0.000 000 000 000 198 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 198(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
0.000 000 000 000 198(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: 0.000 000 000 000 198.

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) 0.000 000 000 000 198 × 2 = 0 + 0.000 000 000 000 396;
  • 2) 0.000 000 000 000 396 × 2 = 0 + 0.000 000 000 000 792;
  • 3) 0.000 000 000 000 792 × 2 = 0 + 0.000 000 000 001 584;
  • 4) 0.000 000 000 001 584 × 2 = 0 + 0.000 000 000 003 168;
  • 5) 0.000 000 000 003 168 × 2 = 0 + 0.000 000 000 006 336;
  • 6) 0.000 000 000 006 336 × 2 = 0 + 0.000 000 000 012 672;
  • 7) 0.000 000 000 012 672 × 2 = 0 + 0.000 000 000 025 344;
  • 8) 0.000 000 000 025 344 × 2 = 0 + 0.000 000 000 050 688;
  • 9) 0.000 000 000 050 688 × 2 = 0 + 0.000 000 000 101 376;
  • 10) 0.000 000 000 101 376 × 2 = 0 + 0.000 000 000 202 752;
  • 11) 0.000 000 000 202 752 × 2 = 0 + 0.000 000 000 405 504;
  • 12) 0.000 000 000 405 504 × 2 = 0 + 0.000 000 000 811 008;
  • 13) 0.000 000 000 811 008 × 2 = 0 + 0.000 000 001 622 016;
  • 14) 0.000 000 001 622 016 × 2 = 0 + 0.000 000 003 244 032;
  • 15) 0.000 000 003 244 032 × 2 = 0 + 0.000 000 006 488 064;
  • 16) 0.000 000 006 488 064 × 2 = 0 + 0.000 000 012 976 128;
  • 17) 0.000 000 012 976 128 × 2 = 0 + 0.000 000 025 952 256;
  • 18) 0.000 000 025 952 256 × 2 = 0 + 0.000 000 051 904 512;
  • 19) 0.000 000 051 904 512 × 2 = 0 + 0.000 000 103 809 024;
  • 20) 0.000 000 103 809 024 × 2 = 0 + 0.000 000 207 618 048;
  • 21) 0.000 000 207 618 048 × 2 = 0 + 0.000 000 415 236 096;
  • 22) 0.000 000 415 236 096 × 2 = 0 + 0.000 000 830 472 192;
  • 23) 0.000 000 830 472 192 × 2 = 0 + 0.000 001 660 944 384;
  • 24) 0.000 001 660 944 384 × 2 = 0 + 0.000 003 321 888 768;
  • 25) 0.000 003 321 888 768 × 2 = 0 + 0.000 006 643 777 536;
  • 26) 0.000 006 643 777 536 × 2 = 0 + 0.000 013 287 555 072;
  • 27) 0.000 013 287 555 072 × 2 = 0 + 0.000 026 575 110 144;
  • 28) 0.000 026 575 110 144 × 2 = 0 + 0.000 053 150 220 288;
  • 29) 0.000 053 150 220 288 × 2 = 0 + 0.000 106 300 440 576;
  • 30) 0.000 106 300 440 576 × 2 = 0 + 0.000 212 600 881 152;
  • 31) 0.000 212 600 881 152 × 2 = 0 + 0.000 425 201 762 304;
  • 32) 0.000 425 201 762 304 × 2 = 0 + 0.000 850 403 524 608;
  • 33) 0.000 850 403 524 608 × 2 = 0 + 0.001 700 807 049 216;
  • 34) 0.001 700 807 049 216 × 2 = 0 + 0.003 401 614 098 432;
  • 35) 0.003 401 614 098 432 × 2 = 0 + 0.006 803 228 196 864;
  • 36) 0.006 803 228 196 864 × 2 = 0 + 0.013 606 456 393 728;
  • 37) 0.013 606 456 393 728 × 2 = 0 + 0.027 212 912 787 456;
  • 38) 0.027 212 912 787 456 × 2 = 0 + 0.054 425 825 574 912;
  • 39) 0.054 425 825 574 912 × 2 = 0 + 0.108 851 651 149 824;
  • 40) 0.108 851 651 149 824 × 2 = 0 + 0.217 703 302 299 648;
  • 41) 0.217 703 302 299 648 × 2 = 0 + 0.435 406 604 599 296;
  • 42) 0.435 406 604 599 296 × 2 = 0 + 0.870 813 209 198 592;
  • 43) 0.870 813 209 198 592 × 2 = 1 + 0.741 626 418 397 184;
  • 44) 0.741 626 418 397 184 × 2 = 1 + 0.483 252 836 794 368;
  • 45) 0.483 252 836 794 368 × 2 = 0 + 0.966 505 673 588 736;
  • 46) 0.966 505 673 588 736 × 2 = 1 + 0.933 011 347 177 472;
  • 47) 0.933 011 347 177 472 × 2 = 1 + 0.866 022 694 354 944;
  • 48) 0.866 022 694 354 944 × 2 = 1 + 0.732 045 388 709 888;
  • 49) 0.732 045 388 709 888 × 2 = 1 + 0.464 090 777 419 776;
  • 50) 0.464 090 777 419 776 × 2 = 0 + 0.928 181 554 839 552;
  • 51) 0.928 181 554 839 552 × 2 = 1 + 0.856 363 109 679 104;
  • 52) 0.856 363 109 679 104 × 2 = 1 + 0.712 726 219 358 208;
  • 53) 0.712 726 219 358 208 × 2 = 1 + 0.425 452 438 716 416;
  • 54) 0.425 452 438 716 416 × 2 = 0 + 0.850 904 877 432 832;
  • 55) 0.850 904 877 432 832 × 2 = 1 + 0.701 809 754 865 664;
  • 56) 0.701 809 754 865 664 × 2 = 1 + 0.403 619 509 731 328;
  • 57) 0.403 619 509 731 328 × 2 = 0 + 0.807 239 019 462 656;
  • 58) 0.807 239 019 462 656 × 2 = 1 + 0.614 478 038 925 312;
  • 59) 0.614 478 038 925 312 × 2 = 1 + 0.228 956 077 850 624;
  • 60) 0.228 956 077 850 624 × 2 = 0 + 0.457 912 155 701 248;
  • 61) 0.457 912 155 701 248 × 2 = 0 + 0.915 824 311 402 496;
  • 62) 0.915 824 311 402 496 × 2 = 1 + 0.831 648 622 804 992;
  • 63) 0.831 648 622 804 992 × 2 = 1 + 0.663 297 245 609 984;
  • 64) 0.663 297 245 609 984 × 2 = 1 + 0.326 594 491 219 968;
  • 65) 0.326 594 491 219 968 × 2 = 0 + 0.653 188 982 439 936;
  • 66) 0.653 188 982 439 936 × 2 = 1 + 0.306 377 964 879 872;
  • 67) 0.306 377 964 879 872 × 2 = 0 + 0.612 755 929 759 744;
  • 68) 0.612 755 929 759 744 × 2 = 1 + 0.225 511 859 519 488;
  • 69) 0.225 511 859 519 488 × 2 = 0 + 0.451 023 719 038 976;
  • 70) 0.451 023 719 038 976 × 2 = 0 + 0.902 047 438 077 952;
  • 71) 0.902 047 438 077 952 × 2 = 1 + 0.804 094 876 155 904;
  • 72) 0.804 094 876 155 904 × 2 = 1 + 0.608 189 752 311 808;
  • 73) 0.608 189 752 311 808 × 2 = 1 + 0.216 379 504 623 616;
  • 74) 0.216 379 504 623 616 × 2 = 0 + 0.432 759 009 247 232;
  • 75) 0.432 759 009 247 232 × 2 = 0 + 0.865 518 018 494 464;
  • 76) 0.865 518 018 494 464 × 2 = 1 + 0.731 036 036 988 928;
  • 77) 0.731 036 036 988 928 × 2 = 1 + 0.462 072 073 977 856;
  • 78) 0.462 072 073 977 856 × 2 = 0 + 0.924 144 147 955 712;
  • 79) 0.924 144 147 955 712 × 2 = 1 + 0.848 288 295 911 424;
  • 80) 0.848 288 295 911 424 × 2 = 1 + 0.696 576 591 822 848;
  • 81) 0.696 576 591 822 848 × 2 = 1 + 0.393 153 183 645 696;
  • 82) 0.393 153 183 645 696 × 2 = 0 + 0.786 306 367 291 392;
  • 83) 0.786 306 367 291 392 × 2 = 1 + 0.572 612 734 582 784;
  • 84) 0.572 612 734 582 784 × 2 = 1 + 0.145 225 469 165 568;
  • 85) 0.145 225 469 165 568 × 2 = 0 + 0.290 450 938 331 136;
  • 86) 0.290 450 938 331 136 × 2 = 0 + 0.580 901 876 662 272;
  • 87) 0.580 901 876 662 272 × 2 = 1 + 0.161 803 753 324 544;
  • 88) 0.161 803 753 324 544 × 2 = 0 + 0.323 607 506 649 088;
  • 89) 0.323 607 506 649 088 × 2 = 0 + 0.647 215 013 298 176;
  • 90) 0.647 215 013 298 176 × 2 = 1 + 0.294 430 026 596 352;
  • 91) 0.294 430 026 596 352 × 2 = 0 + 0.588 860 053 192 704;
  • 92) 0.588 860 053 192 704 × 2 = 1 + 0.177 720 106 385 408;
  • 93) 0.177 720 106 385 408 × 2 = 0 + 0.355 440 212 770 816;
  • 94) 0.355 440 212 770 816 × 2 = 0 + 0.710 880 425 541 632;
  • 95) 0.710 880 425 541 632 × 2 = 1 + 0.421 760 851 083 264;

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:


0.000 000 000 000 198(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 1011 1011 0110 0111 0101 0011 1001 1011 1011 0010 0101 001(2)

5. Positive number before normalization:

0.000 000 000 000 198(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 1011 1011 0110 0111 0101 0011 1001 1011 1011 0010 0101 001(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:


0.000 000 000 000 198(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 1011 1011 0110 0111 0101 0011 1001 1011 1011 0010 0101 001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0111 1011 1011 0110 0111 0101 0011 1001 1011 1011 0010 0101 001(2) × 20 =

1.1011 1101 1101 1011 0011 1010 1001 1100 1101 1101 1001 0010 1001(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):
1.1011 1101 1101 1011 0011 1010 1001 1100 1101 1101 1001 0010 1001


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. 1011 1101 1101 1011 0011 1010 1001 1100 1101 1101 1001 0010 1001 =

1011 1101 1101 1011 0011 1010 1001 1100 1101 1101 1001 0010 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 0100

Mantissa (52 bits) =
1011 1101 1101 1011 0011 1010 1001 1100 1101 1101 1001 0010 1001


Decimal number 0.000 000 000 000 198 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0100 - 1011 1101 1101 1011 0011 1010 1001 1100 1101 1101 1001 0010 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