0.000 000 000 000 133 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 133(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
0.000 000 000 000 133(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: 0.000 000 000 000 133.

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) 0.000 000 000 000 133 × 2 = 0 + 0.000 000 000 000 266;
  • 2) 0.000 000 000 000 266 × 2 = 0 + 0.000 000 000 000 532;
  • 3) 0.000 000 000 000 532 × 2 = 0 + 0.000 000 000 001 064;
  • 4) 0.000 000 000 001 064 × 2 = 0 + 0.000 000 000 002 128;
  • 5) 0.000 000 000 002 128 × 2 = 0 + 0.000 000 000 004 256;
  • 6) 0.000 000 000 004 256 × 2 = 0 + 0.000 000 000 008 512;
  • 7) 0.000 000 000 008 512 × 2 = 0 + 0.000 000 000 017 024;
  • 8) 0.000 000 000 017 024 × 2 = 0 + 0.000 000 000 034 048;
  • 9) 0.000 000 000 034 048 × 2 = 0 + 0.000 000 000 068 096;
  • 10) 0.000 000 000 068 096 × 2 = 0 + 0.000 000 000 136 192;
  • 11) 0.000 000 000 136 192 × 2 = 0 + 0.000 000 000 272 384;
  • 12) 0.000 000 000 272 384 × 2 = 0 + 0.000 000 000 544 768;
  • 13) 0.000 000 000 544 768 × 2 = 0 + 0.000 000 001 089 536;
  • 14) 0.000 000 001 089 536 × 2 = 0 + 0.000 000 002 179 072;
  • 15) 0.000 000 002 179 072 × 2 = 0 + 0.000 000 004 358 144;
  • 16) 0.000 000 004 358 144 × 2 = 0 + 0.000 000 008 716 288;
  • 17) 0.000 000 008 716 288 × 2 = 0 + 0.000 000 017 432 576;
  • 18) 0.000 000 017 432 576 × 2 = 0 + 0.000 000 034 865 152;
  • 19) 0.000 000 034 865 152 × 2 = 0 + 0.000 000 069 730 304;
  • 20) 0.000 000 069 730 304 × 2 = 0 + 0.000 000 139 460 608;
  • 21) 0.000 000 139 460 608 × 2 = 0 + 0.000 000 278 921 216;
  • 22) 0.000 000 278 921 216 × 2 = 0 + 0.000 000 557 842 432;
  • 23) 0.000 000 557 842 432 × 2 = 0 + 0.000 001 115 684 864;
  • 24) 0.000 001 115 684 864 × 2 = 0 + 0.000 002 231 369 728;
  • 25) 0.000 002 231 369 728 × 2 = 0 + 0.000 004 462 739 456;
  • 26) 0.000 004 462 739 456 × 2 = 0 + 0.000 008 925 478 912;
  • 27) 0.000 008 925 478 912 × 2 = 0 + 0.000 017 850 957 824;
  • 28) 0.000 017 850 957 824 × 2 = 0 + 0.000 035 701 915 648;
  • 29) 0.000 035 701 915 648 × 2 = 0 + 0.000 071 403 831 296;
  • 30) 0.000 071 403 831 296 × 2 = 0 + 0.000 142 807 662 592;
  • 31) 0.000 142 807 662 592 × 2 = 0 + 0.000 285 615 325 184;
  • 32) 0.000 285 615 325 184 × 2 = 0 + 0.000 571 230 650 368;
  • 33) 0.000 571 230 650 368 × 2 = 0 + 0.001 142 461 300 736;
  • 34) 0.001 142 461 300 736 × 2 = 0 + 0.002 284 922 601 472;
  • 35) 0.002 284 922 601 472 × 2 = 0 + 0.004 569 845 202 944;
  • 36) 0.004 569 845 202 944 × 2 = 0 + 0.009 139 690 405 888;
  • 37) 0.009 139 690 405 888 × 2 = 0 + 0.018 279 380 811 776;
  • 38) 0.018 279 380 811 776 × 2 = 0 + 0.036 558 761 623 552;
  • 39) 0.036 558 761 623 552 × 2 = 0 + 0.073 117 523 247 104;
  • 40) 0.073 117 523 247 104 × 2 = 0 + 0.146 235 046 494 208;
  • 41) 0.146 235 046 494 208 × 2 = 0 + 0.292 470 092 988 416;
  • 42) 0.292 470 092 988 416 × 2 = 0 + 0.584 940 185 976 832;
  • 43) 0.584 940 185 976 832 × 2 = 1 + 0.169 880 371 953 664;
  • 44) 0.169 880 371 953 664 × 2 = 0 + 0.339 760 743 907 328;
  • 45) 0.339 760 743 907 328 × 2 = 0 + 0.679 521 487 814 656;
  • 46) 0.679 521 487 814 656 × 2 = 1 + 0.359 042 975 629 312;
  • 47) 0.359 042 975 629 312 × 2 = 0 + 0.718 085 951 258 624;
  • 48) 0.718 085 951 258 624 × 2 = 1 + 0.436 171 902 517 248;
  • 49) 0.436 171 902 517 248 × 2 = 0 + 0.872 343 805 034 496;
  • 50) 0.872 343 805 034 496 × 2 = 1 + 0.744 687 610 068 992;
  • 51) 0.744 687 610 068 992 × 2 = 1 + 0.489 375 220 137 984;
  • 52) 0.489 375 220 137 984 × 2 = 0 + 0.978 750 440 275 968;
  • 53) 0.978 750 440 275 968 × 2 = 1 + 0.957 500 880 551 936;
  • 54) 0.957 500 880 551 936 × 2 = 1 + 0.915 001 761 103 872;
  • 55) 0.915 001 761 103 872 × 2 = 1 + 0.830 003 522 207 744;
  • 56) 0.830 003 522 207 744 × 2 = 1 + 0.660 007 044 415 488;
  • 57) 0.660 007 044 415 488 × 2 = 1 + 0.320 014 088 830 976;
  • 58) 0.320 014 088 830 976 × 2 = 0 + 0.640 028 177 661 952;
  • 59) 0.640 028 177 661 952 × 2 = 1 + 0.280 056 355 323 904;
  • 60) 0.280 056 355 323 904 × 2 = 0 + 0.560 112 710 647 808;
  • 61) 0.560 112 710 647 808 × 2 = 1 + 0.120 225 421 295 616;
  • 62) 0.120 225 421 295 616 × 2 = 0 + 0.240 450 842 591 232;
  • 63) 0.240 450 842 591 232 × 2 = 0 + 0.480 901 685 182 464;
  • 64) 0.480 901 685 182 464 × 2 = 0 + 0.961 803 370 364 928;
  • 65) 0.961 803 370 364 928 × 2 = 1 + 0.923 606 740 729 856;
  • 66) 0.923 606 740 729 856 × 2 = 1 + 0.847 213 481 459 712;
  • 67) 0.847 213 481 459 712 × 2 = 1 + 0.694 426 962 919 424;
  • 68) 0.694 426 962 919 424 × 2 = 1 + 0.388 853 925 838 848;
  • 69) 0.388 853 925 838 848 × 2 = 0 + 0.777 707 851 677 696;
  • 70) 0.777 707 851 677 696 × 2 = 1 + 0.555 415 703 355 392;
  • 71) 0.555 415 703 355 392 × 2 = 1 + 0.110 831 406 710 784;
  • 72) 0.110 831 406 710 784 × 2 = 0 + 0.221 662 813 421 568;
  • 73) 0.221 662 813 421 568 × 2 = 0 + 0.443 325 626 843 136;
  • 74) 0.443 325 626 843 136 × 2 = 0 + 0.886 651 253 686 272;
  • 75) 0.886 651 253 686 272 × 2 = 1 + 0.773 302 507 372 544;
  • 76) 0.773 302 507 372 544 × 2 = 1 + 0.546 605 014 745 088;
  • 77) 0.546 605 014 745 088 × 2 = 1 + 0.093 210 029 490 176;
  • 78) 0.093 210 029 490 176 × 2 = 0 + 0.186 420 058 980 352;
  • 79) 0.186 420 058 980 352 × 2 = 0 + 0.372 840 117 960 704;
  • 80) 0.372 840 117 960 704 × 2 = 0 + 0.745 680 235 921 408;
  • 81) 0.745 680 235 921 408 × 2 = 1 + 0.491 360 471 842 816;
  • 82) 0.491 360 471 842 816 × 2 = 0 + 0.982 720 943 685 632;
  • 83) 0.982 720 943 685 632 × 2 = 1 + 0.965 441 887 371 264;
  • 84) 0.965 441 887 371 264 × 2 = 1 + 0.930 883 774 742 528;
  • 85) 0.930 883 774 742 528 × 2 = 1 + 0.861 767 549 485 056;
  • 86) 0.861 767 549 485 056 × 2 = 1 + 0.723 535 098 970 112;
  • 87) 0.723 535 098 970 112 × 2 = 1 + 0.447 070 197 940 224;
  • 88) 0.447 070 197 940 224 × 2 = 0 + 0.894 140 395 880 448;
  • 89) 0.894 140 395 880 448 × 2 = 1 + 0.788 280 791 760 896;
  • 90) 0.788 280 791 760 896 × 2 = 1 + 0.576 561 583 521 792;
  • 91) 0.576 561 583 521 792 × 2 = 1 + 0.153 123 167 043 584;
  • 92) 0.153 123 167 043 584 × 2 = 0 + 0.306 246 334 087 168;
  • 93) 0.306 246 334 087 168 × 2 = 0 + 0.612 492 668 174 336;
  • 94) 0.612 492 668 174 336 × 2 = 1 + 0.224 985 336 348 672;
  • 95) 0.224 985 336 348 672 × 2 = 0 + 0.449 970 672 697 344;

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:


0.000 000 000 000 133(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0101 0110 1111 1010 1000 1111 0110 0011 1000 1011 1110 1110 010(2)

5. Positive number before normalization:

0.000 000 000 000 133(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0101 0110 1111 1010 1000 1111 0110 0011 1000 1011 1110 1110 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 133(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0101 0110 1111 1010 1000 1111 0110 0011 1000 1011 1110 1110 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0101 0110 1111 1010 1000 1111 0110 0011 1000 1011 1110 1110 010(2) × 20 =

1.0010 1011 0111 1101 0100 0111 1011 0001 1100 0101 1111 0111 0010(2) × 2-43


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): -43

Mantissa (not normalized):
1.0010 1011 0111 1101 0100 0111 1011 0001 1100 0101 1111 0111 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 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) =

980(10) =

011 1101 0100(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 1011 0111 1101 0100 0111 1011 0001 1100 0101 1111 0111 0010 =

0010 1011 0111 1101 0100 0111 1011 0001 1100 0101 1111 0111 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 1101 0100

Mantissa (52 bits) =
0010 1011 0111 1101 0100 0111 1011 0001 1100 0101 1111 0111 0010


Decimal number 0.000 000 000 000 133 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0100 - 0010 1011 0111 1101 0100 0111 1011 0001 1100 0101 1111 0111 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