1.004 999 999 999 783 732 995 695 118 006 18 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.004 999 999 999 783 732 995 695 118 006 18(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
1.004 999 999 999 783 732 995 695 118 006 18(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: 0.004 999 999 999 783 732 995 695 118 006 18.

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) 0.004 999 999 999 783 732 995 695 118 006 18 × 2 = 0 + 0.009 999 999 999 567 465 991 390 236 012 36;
  • 2) 0.009 999 999 999 567 465 991 390 236 012 36 × 2 = 0 + 0.019 999 999 999 134 931 982 780 472 024 72;
  • 3) 0.019 999 999 999 134 931 982 780 472 024 72 × 2 = 0 + 0.039 999 999 998 269 863 965 560 944 049 44;
  • 4) 0.039 999 999 998 269 863 965 560 944 049 44 × 2 = 0 + 0.079 999 999 996 539 727 931 121 888 098 88;
  • 5) 0.079 999 999 996 539 727 931 121 888 098 88 × 2 = 0 + 0.159 999 999 993 079 455 862 243 776 197 76;
  • 6) 0.159 999 999 993 079 455 862 243 776 197 76 × 2 = 0 + 0.319 999 999 986 158 911 724 487 552 395 52;
  • 7) 0.319 999 999 986 158 911 724 487 552 395 52 × 2 = 0 + 0.639 999 999 972 317 823 448 975 104 791 04;
  • 8) 0.639 999 999 972 317 823 448 975 104 791 04 × 2 = 1 + 0.279 999 999 944 635 646 897 950 209 582 08;
  • 9) 0.279 999 999 944 635 646 897 950 209 582 08 × 2 = 0 + 0.559 999 999 889 271 293 795 900 419 164 16;
  • 10) 0.559 999 999 889 271 293 795 900 419 164 16 × 2 = 1 + 0.119 999 999 778 542 587 591 800 838 328 32;
  • 11) 0.119 999 999 778 542 587 591 800 838 328 32 × 2 = 0 + 0.239 999 999 557 085 175 183 601 676 656 64;
  • 12) 0.239 999 999 557 085 175 183 601 676 656 64 × 2 = 0 + 0.479 999 999 114 170 350 367 203 353 313 28;
  • 13) 0.479 999 999 114 170 350 367 203 353 313 28 × 2 = 0 + 0.959 999 998 228 340 700 734 406 706 626 56;
  • 14) 0.959 999 998 228 340 700 734 406 706 626 56 × 2 = 1 + 0.919 999 996 456 681 401 468 813 413 253 12;
  • 15) 0.919 999 996 456 681 401 468 813 413 253 12 × 2 = 1 + 0.839 999 992 913 362 802 937 626 826 506 24;
  • 16) 0.839 999 992 913 362 802 937 626 826 506 24 × 2 = 1 + 0.679 999 985 826 725 605 875 253 653 012 48;
  • 17) 0.679 999 985 826 725 605 875 253 653 012 48 × 2 = 1 + 0.359 999 971 653 451 211 750 507 306 024 96;
  • 18) 0.359 999 971 653 451 211 750 507 306 024 96 × 2 = 0 + 0.719 999 943 306 902 423 501 014 612 049 92;
  • 19) 0.719 999 943 306 902 423 501 014 612 049 92 × 2 = 1 + 0.439 999 886 613 804 847 002 029 224 099 84;
  • 20) 0.439 999 886 613 804 847 002 029 224 099 84 × 2 = 0 + 0.879 999 773 227 609 694 004 058 448 199 68;
  • 21) 0.879 999 773 227 609 694 004 058 448 199 68 × 2 = 1 + 0.759 999 546 455 219 388 008 116 896 399 36;
  • 22) 0.759 999 546 455 219 388 008 116 896 399 36 × 2 = 1 + 0.519 999 092 910 438 776 016 233 792 798 72;
  • 23) 0.519 999 092 910 438 776 016 233 792 798 72 × 2 = 1 + 0.039 998 185 820 877 552 032 467 585 597 44;
  • 24) 0.039 998 185 820 877 552 032 467 585 597 44 × 2 = 0 + 0.079 996 371 641 755 104 064 935 171 194 88;
  • 25) 0.079 996 371 641 755 104 064 935 171 194 88 × 2 = 0 + 0.159 992 743 283 510 208 129 870 342 389 76;
  • 26) 0.159 992 743 283 510 208 129 870 342 389 76 × 2 = 0 + 0.319 985 486 567 020 416 259 740 684 779 52;
  • 27) 0.319 985 486 567 020 416 259 740 684 779 52 × 2 = 0 + 0.639 970 973 134 040 832 519 481 369 559 04;
  • 28) 0.639 970 973 134 040 832 519 481 369 559 04 × 2 = 1 + 0.279 941 946 268 081 665 038 962 739 118 08;
  • 29) 0.279 941 946 268 081 665 038 962 739 118 08 × 2 = 0 + 0.559 883 892 536 163 330 077 925 478 236 16;
  • 30) 0.559 883 892 536 163 330 077 925 478 236 16 × 2 = 1 + 0.119 767 785 072 326 660 155 850 956 472 32;
  • 31) 0.119 767 785 072 326 660 155 850 956 472 32 × 2 = 0 + 0.239 535 570 144 653 320 311 701 912 944 64;
  • 32) 0.239 535 570 144 653 320 311 701 912 944 64 × 2 = 0 + 0.479 071 140 289 306 640 623 403 825 889 28;
  • 33) 0.479 071 140 289 306 640 623 403 825 889 28 × 2 = 0 + 0.958 142 280 578 613 281 246 807 651 778 56;
  • 34) 0.958 142 280 578 613 281 246 807 651 778 56 × 2 = 1 + 0.916 284 561 157 226 562 493 615 303 557 12;
  • 35) 0.916 284 561 157 226 562 493 615 303 557 12 × 2 = 1 + 0.832 569 122 314 453 124 987 230 607 114 24;
  • 36) 0.832 569 122 314 453 124 987 230 607 114 24 × 2 = 1 + 0.665 138 244 628 906 249 974 461 214 228 48;
  • 37) 0.665 138 244 628 906 249 974 461 214 228 48 × 2 = 1 + 0.330 276 489 257 812 499 948 922 428 456 96;
  • 38) 0.330 276 489 257 812 499 948 922 428 456 96 × 2 = 0 + 0.660 552 978 515 624 999 897 844 856 913 92;
  • 39) 0.660 552 978 515 624 999 897 844 856 913 92 × 2 = 1 + 0.321 105 957 031 249 999 795 689 713 827 84;
  • 40) 0.321 105 957 031 249 999 795 689 713 827 84 × 2 = 0 + 0.642 211 914 062 499 999 591 379 427 655 68;
  • 41) 0.642 211 914 062 499 999 591 379 427 655 68 × 2 = 1 + 0.284 423 828 124 999 999 182 758 855 311 36;
  • 42) 0.284 423 828 124 999 999 182 758 855 311 36 × 2 = 0 + 0.568 847 656 249 999 998 365 517 710 622 72;
  • 43) 0.568 847 656 249 999 998 365 517 710 622 72 × 2 = 1 + 0.137 695 312 499 999 996 731 035 421 245 44;
  • 44) 0.137 695 312 499 999 996 731 035 421 245 44 × 2 = 0 + 0.275 390 624 999 999 993 462 070 842 490 88;
  • 45) 0.275 390 624 999 999 993 462 070 842 490 88 × 2 = 0 + 0.550 781 249 999 999 986 924 141 684 981 76;
  • 46) 0.550 781 249 999 999 986 924 141 684 981 76 × 2 = 1 + 0.101 562 499 999 999 973 848 283 369 963 52;
  • 47) 0.101 562 499 999 999 973 848 283 369 963 52 × 2 = 0 + 0.203 124 999 999 999 947 696 566 739 927 04;
  • 48) 0.203 124 999 999 999 947 696 566 739 927 04 × 2 = 0 + 0.406 249 999 999 999 895 393 133 479 854 08;
  • 49) 0.406 249 999 999 999 895 393 133 479 854 08 × 2 = 0 + 0.812 499 999 999 999 790 786 266 959 708 16;
  • 50) 0.812 499 999 999 999 790 786 266 959 708 16 × 2 = 1 + 0.624 999 999 999 999 581 572 533 919 416 32;
  • 51) 0.624 999 999 999 999 581 572 533 919 416 32 × 2 = 1 + 0.249 999 999 999 999 163 145 067 838 832 64;
  • 52) 0.249 999 999 999 999 163 145 067 838 832 64 × 2 = 0 + 0.499 999 999 999 998 326 290 135 677 665 28;
  • 53) 0.499 999 999 999 998 326 290 135 677 665 28 × 2 = 0 + 0.999 999 999 999 996 652 580 271 355 330 56;

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:


0.004 999 999 999 783 732 995 695 118 006 18(10) =

0.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 0(2)

5. Positive number before normalization:

1.004 999 999 999 783 732 995 695 118 006 18(10) =

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


1.004 999 999 999 783 732 995 695 118 006 18(10) =

1.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 0(2) =

1.0000 0001 0100 0111 1010 1110 0001 0100 0111 1010 1010 0100 0110 0(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 0


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 0 =

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 1.004 999 999 999 783 732 995 695 118 006 18 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