0 + 0.061 572 651 155 456;
  • 42) 0.061 572 651 155 456 × 2 = 0 + 0.123 145 302 310 912;
  • 43) 0.123 145 302 310 912 × 2 = 0 + 0.246 290 604 621 824;
  • 44) 0.246 290 604 621 824 × 2 = 0 + 0.492 581 209 243 648;
  • 45) 0.492 581 209 243 648 × 2 = 0 + 0.985 162 418 487 296;
  • 46) 0.985 162 418 487 296 × 2 = 1 + 0.970 324 836 974 592;
  • 47) 0.970 324 836 974 592 × 2 = 1 + 0.940 649 673 949 184;
  • 48) 0.940 649 673 949 184 × 2 = 1 + 0.881 299 347 898 368;
  • 49) 0.881 299 347 898 368 × 2 = 1 + 0.762 598 695 796 736;
  • 50) 0.762 598 695 796 736 × 2 = 1 + 0.525 197 391 593 472;
  • 51) 0.525 197 391 593 472 × 2 = 1 + 0.050 394 783 186 944;
  • 52) 0.050 394 783 186 944 × 2 = 0 + 0.100 789 566 373 888;
  • 53) 0.100 789 566 373 888 × 2 = 0 + 0.201 579 132 747 776;
  • 54) 0.201 579 132 747 776 × 2 = 0 + 0.403 158 265 495 552;
  • 55) 0.403 158 265 495 552 × 2 = 0 + 0.806 316 530 991 104;
  • 56) 0.806 316 530 991 104 × 2 = 1 + 0.612 633 061 982 208;
  • 57) 0.612 633 061 982 208 × 2 = 1 + 0.225 266 123 964 416;
  • 58) 0.225 266 123 964 416 × 2 = 0 + 0.450 532 247 928 832;
  • 59) 0.450 532 247 928 832 × 2 = 0 + 0.901 064 495 857 664;
  • 60) 0.901 064 495 857 664 × 2 = 1 + 0.802 128 991 715 328;
  • 61) 0.802 128 991 715 328 × 2 = 1 + 0.604 257 983 430 656;
  • 62) 0.604 257 983 430 656 × 2 = 1 + 0.208 515 966 861 312;
  • 63) 0.208 515 966 861 312 × 2 = 0 + 0.417 031 933 722 624;
  • 64) 0.417 031 933 722 624 × 2 = 0 + 0.834 063 867 445 248;
  • 65) 0.834 063 867 445 248 × 2 = 1 + 0.668 127 734 890 496;
  • 66) 0.668 127 734 890 496 × 2 = 1 + 0.336 255 469 780 992;
  • 67) 0.336 255 469 780 992 × 2 = 0 + 0.672 510 939 561 984;
  • 68) 0.672 510 939 561 984 × 2 = 1 + 0.345 021 879 123 968;
  • 69) 0.345 021 879 123 968 × 2 = 0 + 0.690 043 758 247 936;
  • 70) 0.690 043 758 247 936 × 2 = 1 + 0.380 087 516 495 872;
  • 71) 0.380 087 516 495 872 × 2 = 0 + 0.760 175 032 991 744;
  • 72) 0.760 175 032 991 744 × 2 = 1 + 0.520 350 065 983 488;
  • 73) 0.520 350 065 983 488 × 2 = 1 + 0.040 700 131 966 976;
  • 74) 0.040 700 131 966 976 × 2 = 0 + 0.081 400 263 933 952;
  • 75) 0.081 400 263 933 952 × 2 = 0 + 0.162 800 527 867 904;
  • 76) 0.162 800 527 867 904 × 2 = 0 + 0.325 601 055 735 808;
  • 77) 0.325 601 055 735 808 × 2 = 0 + 0.651 202 111 471 616;
  • 78) 0.651 202 111 471 616 × 2 = 1 + 0.302 404 222 943 232;
  • 79) 0.302 404 222 943 232 × 2 = 0 + 0.604 808 445 886 464;
  • 80) 0.604 808 445 886 464 × 2 = 1 + 0.209 616 891 772 928;
  • 81) 0.209 616 891 772 928 × 2 = 0 + 0.419 233 783 545 856;
  • 82) 0.419 233 783 545 856 × 2 = 0 + 0.838 467 567 091 712;
  • 83) 0.838 467 567 091 712 × 2 = 1 + 0.676 935 134 183 424;
  • 84) 0.676 935 134 183 424 × 2 = 1 + 0.353 870 268 366 848;
  • 85) 0.353 870 268 366 848 × 2 = 0 + 0.707 740 536 733 696;
  • 86) 0.707 740 536 733 696 × 2 = 1 + 0.415 481 073 467 392;
  • 87) 0.415 481 073 467 392 × 2 = 0 + 0.830 962 146 934 784;
  • 88) 0.830 962 146 934 784 × 2 = 1 + 0.661 924 293 869 568;
  • 89) 0.661 924 293 869 568 × 2 = 1 + 0.323 848 587 739 136;
  • 90) 0.323 848 587 739 136 × 2 = 0 + 0.647 697 175 478 272;
  • 91) 0.647 697 175 478 272 × 2 = 1 + 0.295 394 350 956 544;
  • 92) 0.295 394 350 956 544 × 2 = 0 + 0.590 788 701 913 088;
  • 93) 0.590 788 701 913 088 × 2 = 1 + 0.181 577 403 826 176;
  • 94) 0.181 577 403 826 176 × 2 = 0 + 0.363 154 807 652 352;
  • 95) 0.363 154 807 652 352 × 2 = 0 + 0.726 309 615 304 704;
  • 96) 0.726 309 615 304 704 × 2 = 1 + 0.452 619 230 609 408;
  • 97) 0.452 619 230 609 408 × 2 = 0 + 0.905 238 461 218 816;
  • 98) 0.905 238 461 218 816 × 2 = 1 + 0.810 476 922 437 632;

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:


0.000 000 000 000 028(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1110 0001 1001 1100 1101 0101 1000 0101 0011 0101 1010 1001 01(2)

5. Positive number before normalization:

0.000 000 000 000 028(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1110 0001 1001 1100 1101 0101 1000 0101 0011 0101 1010 1001 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 028(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1110 0001 1001 1100 1101 0101 1000 0101 0011 0101 1010 1001 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 1110 0001 1001 1100 1101 0101 1000 0101 0011 0101 1010 1001 01(2) × 20 =

1.1111 1000 0110 0111 0011 0101 0110 0001 0100 1101 0110 1010 0101(2) × 2-46


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): -46

Mantissa (not normalized):
1.1111 1000 0110 0111 0011 0101 0110 0001 0100 1101 0110 1010 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

977(10) =

011 1101 0001(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 0110 0111 0011 0101 0110 0001 0100 1101 0110 1010 0101 =

1111 1000 0110 0111 0011 0101 0110 0001 0100 1101 0110 1010 0101


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 1101 0001

Mantissa (52 bits) =
1111 1000 0110 0111 0011 0101 0110 0001 0100 1101 0110 1010 0101


Decimal number 0.000 000 000 000 028 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0001 - 1111 1000 0110 0111 0011 0101 0110 0001 0100 1101 0110 1010 0101

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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