0.000 000 000 000 199 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 199(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
0.000 000 000 000 199(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: 0.000 000 000 000 199.

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) 0.000 000 000 000 199 × 2 = 0 + 0.000 000 000 000 398;
  • 2) 0.000 000 000 000 398 × 2 = 0 + 0.000 000 000 000 796;
  • 3) 0.000 000 000 000 796 × 2 = 0 + 0.000 000 000 001 592;
  • 4) 0.000 000 000 001 592 × 2 = 0 + 0.000 000 000 003 184;
  • 5) 0.000 000 000 003 184 × 2 = 0 + 0.000 000 000 006 368;
  • 6) 0.000 000 000 006 368 × 2 = 0 + 0.000 000 000 012 736;
  • 7) 0.000 000 000 012 736 × 2 = 0 + 0.000 000 000 025 472;
  • 8) 0.000 000 000 025 472 × 2 = 0 + 0.000 000 000 050 944;
  • 9) 0.000 000 000 050 944 × 2 = 0 + 0.000 000 000 101 888;
  • 10) 0.000 000 000 101 888 × 2 = 0 + 0.000 000 000 203 776;
  • 11) 0.000 000 000 203 776 × 2 = 0 + 0.000 000 000 407 552;
  • 12) 0.000 000 000 407 552 × 2 = 0 + 0.000 000 000 815 104;
  • 13) 0.000 000 000 815 104 × 2 = 0 + 0.000 000 001 630 208;
  • 14) 0.000 000 001 630 208 × 2 = 0 + 0.000 000 003 260 416;
  • 15) 0.000 000 003 260 416 × 2 = 0 + 0.000 000 006 520 832;
  • 16) 0.000 000 006 520 832 × 2 = 0 + 0.000 000 013 041 664;
  • 17) 0.000 000 013 041 664 × 2 = 0 + 0.000 000 026 083 328;
  • 18) 0.000 000 026 083 328 × 2 = 0 + 0.000 000 052 166 656;
  • 19) 0.000 000 052 166 656 × 2 = 0 + 0.000 000 104 333 312;
  • 20) 0.000 000 104 333 312 × 2 = 0 + 0.000 000 208 666 624;
  • 21) 0.000 000 208 666 624 × 2 = 0 + 0.000 000 417 333 248;
  • 22) 0.000 000 417 333 248 × 2 = 0 + 0.000 000 834 666 496;
  • 23) 0.000 000 834 666 496 × 2 = 0 + 0.000 001 669 332 992;
  • 24) 0.000 001 669 332 992 × 2 = 0 + 0.000 003 338 665 984;
  • 25) 0.000 003 338 665 984 × 2 = 0 + 0.000 006 677 331 968;
  • 26) 0.000 006 677 331 968 × 2 = 0 + 0.000 013 354 663 936;
  • 27) 0.000 013 354 663 936 × 2 = 0 + 0.000 026 709 327 872;
  • 28) 0.000 026 709 327 872 × 2 = 0 + 0.000 053 418 655 744;
  • 29) 0.000 053 418 655 744 × 2 = 0 + 0.000 106 837 311 488;
  • 30) 0.000 106 837 311 488 × 2 = 0 + 0.000 213 674 622 976;
  • 31) 0.000 213 674 622 976 × 2 = 0 + 0.000 427 349 245 952;
  • 32) 0.000 427 349 245 952 × 2 = 0 + 0.000 854 698 491 904;
  • 33) 0.000 854 698 491 904 × 2 = 0 + 0.001 709 396 983 808;
  • 34) 0.001 709 396 983 808 × 2 = 0 + 0.003 418 793 967 616;
  • 35) 0.003 418 793 967 616 × 2 = 0 + 0.006 837 587 935 232;
  • 36) 0.006 837 587 935 232 × 2 = 0 + 0.013 675 175 870 464;
  • 37) 0.013 675 175 870 464 × 2 = 0 + 0.027 350 351 740 928;
  • 38) 0.027 350 351 740 928 × 2 = 0 + 0.054 700 703 481 856;
  • 39) 0.054 700 703 481 856 × 2 = 0 + 0.109 401 406 963 712;
  • 40) 0.109 401 406 963 712 × 2 = 0 + 0.218 802 813 927 424;
  • 41) 0.218 802 813 927 424 × 2 = 0 + 0.437 605 627 854 848;
  • 42) 0.437 605 627 854 848 × 2 = 0 + 0.875 211 255 709 696;
  • 43) 0.875 211 255 709 696 × 2 = 1 + 0.750 422 511 419 392;
  • 44) 0.750 422 511 419 392 × 2 = 1 + 0.500 845 022 838 784;
  • 45) 0.500 845 022 838 784 × 2 = 1 + 0.001 690 045 677 568;
  • 46) 0.001 690 045 677 568 × 2 = 0 + 0.003 380 091 355 136;
  • 47) 0.003 380 091 355 136 × 2 = 0 + 0.006 760 182 710 272;
  • 48) 0.006 760 182 710 272 × 2 = 0 + 0.013 520 365 420 544;
  • 49) 0.013 520 365 420 544 × 2 = 0 + 0.027 040 730 841 088;
  • 50) 0.027 040 730 841 088 × 2 = 0 + 0.054 081 461 682 176;
  • 51) 0.054 081 461 682 176 × 2 = 0 + 0.108 162 923 364 352;
  • 52) 0.108 162 923 364 352 × 2 = 0 + 0.216 325 846 728 704;
  • 53) 0.216 325 846 728 704 × 2 = 0 + 0.432 651 693 457 408;
  • 54) 0.432 651 693 457 408 × 2 = 0 + 0.865 303 386 914 816;
  • 55) 0.865 303 386 914 816 × 2 = 1 + 0.730 606 773 829 632;
  • 56) 0.730 606 773 829 632 × 2 = 1 + 0.461 213 547 659 264;
  • 57) 0.461 213 547 659 264 × 2 = 0 + 0.922 427 095 318 528;
  • 58) 0.922 427 095 318 528 × 2 = 1 + 0.844 854 190 637 056;
  • 59) 0.844 854 190 637 056 × 2 = 1 + 0.689 708 381 274 112;
  • 60) 0.689 708 381 274 112 × 2 = 1 + 0.379 416 762 548 224;
  • 61) 0.379 416 762 548 224 × 2 = 0 + 0.758 833 525 096 448;
  • 62) 0.758 833 525 096 448 × 2 = 1 + 0.517 667 050 192 896;
  • 63) 0.517 667 050 192 896 × 2 = 1 + 0.035 334 100 385 792;
  • 64) 0.035 334 100 385 792 × 2 = 0 + 0.070 668 200 771 584;
  • 65) 0.070 668 200 771 584 × 2 = 0 + 0.141 336 401 543 168;
  • 66) 0.141 336 401 543 168 × 2 = 0 + 0.282 672 803 086 336;
  • 67) 0.282 672 803 086 336 × 2 = 0 + 0.565 345 606 172 672;
  • 68) 0.565 345 606 172 672 × 2 = 1 + 0.130 691 212 345 344;
  • 69) 0.130 691 212 345 344 × 2 = 0 + 0.261 382 424 690 688;
  • 70) 0.261 382 424 690 688 × 2 = 0 + 0.522 764 849 381 376;
  • 71) 0.522 764 849 381 376 × 2 = 1 + 0.045 529 698 762 752;
  • 72) 0.045 529 698 762 752 × 2 = 0 + 0.091 059 397 525 504;
  • 73) 0.091 059 397 525 504 × 2 = 0 + 0.182 118 795 051 008;
  • 74) 0.182 118 795 051 008 × 2 = 0 + 0.364 237 590 102 016;
  • 75) 0.364 237 590 102 016 × 2 = 0 + 0.728 475 180 204 032;
  • 76) 0.728 475 180 204 032 × 2 = 1 + 0.456 950 360 408 064;
  • 77) 0.456 950 360 408 064 × 2 = 0 + 0.913 900 720 816 128;
  • 78) 0.913 900 720 816 128 × 2 = 1 + 0.827 801 441 632 256;
  • 79) 0.827 801 441 632 256 × 2 = 1 + 0.655 602 883 264 512;
  • 80) 0.655 602 883 264 512 × 2 = 1 + 0.311 205 766 529 024;
  • 81) 0.311 205 766 529 024 × 2 = 0 + 0.622 411 533 058 048;
  • 82) 0.622 411 533 058 048 × 2 = 1 + 0.244 823 066 116 096;
  • 83) 0.244 823 066 116 096 × 2 = 0 + 0.489 646 132 232 192;
  • 84) 0.489 646 132 232 192 × 2 = 0 + 0.979 292 264 464 384;
  • 85) 0.979 292 264 464 384 × 2 = 1 + 0.958 584 528 928 768;
  • 86) 0.958 584 528 928 768 × 2 = 1 + 0.917 169 057 857 536;
  • 87) 0.917 169 057 857 536 × 2 = 1 + 0.834 338 115 715 072;
  • 88) 0.834 338 115 715 072 × 2 = 1 + 0.668 676 231 430 144;
  • 89) 0.668 676 231 430 144 × 2 = 1 + 0.337 352 462 860 288;
  • 90) 0.337 352 462 860 288 × 2 = 0 + 0.674 704 925 720 576;
  • 91) 0.674 704 925 720 576 × 2 = 1 + 0.349 409 851 441 152;
  • 92) 0.349 409 851 441 152 × 2 = 0 + 0.698 819 702 882 304;
  • 93) 0.698 819 702 882 304 × 2 = 1 + 0.397 639 405 764 608;
  • 94) 0.397 639 405 764 608 × 2 = 0 + 0.795 278 811 529 216;
  • 95) 0.795 278 811 529 216 × 2 = 1 + 0.590 557 623 058 432;

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:


0.000 000 000 000 199(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1000 0000 0011 0111 0110 0001 0010 0001 0111 0100 1111 1010 101(2)

5. Positive number before normalization:

0.000 000 000 000 199(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1000 0000 0011 0111 0110 0001 0010 0001 0111 0100 1111 1010 101(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:


0.000 000 000 000 199(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1000 0000 0011 0111 0110 0001 0010 0001 0111 0100 1111 1010 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1000 0000 0011 0111 0110 0001 0010 0001 0111 0100 1111 1010 101(2) × 20 =

1.1100 0000 0001 1011 1011 0000 1001 0000 1011 1010 0111 1101 0101(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):
1.1100 0000 0001 1011 1011 0000 1001 0000 1011 1010 0111 1101 0101


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. 1100 0000 0001 1011 1011 0000 1001 0000 1011 1010 0111 1101 0101 =

1100 0000 0001 1011 1011 0000 1001 0000 1011 1010 0111 1101 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 0100

Mantissa (52 bits) =
1100 0000 0001 1011 1011 0000 1001 0000 1011 1010 0111 1101 0101


Decimal number 0.000 000 000 000 199 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0100 - 1100 0000 0001 1011 1011 0000 1001 0000 1011 1010 0111 1101 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