0.000 000 000 000 189 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 189(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
0.000 000 000 000 189(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: 0.000 000 000 000 189.

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) 0.000 000 000 000 189 × 2 = 0 + 0.000 000 000 000 378;
  • 2) 0.000 000 000 000 378 × 2 = 0 + 0.000 000 000 000 756;
  • 3) 0.000 000 000 000 756 × 2 = 0 + 0.000 000 000 001 512;
  • 4) 0.000 000 000 001 512 × 2 = 0 + 0.000 000 000 003 024;
  • 5) 0.000 000 000 003 024 × 2 = 0 + 0.000 000 000 006 048;
  • 6) 0.000 000 000 006 048 × 2 = 0 + 0.000 000 000 012 096;
  • 7) 0.000 000 000 012 096 × 2 = 0 + 0.000 000 000 024 192;
  • 8) 0.000 000 000 024 192 × 2 = 0 + 0.000 000 000 048 384;
  • 9) 0.000 000 000 048 384 × 2 = 0 + 0.000 000 000 096 768;
  • 10) 0.000 000 000 096 768 × 2 = 0 + 0.000 000 000 193 536;
  • 11) 0.000 000 000 193 536 × 2 = 0 + 0.000 000 000 387 072;
  • 12) 0.000 000 000 387 072 × 2 = 0 + 0.000 000 000 774 144;
  • 13) 0.000 000 000 774 144 × 2 = 0 + 0.000 000 001 548 288;
  • 14) 0.000 000 001 548 288 × 2 = 0 + 0.000 000 003 096 576;
  • 15) 0.000 000 003 096 576 × 2 = 0 + 0.000 000 006 193 152;
  • 16) 0.000 000 006 193 152 × 2 = 0 + 0.000 000 012 386 304;
  • 17) 0.000 000 012 386 304 × 2 = 0 + 0.000 000 024 772 608;
  • 18) 0.000 000 024 772 608 × 2 = 0 + 0.000 000 049 545 216;
  • 19) 0.000 000 049 545 216 × 2 = 0 + 0.000 000 099 090 432;
  • 20) 0.000 000 099 090 432 × 2 = 0 + 0.000 000 198 180 864;
  • 21) 0.000 000 198 180 864 × 2 = 0 + 0.000 000 396 361 728;
  • 22) 0.000 000 396 361 728 × 2 = 0 + 0.000 000 792 723 456;
  • 23) 0.000 000 792 723 456 × 2 = 0 + 0.000 001 585 446 912;
  • 24) 0.000 001 585 446 912 × 2 = 0 + 0.000 003 170 893 824;
  • 25) 0.000 003 170 893 824 × 2 = 0 + 0.000 006 341 787 648;
  • 26) 0.000 006 341 787 648 × 2 = 0 + 0.000 012 683 575 296;
  • 27) 0.000 012 683 575 296 × 2 = 0 + 0.000 025 367 150 592;
  • 28) 0.000 025 367 150 592 × 2 = 0 + 0.000 050 734 301 184;
  • 29) 0.000 050 734 301 184 × 2 = 0 + 0.000 101 468 602 368;
  • 30) 0.000 101 468 602 368 × 2 = 0 + 0.000 202 937 204 736;
  • 31) 0.000 202 937 204 736 × 2 = 0 + 0.000 405 874 409 472;
  • 32) 0.000 405 874 409 472 × 2 = 0 + 0.000 811 748 818 944;
  • 33) 0.000 811 748 818 944 × 2 = 0 + 0.001 623 497 637 888;
  • 34) 0.001 623 497 637 888 × 2 = 0 + 0.003 246 995 275 776;
  • 35) 0.003 246 995 275 776 × 2 = 0 + 0.006 493 990 551 552;
  • 36) 0.006 493 990 551 552 × 2 = 0 + 0.012 987 981 103 104;
  • 37) 0.012 987 981 103 104 × 2 = 0 + 0.025 975 962 206 208;
  • 38) 0.025 975 962 206 208 × 2 = 0 + 0.051 951 924 412 416;
  • 39) 0.051 951 924 412 416 × 2 = 0 + 0.103 903 848 824 832;
  • 40) 0.103 903 848 824 832 × 2 = 0 + 0.207 807 697 649 664;
  • 41) 0.207 807 697 649 664 × 2 = 0 + 0.415 615 395 299 328;
  • 42) 0.415 615 395 299 328 × 2 = 0 + 0.831 230 790 598 656;
  • 43) 0.831 230 790 598 656 × 2 = 1 + 0.662 461 581 197 312;
  • 44) 0.662 461 581 197 312 × 2 = 1 + 0.324 923 162 394 624;
  • 45) 0.324 923 162 394 624 × 2 = 0 + 0.649 846 324 789 248;
  • 46) 0.649 846 324 789 248 × 2 = 1 + 0.299 692 649 578 496;
  • 47) 0.299 692 649 578 496 × 2 = 0 + 0.599 385 299 156 992;
  • 48) 0.599 385 299 156 992 × 2 = 1 + 0.198 770 598 313 984;
  • 49) 0.198 770 598 313 984 × 2 = 0 + 0.397 541 196 627 968;
  • 50) 0.397 541 196 627 968 × 2 = 0 + 0.795 082 393 255 936;
  • 51) 0.795 082 393 255 936 × 2 = 1 + 0.590 164 786 511 872;
  • 52) 0.590 164 786 511 872 × 2 = 1 + 0.180 329 573 023 744;
  • 53) 0.180 329 573 023 744 × 2 = 0 + 0.360 659 146 047 488;
  • 54) 0.360 659 146 047 488 × 2 = 0 + 0.721 318 292 094 976;
  • 55) 0.721 318 292 094 976 × 2 = 1 + 0.442 636 584 189 952;
  • 56) 0.442 636 584 189 952 × 2 = 0 + 0.885 273 168 379 904;
  • 57) 0.885 273 168 379 904 × 2 = 1 + 0.770 546 336 759 808;
  • 58) 0.770 546 336 759 808 × 2 = 1 + 0.541 092 673 519 616;
  • 59) 0.541 092 673 519 616 × 2 = 1 + 0.082 185 347 039 232;
  • 60) 0.082 185 347 039 232 × 2 = 0 + 0.164 370 694 078 464;
  • 61) 0.164 370 694 078 464 × 2 = 0 + 0.328 741 388 156 928;
  • 62) 0.328 741 388 156 928 × 2 = 0 + 0.657 482 776 313 856;
  • 63) 0.657 482 776 313 856 × 2 = 1 + 0.314 965 552 627 712;
  • 64) 0.314 965 552 627 712 × 2 = 0 + 0.629 931 105 255 424;
  • 65) 0.629 931 105 255 424 × 2 = 1 + 0.259 862 210 510 848;
  • 66) 0.259 862 210 510 848 × 2 = 0 + 0.519 724 421 021 696;
  • 67) 0.519 724 421 021 696 × 2 = 1 + 0.039 448 842 043 392;
  • 68) 0.039 448 842 043 392 × 2 = 0 + 0.078 897 684 086 784;
  • 69) 0.078 897 684 086 784 × 2 = 0 + 0.157 795 368 173 568;
  • 70) 0.157 795 368 173 568 × 2 = 0 + 0.315 590 736 347 136;
  • 71) 0.315 590 736 347 136 × 2 = 0 + 0.631 181 472 694 272;
  • 72) 0.631 181 472 694 272 × 2 = 1 + 0.262 362 945 388 544;
  • 73) 0.262 362 945 388 544 × 2 = 0 + 0.524 725 890 777 088;
  • 74) 0.524 725 890 777 088 × 2 = 1 + 0.049 451 781 554 176;
  • 75) 0.049 451 781 554 176 × 2 = 0 + 0.098 903 563 108 352;
  • 76) 0.098 903 563 108 352 × 2 = 0 + 0.197 807 126 216 704;
  • 77) 0.197 807 126 216 704 × 2 = 0 + 0.395 614 252 433 408;
  • 78) 0.395 614 252 433 408 × 2 = 0 + 0.791 228 504 866 816;
  • 79) 0.791 228 504 866 816 × 2 = 1 + 0.582 457 009 733 632;
  • 80) 0.582 457 009 733 632 × 2 = 1 + 0.164 914 019 467 264;
  • 81) 0.164 914 019 467 264 × 2 = 0 + 0.329 828 038 934 528;
  • 82) 0.329 828 038 934 528 × 2 = 0 + 0.659 656 077 869 056;
  • 83) 0.659 656 077 869 056 × 2 = 1 + 0.319 312 155 738 112;
  • 84) 0.319 312 155 738 112 × 2 = 0 + 0.638 624 311 476 224;
  • 85) 0.638 624 311 476 224 × 2 = 1 + 0.277 248 622 952 448;
  • 86) 0.277 248 622 952 448 × 2 = 0 + 0.554 497 245 904 896;
  • 87) 0.554 497 245 904 896 × 2 = 1 + 0.108 994 491 809 792;
  • 88) 0.108 994 491 809 792 × 2 = 0 + 0.217 988 983 619 584;
  • 89) 0.217 988 983 619 584 × 2 = 0 + 0.435 977 967 239 168;
  • 90) 0.435 977 967 239 168 × 2 = 0 + 0.871 955 934 478 336;
  • 91) 0.871 955 934 478 336 × 2 = 1 + 0.743 911 868 956 672;
  • 92) 0.743 911 868 956 672 × 2 = 1 + 0.487 823 737 913 344;
  • 93) 0.487 823 737 913 344 × 2 = 0 + 0.975 647 475 826 688;
  • 94) 0.975 647 475 826 688 × 2 = 1 + 0.951 294 951 653 376;
  • 95) 0.951 294 951 653 376 × 2 = 1 + 0.902 589 903 306 752;

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:


0.000 000 000 000 189(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0010 1110 0010 1010 0001 0100 0011 0010 1010 0011 011(2)

5. Positive number before normalization:

0.000 000 000 000 189(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0010 1110 0010 1010 0001 0100 0011 0010 1010 0011 011(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:


0.000 000 000 000 189(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0010 1110 0010 1010 0001 0100 0011 0010 1010 0011 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0010 1110 0010 1010 0001 0100 0011 0010 1010 0011 011(2) × 20 =

1.1010 1001 1001 0111 0001 0101 0000 1010 0001 1001 0101 0001 1011(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):
1.1010 1001 1001 0111 0001 0101 0000 1010 0001 1001 0101 0001 1011


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. 1010 1001 1001 0111 0001 0101 0000 1010 0001 1001 0101 0001 1011 =

1010 1001 1001 0111 0001 0101 0000 1010 0001 1001 0101 0001 1011


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) =
1010 1001 1001 0111 0001 0101 0000 1010 0001 1001 0101 0001 1011


Decimal number 0.000 000 000 000 189 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0100 - 1010 1001 1001 0111 0001 0101 0000 1010 0001 1001 0101 0001 1011

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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