0.000 000 000 000 019 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 019(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
0.000 000 000 000 019(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: 0.000 000 000 000 019.

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) 0.000 000 000 000 019 × 2 = 0 + 0.000 000 000 000 038;
  • 2) 0.000 000 000 000 038 × 2 = 0 + 0.000 000 000 000 076;
  • 3) 0.000 000 000 000 076 × 2 = 0 + 0.000 000 000 000 152;
  • 4) 0.000 000 000 000 152 × 2 = 0 + 0.000 000 000 000 304;
  • 5) 0.000 000 000 000 304 × 2 = 0 + 0.000 000 000 000 608;
  • 6) 0.000 000 000 000 608 × 2 = 0 + 0.000 000 000 001 216;
  • 7) 0.000 000 000 001 216 × 2 = 0 + 0.000 000 000 002 432;
  • 8) 0.000 000 000 002 432 × 2 = 0 + 0.000 000 000 004 864;
  • 9) 0.000 000 000 004 864 × 2 = 0 + 0.000 000 000 009 728;
  • 10) 0.000 000 000 009 728 × 2 = 0 + 0.000 000 000 019 456;
  • 11) 0.000 000 000 019 456 × 2 = 0 + 0.000 000 000 038 912;
  • 12) 0.000 000 000 038 912 × 2 = 0 + 0.000 000 000 077 824;
  • 13) 0.000 000 000 077 824 × 2 = 0 + 0.000 000 000 155 648;
  • 14) 0.000 000 000 155 648 × 2 = 0 + 0.000 000 000 311 296;
  • 15) 0.000 000 000 311 296 × 2 = 0 + 0.000 000 000 622 592;
  • 16) 0.000 000 000 622 592 × 2 = 0 + 0.000 000 001 245 184;
  • 17) 0.000 000 001 245 184 × 2 = 0 + 0.000 000 002 490 368;
  • 18) 0.000 000 002 490 368 × 2 = 0 + 0.000 000 004 980 736;
  • 19) 0.000 000 004 980 736 × 2 = 0 + 0.000 000 009 961 472;
  • 20) 0.000 000 009 961 472 × 2 = 0 + 0.000 000 019 922 944;
  • 21) 0.000 000 019 922 944 × 2 = 0 + 0.000 000 039 845 888;
  • 22) 0.000 000 039 845 888 × 2 = 0 + 0.000 000 079 691 776;
  • 23) 0.000 000 079 691 776 × 2 = 0 + 0.000 000 159 383 552;
  • 24) 0.000 000 159 383 552 × 2 = 0 + 0.000 000 318 767 104;
  • 25) 0.000 000 318 767 104 × 2 = 0 + 0.000 000 637 534 208;
  • 26) 0.000 000 637 534 208 × 2 = 0 + 0.000 001 275 068 416;
  • 27) 0.000 001 275 068 416 × 2 = 0 + 0.000 002 550 136 832;
  • 28) 0.000 002 550 136 832 × 2 = 0 + 0.000 005 100 273 664;
  • 29) 0.000 005 100 273 664 × 2 = 0 + 0.000 010 200 547 328;
  • 30) 0.000 010 200 547 328 × 2 = 0 + 0.000 020 401 094 656;
  • 31) 0.000 020 401 094 656 × 2 = 0 + 0.000 040 802 189 312;
  • 32) 0.000 040 802 189 312 × 2 = 0 + 0.000 081 604 378 624;
  • 33) 0.000 081 604 378 624 × 2 = 0 + 0.000 163 208 757 248;
  • 34) 0.000 163 208 757 248 × 2 = 0 + 0.000 326 417 514 496;
  • 35) 0.000 326 417 514 496 × 2 = 0 + 0.000 652 835 028 992;
  • 36) 0.000 652 835 028 992 × 2 = 0 + 0.001 305 670 057 984;
  • 37) 0.001 305 670 057 984 × 2 = 0 + 0.002 611 340 115 968;
  • 38) 0.002 611 340 115 968 × 2 = 0 + 0.005 222 680 231 936;
  • 39) 0.005 222 680 231 936 × 2 = 0 + 0.010 445 360 463 872;
  • 40) 0.010 445 360 463 872 × 2 = 0 + 0.020 890 720 927 744;
  • 41) 0.020 890 720 927 744 × 2 = 0 + 0.041 781 441 855 488;
  • 42) 0.041 781 441 855 488 × 2 = 0 + 0.083 562 883 710 976;
  • 43) 0.083 562 883 710 976 × 2 = 0 + 0.167 125 767 421 952;
  • 44) 0.167 125 767 421 952 × 2 = 0 + 0.334 251 534 843 904;
  • 45) 0.334 251 534 843 904 × 2 = 0 + 0.668 503 069 687 808;
  • 46) 0.668 503 069 687 808 × 2 = 1 + 0.337 006 139 375 616;
  • 47) 0.337 006 139 375 616 × 2 = 0 + 0.674 012 278 751 232;
  • 48) 0.674 012 278 751 232 × 2 = 1 + 0.348 024 557 502 464;
  • 49) 0.348 024 557 502 464 × 2 = 0 + 0.696 049 115 004 928;
  • 50) 0.696 049 115 004 928 × 2 = 1 + 0.392 098 230 009 856;
  • 51) 0.392 098 230 009 856 × 2 = 0 + 0.784 196 460 019 712;
  • 52) 0.784 196 460 019 712 × 2 = 1 + 0.568 392 920 039 424;
  • 53) 0.568 392 920 039 424 × 2 = 1 + 0.136 785 840 078 848;
  • 54) 0.136 785 840 078 848 × 2 = 0 + 0.273 571 680 157 696;
  • 55) 0.273 571 680 157 696 × 2 = 0 + 0.547 143 360 315 392;
  • 56) 0.547 143 360 315 392 × 2 = 1 + 0.094 286 720 630 784;
  • 57) 0.094 286 720 630 784 × 2 = 0 + 0.188 573 441 261 568;
  • 58) 0.188 573 441 261 568 × 2 = 0 + 0.377 146 882 523 136;
  • 59) 0.377 146 882 523 136 × 2 = 0 + 0.754 293 765 046 272;
  • 60) 0.754 293 765 046 272 × 2 = 1 + 0.508 587 530 092 544;
  • 61) 0.508 587 530 092 544 × 2 = 1 + 0.017 175 060 185 088;
  • 62) 0.017 175 060 185 088 × 2 = 0 + 0.034 350 120 370 176;
  • 63) 0.034 350 120 370 176 × 2 = 0 + 0.068 700 240 740 352;
  • 64) 0.068 700 240 740 352 × 2 = 0 + 0.137 400 481 480 704;
  • 65) 0.137 400 481 480 704 × 2 = 0 + 0.274 800 962 961 408;
  • 66) 0.274 800 962 961 408 × 2 = 0 + 0.549 601 925 922 816;
  • 67) 0.549 601 925 922 816 × 2 = 1 + 0.099 203 851 845 632;
  • 68) 0.099 203 851 845 632 × 2 = 0 + 0.198 407 703 691 264;
  • 69) 0.198 407 703 691 264 × 2 = 0 + 0.396 815 407 382 528;
  • 70) 0.396 815 407 382 528 × 2 = 0 + 0.793 630 814 765 056;
  • 71) 0.793 630 814 765 056 × 2 = 1 + 0.587 261 629 530 112;
  • 72) 0.587 261 629 530 112 × 2 = 1 + 0.174 523 259 060 224;
  • 73) 0.174 523 259 060 224 × 2 = 0 + 0.349 046 518 120 448;
  • 74) 0.349 046 518 120 448 × 2 = 0 + 0.698 093 036 240 896;
  • 75) 0.698 093 036 240 896 × 2 = 1 + 0.396 186 072 481 792;
  • 76) 0.396 186 072 481 792 × 2 = 0 + 0.792 372 144 963 584;
  • 77) 0.792 372 144 963 584 × 2 = 1 + 0.584 744 289 927 168;
  • 78) 0.584 744 289 927 168 × 2 = 1 + 0.169 488 579 854 336;
  • 79) 0.169 488 579 854 336 × 2 = 0 + 0.338 977 159 708 672;
  • 80) 0.338 977 159 708 672 × 2 = 0 + 0.677 954 319 417 344;
  • 81) 0.677 954 319 417 344 × 2 = 1 + 0.355 908 638 834 688;
  • 82) 0.355 908 638 834 688 × 2 = 0 + 0.711 817 277 669 376;
  • 83) 0.711 817 277 669 376 × 2 = 1 + 0.423 634 555 338 752;
  • 84) 0.423 634 555 338 752 × 2 = 0 + 0.847 269 110 677 504;
  • 85) 0.847 269 110 677 504 × 2 = 1 + 0.694 538 221 355 008;
  • 86) 0.694 538 221 355 008 × 2 = 1 + 0.389 076 442 710 016;
  • 87) 0.389 076 442 710 016 × 2 = 0 + 0.778 152 885 420 032;
  • 88) 0.778 152 885 420 032 × 2 = 1 + 0.556 305 770 840 064;
  • 89) 0.556 305 770 840 064 × 2 = 1 + 0.112 611 541 680 128;
  • 90) 0.112 611 541 680 128 × 2 = 0 + 0.225 223 083 360 256;
  • 91) 0.225 223 083 360 256 × 2 = 0 + 0.450 446 166 720 512;
  • 92) 0.450 446 166 720 512 × 2 = 0 + 0.900 892 333 441 024;
  • 93) 0.900 892 333 441 024 × 2 = 1 + 0.801 784 666 882 048;
  • 94) 0.801 784 666 882 048 × 2 = 1 + 0.603 569 333 764 096;
  • 95) 0.603 569 333 764 096 × 2 = 1 + 0.207 138 667 528 192;
  • 96) 0.207 138 667 528 192 × 2 = 0 + 0.414 277 335 056 384;
  • 97) 0.414 277 335 056 384 × 2 = 0 + 0.828 554 670 112 768;
  • 98) 0.828 554 670 112 768 × 2 = 1 + 0.657 109 340 225 536;

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:


0.000 000 000 000 019(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0101 1001 0001 1000 0010 0011 0010 1100 1010 1101 1000 1110 01(2)

5. Positive number before normalization:

0.000 000 000 000 019(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0101 1001 0001 1000 0010 0011 0010 1100 1010 1101 1000 1110 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:


0.000 000 000 000 019(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0101 1001 0001 1000 0010 0011 0010 1100 1010 1101 1000 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0101 1001 0001 1000 0010 0011 0010 1100 1010 1101 1000 1110 01(2) × 20 =

1.0101 0110 0100 0110 0000 1000 1100 1011 0010 1011 0110 0011 1001(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):
1.0101 0110 0100 0110 0000 1000 1100 1011 0010 1011 0110 0011 1001


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. 0101 0110 0100 0110 0000 1000 1100 1011 0010 1011 0110 0011 1001 =

0101 0110 0100 0110 0000 1000 1100 1011 0010 1011 0110 0011 1001


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) =
0101 0110 0100 0110 0000 1000 1100 1011 0010 1011 0110 0011 1001


Decimal number 0.000 000 000 000 019 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0001 - 0101 0110 0100 0110 0000 1000 1100 1011 0010 1011 0110 0011 1001

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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