1.004 999 999 999 783 732 995 695 131 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.004 999 999 999 783 732 995 695 131(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
1.004 999 999 999 783 732 995 695 131(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: 1.
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;
  • 1 ÷ 2 = 0 + 1;

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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.004 999 999 999 783 732 995 695 131.

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) 0.004 999 999 999 783 732 995 695 131 × 2 = 0 + 0.009 999 999 999 567 465 991 390 262;
  • 2) 0.009 999 999 999 567 465 991 390 262 × 2 = 0 + 0.019 999 999 999 134 931 982 780 524;
  • 3) 0.019 999 999 999 134 931 982 780 524 × 2 = 0 + 0.039 999 999 998 269 863 965 561 048;
  • 4) 0.039 999 999 998 269 863 965 561 048 × 2 = 0 + 0.079 999 999 996 539 727 931 122 096;
  • 5) 0.079 999 999 996 539 727 931 122 096 × 2 = 0 + 0.159 999 999 993 079 455 862 244 192;
  • 6) 0.159 999 999 993 079 455 862 244 192 × 2 = 0 + 0.319 999 999 986 158 911 724 488 384;
  • 7) 0.319 999 999 986 158 911 724 488 384 × 2 = 0 + 0.639 999 999 972 317 823 448 976 768;
  • 8) 0.639 999 999 972 317 823 448 976 768 × 2 = 1 + 0.279 999 999 944 635 646 897 953 536;
  • 9) 0.279 999 999 944 635 646 897 953 536 × 2 = 0 + 0.559 999 999 889 271 293 795 907 072;
  • 10) 0.559 999 999 889 271 293 795 907 072 × 2 = 1 + 0.119 999 999 778 542 587 591 814 144;
  • 11) 0.119 999 999 778 542 587 591 814 144 × 2 = 0 + 0.239 999 999 557 085 175 183 628 288;
  • 12) 0.239 999 999 557 085 175 183 628 288 × 2 = 0 + 0.479 999 999 114 170 350 367 256 576;
  • 13) 0.479 999 999 114 170 350 367 256 576 × 2 = 0 + 0.959 999 998 228 340 700 734 513 152;
  • 14) 0.959 999 998 228 340 700 734 513 152 × 2 = 1 + 0.919 999 996 456 681 401 469 026 304;
  • 15) 0.919 999 996 456 681 401 469 026 304 × 2 = 1 + 0.839 999 992 913 362 802 938 052 608;
  • 16) 0.839 999 992 913 362 802 938 052 608 × 2 = 1 + 0.679 999 985 826 725 605 876 105 216;
  • 17) 0.679 999 985 826 725 605 876 105 216 × 2 = 1 + 0.359 999 971 653 451 211 752 210 432;
  • 18) 0.359 999 971 653 451 211 752 210 432 × 2 = 0 + 0.719 999 943 306 902 423 504 420 864;
  • 19) 0.719 999 943 306 902 423 504 420 864 × 2 = 1 + 0.439 999 886 613 804 847 008 841 728;
  • 20) 0.439 999 886 613 804 847 008 841 728 × 2 = 0 + 0.879 999 773 227 609 694 017 683 456;
  • 21) 0.879 999 773 227 609 694 017 683 456 × 2 = 1 + 0.759 999 546 455 219 388 035 366 912;
  • 22) 0.759 999 546 455 219 388 035 366 912 × 2 = 1 + 0.519 999 092 910 438 776 070 733 824;
  • 23) 0.519 999 092 910 438 776 070 733 824 × 2 = 1 + 0.039 998 185 820 877 552 141 467 648;
  • 24) 0.039 998 185 820 877 552 141 467 648 × 2 = 0 + 0.079 996 371 641 755 104 282 935 296;
  • 25) 0.079 996 371 641 755 104 282 935 296 × 2 = 0 + 0.159 992 743 283 510 208 565 870 592;
  • 26) 0.159 992 743 283 510 208 565 870 592 × 2 = 0 + 0.319 985 486 567 020 417 131 741 184;
  • 27) 0.319 985 486 567 020 417 131 741 184 × 2 = 0 + 0.639 970 973 134 040 834 263 482 368;
  • 28) 0.639 970 973 134 040 834 263 482 368 × 2 = 1 + 0.279 941 946 268 081 668 526 964 736;
  • 29) 0.279 941 946 268 081 668 526 964 736 × 2 = 0 + 0.559 883 892 536 163 337 053 929 472;
  • 30) 0.559 883 892 536 163 337 053 929 472 × 2 = 1 + 0.119 767 785 072 326 674 107 858 944;
  • 31) 0.119 767 785 072 326 674 107 858 944 × 2 = 0 + 0.239 535 570 144 653 348 215 717 888;
  • 32) 0.239 535 570 144 653 348 215 717 888 × 2 = 0 + 0.479 071 140 289 306 696 431 435 776;
  • 33) 0.479 071 140 289 306 696 431 435 776 × 2 = 0 + 0.958 142 280 578 613 392 862 871 552;
  • 34) 0.958 142 280 578 613 392 862 871 552 × 2 = 1 + 0.916 284 561 157 226 785 725 743 104;
  • 35) 0.916 284 561 157 226 785 725 743 104 × 2 = 1 + 0.832 569 122 314 453 571 451 486 208;
  • 36) 0.832 569 122 314 453 571 451 486 208 × 2 = 1 + 0.665 138 244 628 907 142 902 972 416;
  • 37) 0.665 138 244 628 907 142 902 972 416 × 2 = 1 + 0.330 276 489 257 814 285 805 944 832;
  • 38) 0.330 276 489 257 814 285 805 944 832 × 2 = 0 + 0.660 552 978 515 628 571 611 889 664;
  • 39) 0.660 552 978 515 628 571 611 889 664 × 2 = 1 + 0.321 105 957 031 257 143 223 779 328;
  • 40) 0.321 105 957 031 257 143 223 779 328 × 2 = 0 + 0.642 211 914 062 514 286 447 558 656;
  • 41) 0.642 211 914 062 514 286 447 558 656 × 2 = 1 + 0.284 423 828 125 028 572 895 117 312;
  • 42) 0.284 423 828 125 028 572 895 117 312 × 2 = 0 + 0.568 847 656 250 057 145 790 234 624;
  • 43) 0.568 847 656 250 057 145 790 234 624 × 2 = 1 + 0.137 695 312 500 114 291 580 469 248;
  • 44) 0.137 695 312 500 114 291 580 469 248 × 2 = 0 + 0.275 390 625 000 228 583 160 938 496;
  • 45) 0.275 390 625 000 228 583 160 938 496 × 2 = 0 + 0.550 781 250 000 457 166 321 876 992;
  • 46) 0.550 781 250 000 457 166 321 876 992 × 2 = 1 + 0.101 562 500 000 914 332 643 753 984;
  • 47) 0.101 562 500 000 914 332 643 753 984 × 2 = 0 + 0.203 125 000 001 828 665 287 507 968;
  • 48) 0.203 125 000 001 828 665 287 507 968 × 2 = 0 + 0.406 250 000 003 657 330 575 015 936;
  • 49) 0.406 250 000 003 657 330 575 015 936 × 2 = 0 + 0.812 500 000 007 314 661 150 031 872;
  • 50) 0.812 500 000 007 314 661 150 031 872 × 2 = 1 + 0.625 000 000 014 629 322 300 063 744;
  • 51) 0.625 000 000 014 629 322 300 063 744 × 2 = 1 + 0.250 000 000 029 258 644 600 127 488;
  • 52) 0.250 000 000 029 258 644 600 127 488 × 2 = 0 + 0.500 000 000 058 517 289 200 254 976;
  • 53) 0.500 000 000 058 517 289 200 254 976 × 2 = 1 + 0.000 000 000 117 034 578 400 509 952;

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:


0.004 999 999 999 783 732 995 695 131(10) =

0.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 1(2)

5. Positive number before normalization:

1.004 999 999 999 783 732 995 695 131(10) =

1.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.004 999 999 999 783 732 995 695 131(10) =

1.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 1(2) =

1.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 1(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): 0

Mantissa (not normalized):
1.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 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;

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) =

1023(10) =

011 1111 1111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 1 =

0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110


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 1111 1111

Mantissa (52 bits) =
0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110


Decimal number 1.004 999 999 999 783 732 995 695 131 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110

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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