Base ten decimal number 0.785 398 163 397 448 163 531 6 converted to 64 bit double precision IEEE 754 binary floating point standard

How to convert the decimal number 0.785 398 163 397 448 163 531 6(10)
to
64 bit double precision IEEE 754 binary floating point
(1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (base 2) the integer part: 0. Divide the number repeatedly by 2, keeping track of each remainder, until 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, by taking 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: 0.785 398 163 397 448 163 531 6. Multiply it 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) 0.785 398 163 397 448 163 531 6 × 2 = 1 + 0.570 796 326 794 896 327 063 2;
  • 2) 0.570 796 326 794 896 327 063 2 × 2 = 1 + 0.141 592 653 589 792 654 126 4;
  • 3) 0.141 592 653 589 792 654 126 4 × 2 = 0 + 0.283 185 307 179 585 308 252 8;
  • 4) 0.283 185 307 179 585 308 252 8 × 2 = 0 + 0.566 370 614 359 170 616 505 6;
  • 5) 0.566 370 614 359 170 616 505 6 × 2 = 1 + 0.132 741 228 718 341 233 011 2;
  • 6) 0.132 741 228 718 341 233 011 2 × 2 = 0 + 0.265 482 457 436 682 466 022 4;
  • 7) 0.265 482 457 436 682 466 022 4 × 2 = 0 + 0.530 964 914 873 364 932 044 8;
  • 8) 0.530 964 914 873 364 932 044 8 × 2 = 1 + 0.061 929 829 746 729 864 089 6;
  • 9) 0.061 929 829 746 729 864 089 6 × 2 = 0 + 0.123 859 659 493 459 728 179 2;
  • 10) 0.123 859 659 493 459 728 179 2 × 2 = 0 + 0.247 719 318 986 919 456 358 4;
  • 11) 0.247 719 318 986 919 456 358 4 × 2 = 0 + 0.495 438 637 973 838 912 716 8;
  • 12) 0.495 438 637 973 838 912 716 8 × 2 = 0 + 0.990 877 275 947 677 825 433 6;
  • 13) 0.990 877 275 947 677 825 433 6 × 2 = 1 + 0.981 754 551 895 355 650 867 2;
  • 14) 0.981 754 551 895 355 650 867 2 × 2 = 1 + 0.963 509 103 790 711 301 734 4;
  • 15) 0.963 509 103 790 711 301 734 4 × 2 = 1 + 0.927 018 207 581 422 603 468 8;
  • 16) 0.927 018 207 581 422 603 468 8 × 2 = 1 + 0.854 036 415 162 845 206 937 6;
  • 17) 0.854 036 415 162 845 206 937 6 × 2 = 1 + 0.708 072 830 325 690 413 875 2;
  • 18) 0.708 072 830 325 690 413 875 2 × 2 = 1 + 0.416 145 660 651 380 827 750 4;
  • 19) 0.416 145 660 651 380 827 750 4 × 2 = 0 + 0.832 291 321 302 761 655 500 8;
  • 20) 0.832 291 321 302 761 655 500 8 × 2 = 1 + 0.664 582 642 605 523 311 001 6;
  • 21) 0.664 582 642 605 523 311 001 6 × 2 = 1 + 0.329 165 285 211 046 622 003 2;
  • 22) 0.329 165 285 211 046 622 003 2 × 2 = 0 + 0.658 330 570 422 093 244 006 4;
  • 23) 0.658 330 570 422 093 244 006 4 × 2 = 1 + 0.316 661 140 844 186 488 012 8;
  • 24) 0.316 661 140 844 186 488 012 8 × 2 = 0 + 0.633 322 281 688 372 976 025 6;
  • 25) 0.633 322 281 688 372 976 025 6 × 2 = 1 + 0.266 644 563 376 745 952 051 2;
  • 26) 0.266 644 563 376 745 952 051 2 × 2 = 0 + 0.533 289 126 753 491 904 102 4;
  • 27) 0.533 289 126 753 491 904 102 4 × 2 = 1 + 0.066 578 253 506 983 808 204 8;
  • 28) 0.066 578 253 506 983 808 204 8 × 2 = 0 + 0.133 156 507 013 967 616 409 6;
  • 29) 0.133 156 507 013 967 616 409 6 × 2 = 0 + 0.266 313 014 027 935 232 819 2;
  • 30) 0.266 313 014 027 935 232 819 2 × 2 = 0 + 0.532 626 028 055 870 465 638 4;
  • 31) 0.532 626 028 055 870 465 638 4 × 2 = 1 + 0.065 252 056 111 740 931 276 8;
  • 32) 0.065 252 056 111 740 931 276 8 × 2 = 0 + 0.130 504 112 223 481 862 553 6;
  • 33) 0.130 504 112 223 481 862 553 6 × 2 = 0 + 0.261 008 224 446 963 725 107 2;
  • 34) 0.261 008 224 446 963 725 107 2 × 2 = 0 + 0.522 016 448 893 927 450 214 4;
  • 35) 0.522 016 448 893 927 450 214 4 × 2 = 1 + 0.044 032 897 787 854 900 428 8;
  • 36) 0.044 032 897 787 854 900 428 8 × 2 = 0 + 0.088 065 795 575 709 800 857 6;
  • 37) 0.088 065 795 575 709 800 857 6 × 2 = 0 + 0.176 131 591 151 419 601 715 2;
  • 38) 0.176 131 591 151 419 601 715 2 × 2 = 0 + 0.352 263 182 302 839 203 430 4;
  • 39) 0.352 263 182 302 839 203 430 4 × 2 = 0 + 0.704 526 364 605 678 406 860 8;
  • 40) 0.704 526 364 605 678 406 860 8 × 2 = 1 + 0.409 052 729 211 356 813 721 6;
  • 41) 0.409 052 729 211 356 813 721 6 × 2 = 0 + 0.818 105 458 422 713 627 443 2;
  • 42) 0.818 105 458 422 713 627 443 2 × 2 = 1 + 0.636 210 916 845 427 254 886 4;
  • 43) 0.636 210 916 845 427 254 886 4 × 2 = 1 + 0.272 421 833 690 854 509 772 8;
  • 44) 0.272 421 833 690 854 509 772 8 × 2 = 0 + 0.544 843 667 381 709 019 545 6;
  • 45) 0.544 843 667 381 709 019 545 6 × 2 = 1 + 0.089 687 334 763 418 039 091 2;
  • 46) 0.089 687 334 763 418 039 091 2 × 2 = 0 + 0.179 374 669 526 836 078 182 4;
  • 47) 0.179 374 669 526 836 078 182 4 × 2 = 0 + 0.358 749 339 053 672 156 364 8;
  • 48) 0.358 749 339 053 672 156 364 8 × 2 = 0 + 0.717 498 678 107 344 312 729 6;
  • 49) 0.717 498 678 107 344 312 729 6 × 2 = 1 + 0.434 997 356 214 688 625 459 2;
  • 50) 0.434 997 356 214 688 625 459 2 × 2 = 0 + 0.869 994 712 429 377 250 918 4;
  • 51) 0.869 994 712 429 377 250 918 4 × 2 = 1 + 0.739 989 424 858 754 501 836 8;
  • 52) 0.739 989 424 858 754 501 836 8 × 2 = 1 + 0.479 978 849 717 509 003 673 6;
  • 53) 0.479 978 849 717 509 003 673 6 × 2 = 0 + 0.959 957 699 435 018 007 347 2;

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...)

4. 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 constructed list above:

0.785 398 163 397 448 163 531 6(10) =


0.1100 1001 0000 1111 1101 1010 1010 0010 0010 0001 0110 1000 1011 0(2)

Positive number before normalization:

0.785 398 163 397 448 163 531 6(10) =


0.1100 1001 0000 1111 1101 1010 1010 0010 0010 0001 0110 1000 1011 0(2)

5. Normalize the binary representation of the number, shifting the decimal mark 1 positions to the right so that only one non zero digit remains to the left of it:

0.785 398 163 397 448 163 531 6(10) =


0.1100 1001 0000 1111 1101 1010 1010 0010 0010 0001 0110 1000 1011 0(2) =


0.1100 1001 0000 1111 1101 1010 1010 0010 0010 0001 0110 1000 1011 0(2) × 20 =


1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110(2) × 2-1

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): -1


Mantissa (not normalized): 1.1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110

6. 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:

Exponent (adjusted) =


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


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


(-1 + 1 023)(10) =


1 022(10)


  • division = quotient + remainder;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


1022(10) =


011 1111 1110(2)

7. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 52 bits, only if necessary (not the case here):

Mantissa (normalized) =


1. 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110 =


1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110

Conclusion:

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 1111 1110


Mantissa (52 bits) =
1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110

Number 0.785 398 163 397 448 163 531 6, a decimal, converted from decimal system (base 10)
to
64 bit double precision IEEE 754 binary floating point:


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

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 0

      49
    • 1

      48
    • 0

      47
    • 0

      46
    • 1

      45
    • 0

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 1

      40
    • 1

      39
    • 1

      38
    • 1

      37
    • 1

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 1

      32
    • 0

      31
    • 1

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 1

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 0

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 1

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 1

      1
    • 0

      0

Convert decimal numbers from base ten to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

0.785 398 163 397 448 163 531 6 = 0 - 011 1111 1110 - 1001 0010 0001 1111 1011 0101 0100 0100 0100 0010 1101 0001 0110 Oct 20 18:55 UTC (GMT)
0.975 = 0 - 011 1111 1110 - 1111 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 Oct 20 18:53 UTC (GMT)
0.120 000 000 000 000 01 = 0 - 011 1111 1011 - 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1001 Oct 20 18:53 UTC (GMT)
-213.45 = 1 - 100 0000 0110 - 1010 1010 1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 Oct 20 18:51 UTC (GMT)
157.079 632 67 = 0 - 100 0000 0110 - 0011 1010 0010 1000 1100 0101 1001 1101 0000 0010 1010 1111 1011 Oct 20 18:50 UTC (GMT)
-127 = 1 - 100 0000 0101 - 1111 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Oct 20 18:50 UTC (GMT)
50.24 = 0 - 100 0000 0100 - 1001 0001 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 Oct 20 18:50 UTC (GMT)
0.347 = 0 - 011 1111 1101 - 0110 0011 0101 0011 1111 0111 1100 1110 1101 1001 0001 0110 1000 Oct 20 18:48 UTC (GMT)
865 = 0 - 100 0000 1000 - 1011 0000 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Oct 20 18:46 UTC (GMT)
-43.625 = 1 - 100 0000 0100 - 0101 1101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Oct 20 18:46 UTC (GMT)
-3.516 613 304 435 003 = 1 - 100 0000 0000 - 1100 0010 0010 0000 0110 0010 0111 1111 1001 1101 0000 1000 1100 Oct 20 18:45 UTC (GMT)
-19.382 = 1 - 100 0000 0011 - 0011 0110 0001 1100 1010 1100 0000 1000 0011 0001 0010 0110 1110 Oct 20 18:45 UTC (GMT)
13.125 = 0 - 100 0000 0010 - 1010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Oct 20 18:44 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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