0.000 001 506 198 729 391 68 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 001 506 198 729 391 68(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
0.000 001 506 198 729 391 68(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: 0.000 001 506 198 729 391 68.

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) 0.000 001 506 198 729 391 68 × 2 = 0 + 0.000 003 012 397 458 783 36;
  • 2) 0.000 003 012 397 458 783 36 × 2 = 0 + 0.000 006 024 794 917 566 72;
  • 3) 0.000 006 024 794 917 566 72 × 2 = 0 + 0.000 012 049 589 835 133 44;
  • 4) 0.000 012 049 589 835 133 44 × 2 = 0 + 0.000 024 099 179 670 266 88;
  • 5) 0.000 024 099 179 670 266 88 × 2 = 0 + 0.000 048 198 359 340 533 76;
  • 6) 0.000 048 198 359 340 533 76 × 2 = 0 + 0.000 096 396 718 681 067 52;
  • 7) 0.000 096 396 718 681 067 52 × 2 = 0 + 0.000 192 793 437 362 135 04;
  • 8) 0.000 192 793 437 362 135 04 × 2 = 0 + 0.000 385 586 874 724 270 08;
  • 9) 0.000 385 586 874 724 270 08 × 2 = 0 + 0.000 771 173 749 448 540 16;
  • 10) 0.000 771 173 749 448 540 16 × 2 = 0 + 0.001 542 347 498 897 080 32;
  • 11) 0.001 542 347 498 897 080 32 × 2 = 0 + 0.003 084 694 997 794 160 64;
  • 12) 0.003 084 694 997 794 160 64 × 2 = 0 + 0.006 169 389 995 588 321 28;
  • 13) 0.006 169 389 995 588 321 28 × 2 = 0 + 0.012 338 779 991 176 642 56;
  • 14) 0.012 338 779 991 176 642 56 × 2 = 0 + 0.024 677 559 982 353 285 12;
  • 15) 0.024 677 559 982 353 285 12 × 2 = 0 + 0.049 355 119 964 706 570 24;
  • 16) 0.049 355 119 964 706 570 24 × 2 = 0 + 0.098 710 239 929 413 140 48;
  • 17) 0.098 710 239 929 413 140 48 × 2 = 0 + 0.197 420 479 858 826 280 96;
  • 18) 0.197 420 479 858 826 280 96 × 2 = 0 + 0.394 840 959 717 652 561 92;
  • 19) 0.394 840 959 717 652 561 92 × 2 = 0 + 0.789 681 919 435 305 123 84;
  • 20) 0.789 681 919 435 305 123 84 × 2 = 1 + 0.579 363 838 870 610 247 68;
  • 21) 0.579 363 838 870 610 247 68 × 2 = 1 + 0.158 727 677 741 220 495 36;
  • 22) 0.158 727 677 741 220 495 36 × 2 = 0 + 0.317 455 355 482 440 990 72;
  • 23) 0.317 455 355 482 440 990 72 × 2 = 0 + 0.634 910 710 964 881 981 44;
  • 24) 0.634 910 710 964 881 981 44 × 2 = 1 + 0.269 821 421 929 763 962 88;
  • 25) 0.269 821 421 929 763 962 88 × 2 = 0 + 0.539 642 843 859 527 925 76;
  • 26) 0.539 642 843 859 527 925 76 × 2 = 1 + 0.079 285 687 719 055 851 52;
  • 27) 0.079 285 687 719 055 851 52 × 2 = 0 + 0.158 571 375 438 111 703 04;
  • 28) 0.158 571 375 438 111 703 04 × 2 = 0 + 0.317 142 750 876 223 406 08;
  • 29) 0.317 142 750 876 223 406 08 × 2 = 0 + 0.634 285 501 752 446 812 16;
  • 30) 0.634 285 501 752 446 812 16 × 2 = 1 + 0.268 571 003 504 893 624 32;
  • 31) 0.268 571 003 504 893 624 32 × 2 = 0 + 0.537 142 007 009 787 248 64;
  • 32) 0.537 142 007 009 787 248 64 × 2 = 1 + 0.074 284 014 019 574 497 28;
  • 33) 0.074 284 014 019 574 497 28 × 2 = 0 + 0.148 568 028 039 148 994 56;
  • 34) 0.148 568 028 039 148 994 56 × 2 = 0 + 0.297 136 056 078 297 989 12;
  • 35) 0.297 136 056 078 297 989 12 × 2 = 0 + 0.594 272 112 156 595 978 24;
  • 36) 0.594 272 112 156 595 978 24 × 2 = 1 + 0.188 544 224 313 191 956 48;
  • 37) 0.188 544 224 313 191 956 48 × 2 = 0 + 0.377 088 448 626 383 912 96;
  • 38) 0.377 088 448 626 383 912 96 × 2 = 0 + 0.754 176 897 252 767 825 92;
  • 39) 0.754 176 897 252 767 825 92 × 2 = 1 + 0.508 353 794 505 535 651 84;
  • 40) 0.508 353 794 505 535 651 84 × 2 = 1 + 0.016 707 589 011 071 303 68;
  • 41) 0.016 707 589 011 071 303 68 × 2 = 0 + 0.033 415 178 022 142 607 36;
  • 42) 0.033 415 178 022 142 607 36 × 2 = 0 + 0.066 830 356 044 285 214 72;
  • 43) 0.066 830 356 044 285 214 72 × 2 = 0 + 0.133 660 712 088 570 429 44;
  • 44) 0.133 660 712 088 570 429 44 × 2 = 0 + 0.267 321 424 177 140 858 88;
  • 45) 0.267 321 424 177 140 858 88 × 2 = 0 + 0.534 642 848 354 281 717 76;
  • 46) 0.534 642 848 354 281 717 76 × 2 = 1 + 0.069 285 696 708 563 435 52;
  • 47) 0.069 285 696 708 563 435 52 × 2 = 0 + 0.138 571 393 417 126 871 04;
  • 48) 0.138 571 393 417 126 871 04 × 2 = 0 + 0.277 142 786 834 253 742 08;
  • 49) 0.277 142 786 834 253 742 08 × 2 = 0 + 0.554 285 573 668 507 484 16;
  • 50) 0.554 285 573 668 507 484 16 × 2 = 1 + 0.108 571 147 337 014 968 32;
  • 51) 0.108 571 147 337 014 968 32 × 2 = 0 + 0.217 142 294 674 029 936 64;
  • 52) 0.217 142 294 674 029 936 64 × 2 = 0 + 0.434 284 589 348 059 873 28;
  • 53) 0.434 284 589 348 059 873 28 × 2 = 0 + 0.868 569 178 696 119 746 56;
  • 54) 0.868 569 178 696 119 746 56 × 2 = 1 + 0.737 138 357 392 239 493 12;
  • 55) 0.737 138 357 392 239 493 12 × 2 = 1 + 0.474 276 714 784 478 986 24;
  • 56) 0.474 276 714 784 478 986 24 × 2 = 0 + 0.948 553 429 568 957 972 48;
  • 57) 0.948 553 429 568 957 972 48 × 2 = 1 + 0.897 106 859 137 915 944 96;
  • 58) 0.897 106 859 137 915 944 96 × 2 = 1 + 0.794 213 718 275 831 889 92;
  • 59) 0.794 213 718 275 831 889 92 × 2 = 1 + 0.588 427 436 551 663 779 84;
  • 60) 0.588 427 436 551 663 779 84 × 2 = 1 + 0.176 854 873 103 327 559 68;
  • 61) 0.176 854 873 103 327 559 68 × 2 = 0 + 0.353 709 746 206 655 119 36;
  • 62) 0.353 709 746 206 655 119 36 × 2 = 0 + 0.707 419 492 413 310 238 72;
  • 63) 0.707 419 492 413 310 238 72 × 2 = 1 + 0.414 838 984 826 620 477 44;
  • 64) 0.414 838 984 826 620 477 44 × 2 = 0 + 0.829 677 969 653 240 954 88;
  • 65) 0.829 677 969 653 240 954 88 × 2 = 1 + 0.659 355 939 306 481 909 76;
  • 66) 0.659 355 939 306 481 909 76 × 2 = 1 + 0.318 711 878 612 963 819 52;
  • 67) 0.318 711 878 612 963 819 52 × 2 = 0 + 0.637 423 757 225 927 639 04;
  • 68) 0.637 423 757 225 927 639 04 × 2 = 1 + 0.274 847 514 451 855 278 08;
  • 69) 0.274 847 514 451 855 278 08 × 2 = 0 + 0.549 695 028 903 710 556 16;
  • 70) 0.549 695 028 903 710 556 16 × 2 = 1 + 0.099 390 057 807 421 112 32;
  • 71) 0.099 390 057 807 421 112 32 × 2 = 0 + 0.198 780 115 614 842 224 64;
  • 72) 0.198 780 115 614 842 224 64 × 2 = 0 + 0.397 560 231 229 684 449 28;

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:


0.000 001 506 198 729 391 68(10) =

0.0000 0000 0000 0000 0001 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0010 1101 0100(2)

5. Positive number before normalization:

0.000 001 506 198 729 391 68(10) =

0.0000 0000 0000 0000 0001 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0010 1101 0100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 506 198 729 391 68(10) =

0.0000 0000 0000 0000 0001 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0010 1101 0100(2) =

0.0000 0000 0000 0000 0001 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0010 1101 0100(2) × 20 =

1.1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0010 1101 0100(2) × 2-20


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): -20

Mantissa (not normalized):
1.1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0010 1101 0100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(11-1) - 1 =

(-20 + 1 023)(10) =

1 003(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;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

1003(10) =

011 1110 1011(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. 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0010 1101 0100 =

1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0010 1101 0100


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 1110 1011

Mantissa (52 bits) =
1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0010 1101 0100


Decimal number 0.000 001 506 198 729 391 68 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1011 - 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0010 1101 0100

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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