1.004 999 999 999 783 732 995 773 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.004 999 999 999 783 732 995 773(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
1.004 999 999 999 783 732 995 773(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: 0.004 999 999 999 783 732 995 773.

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) 0.004 999 999 999 783 732 995 773 × 2 = 0 + 0.009 999 999 999 567 465 991 546;
  • 2) 0.009 999 999 999 567 465 991 546 × 2 = 0 + 0.019 999 999 999 134 931 983 092;
  • 3) 0.019 999 999 999 134 931 983 092 × 2 = 0 + 0.039 999 999 998 269 863 966 184;
  • 4) 0.039 999 999 998 269 863 966 184 × 2 = 0 + 0.079 999 999 996 539 727 932 368;
  • 5) 0.079 999 999 996 539 727 932 368 × 2 = 0 + 0.159 999 999 993 079 455 864 736;
  • 6) 0.159 999 999 993 079 455 864 736 × 2 = 0 + 0.319 999 999 986 158 911 729 472;
  • 7) 0.319 999 999 986 158 911 729 472 × 2 = 0 + 0.639 999 999 972 317 823 458 944;
  • 8) 0.639 999 999 972 317 823 458 944 × 2 = 1 + 0.279 999 999 944 635 646 917 888;
  • 9) 0.279 999 999 944 635 646 917 888 × 2 = 0 + 0.559 999 999 889 271 293 835 776;
  • 10) 0.559 999 999 889 271 293 835 776 × 2 = 1 + 0.119 999 999 778 542 587 671 552;
  • 11) 0.119 999 999 778 542 587 671 552 × 2 = 0 + 0.239 999 999 557 085 175 343 104;
  • 12) 0.239 999 999 557 085 175 343 104 × 2 = 0 + 0.479 999 999 114 170 350 686 208;
  • 13) 0.479 999 999 114 170 350 686 208 × 2 = 0 + 0.959 999 998 228 340 701 372 416;
  • 14) 0.959 999 998 228 340 701 372 416 × 2 = 1 + 0.919 999 996 456 681 402 744 832;
  • 15) 0.919 999 996 456 681 402 744 832 × 2 = 1 + 0.839 999 992 913 362 805 489 664;
  • 16) 0.839 999 992 913 362 805 489 664 × 2 = 1 + 0.679 999 985 826 725 610 979 328;
  • 17) 0.679 999 985 826 725 610 979 328 × 2 = 1 + 0.359 999 971 653 451 221 958 656;
  • 18) 0.359 999 971 653 451 221 958 656 × 2 = 0 + 0.719 999 943 306 902 443 917 312;
  • 19) 0.719 999 943 306 902 443 917 312 × 2 = 1 + 0.439 999 886 613 804 887 834 624;
  • 20) 0.439 999 886 613 804 887 834 624 × 2 = 0 + 0.879 999 773 227 609 775 669 248;
  • 21) 0.879 999 773 227 609 775 669 248 × 2 = 1 + 0.759 999 546 455 219 551 338 496;
  • 22) 0.759 999 546 455 219 551 338 496 × 2 = 1 + 0.519 999 092 910 439 102 676 992;
  • 23) 0.519 999 092 910 439 102 676 992 × 2 = 1 + 0.039 998 185 820 878 205 353 984;
  • 24) 0.039 998 185 820 878 205 353 984 × 2 = 0 + 0.079 996 371 641 756 410 707 968;
  • 25) 0.079 996 371 641 756 410 707 968 × 2 = 0 + 0.159 992 743 283 512 821 415 936;
  • 26) 0.159 992 743 283 512 821 415 936 × 2 = 0 + 0.319 985 486 567 025 642 831 872;
  • 27) 0.319 985 486 567 025 642 831 872 × 2 = 0 + 0.639 970 973 134 051 285 663 744;
  • 28) 0.639 970 973 134 051 285 663 744 × 2 = 1 + 0.279 941 946 268 102 571 327 488;
  • 29) 0.279 941 946 268 102 571 327 488 × 2 = 0 + 0.559 883 892 536 205 142 654 976;
  • 30) 0.559 883 892 536 205 142 654 976 × 2 = 1 + 0.119 767 785 072 410 285 309 952;
  • 31) 0.119 767 785 072 410 285 309 952 × 2 = 0 + 0.239 535 570 144 820 570 619 904;
  • 32) 0.239 535 570 144 820 570 619 904 × 2 = 0 + 0.479 071 140 289 641 141 239 808;
  • 33) 0.479 071 140 289 641 141 239 808 × 2 = 0 + 0.958 142 280 579 282 282 479 616;
  • 34) 0.958 142 280 579 282 282 479 616 × 2 = 1 + 0.916 284 561 158 564 564 959 232;
  • 35) 0.916 284 561 158 564 564 959 232 × 2 = 1 + 0.832 569 122 317 129 129 918 464;
  • 36) 0.832 569 122 317 129 129 918 464 × 2 = 1 + 0.665 138 244 634 258 259 836 928;
  • 37) 0.665 138 244 634 258 259 836 928 × 2 = 1 + 0.330 276 489 268 516 519 673 856;
  • 38) 0.330 276 489 268 516 519 673 856 × 2 = 0 + 0.660 552 978 537 033 039 347 712;
  • 39) 0.660 552 978 537 033 039 347 712 × 2 = 1 + 0.321 105 957 074 066 078 695 424;
  • 40) 0.321 105 957 074 066 078 695 424 × 2 = 0 + 0.642 211 914 148 132 157 390 848;
  • 41) 0.642 211 914 148 132 157 390 848 × 2 = 1 + 0.284 423 828 296 264 314 781 696;
  • 42) 0.284 423 828 296 264 314 781 696 × 2 = 0 + 0.568 847 656 592 528 629 563 392;
  • 43) 0.568 847 656 592 528 629 563 392 × 2 = 1 + 0.137 695 313 185 057 259 126 784;
  • 44) 0.137 695 313 185 057 259 126 784 × 2 = 0 + 0.275 390 626 370 114 518 253 568;
  • 45) 0.275 390 626 370 114 518 253 568 × 2 = 0 + 0.550 781 252 740 229 036 507 136;
  • 46) 0.550 781 252 740 229 036 507 136 × 2 = 1 + 0.101 562 505 480 458 073 014 272;
  • 47) 0.101 562 505 480 458 073 014 272 × 2 = 0 + 0.203 125 010 960 916 146 028 544;
  • 48) 0.203 125 010 960 916 146 028 544 × 2 = 0 + 0.406 250 021 921 832 292 057 088;
  • 49) 0.406 250 021 921 832 292 057 088 × 2 = 0 + 0.812 500 043 843 664 584 114 176;
  • 50) 0.812 500 043 843 664 584 114 176 × 2 = 1 + 0.625 000 087 687 329 168 228 352;
  • 51) 0.625 000 087 687 329 168 228 352 × 2 = 1 + 0.250 000 175 374 658 336 456 704;
  • 52) 0.250 000 175 374 658 336 456 704 × 2 = 0 + 0.500 000 350 749 316 672 913 408;
  • 53) 0.500 000 350 749 316 672 913 408 × 2 = 1 + 0.000 000 701 498 633 345 826 816;

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:


0.004 999 999 999 783 732 995 773(10) =

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

5. Positive number before normalization:

1.004 999 999 999 783 732 995 773(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:


1.004 999 999 999 783 732 995 773(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 1.004 999 999 999 783 732 995 773 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