-2.211 829 052 383 358 300 119 548 660 47 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 300 119 548 660 47(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
-2.211 829 052 383 358 300 119 548 660 47(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. Start with the positive version of the number:

|-2.211 829 052 383 358 300 119 548 660 47| = 2.211 829 052 383 358 300 119 548 660 47


2. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. 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.

2(10) =

10(2)


4. Convert to binary (base 2) the fractional part: 0.211 829 052 383 358 300 119 548 660 47.

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) 0.211 829 052 383 358 300 119 548 660 47 × 2 = 0 + 0.423 658 104 766 716 600 239 097 320 94;
  • 2) 0.423 658 104 766 716 600 239 097 320 94 × 2 = 0 + 0.847 316 209 533 433 200 478 194 641 88;
  • 3) 0.847 316 209 533 433 200 478 194 641 88 × 2 = 1 + 0.694 632 419 066 866 400 956 389 283 76;
  • 4) 0.694 632 419 066 866 400 956 389 283 76 × 2 = 1 + 0.389 264 838 133 732 801 912 778 567 52;
  • 5) 0.389 264 838 133 732 801 912 778 567 52 × 2 = 0 + 0.778 529 676 267 465 603 825 557 135 04;
  • 6) 0.778 529 676 267 465 603 825 557 135 04 × 2 = 1 + 0.557 059 352 534 931 207 651 114 270 08;
  • 7) 0.557 059 352 534 931 207 651 114 270 08 × 2 = 1 + 0.114 118 705 069 862 415 302 228 540 16;
  • 8) 0.114 118 705 069 862 415 302 228 540 16 × 2 = 0 + 0.228 237 410 139 724 830 604 457 080 32;
  • 9) 0.228 237 410 139 724 830 604 457 080 32 × 2 = 0 + 0.456 474 820 279 449 661 208 914 160 64;
  • 10) 0.456 474 820 279 449 661 208 914 160 64 × 2 = 0 + 0.912 949 640 558 899 322 417 828 321 28;
  • 11) 0.912 949 640 558 899 322 417 828 321 28 × 2 = 1 + 0.825 899 281 117 798 644 835 656 642 56;
  • 12) 0.825 899 281 117 798 644 835 656 642 56 × 2 = 1 + 0.651 798 562 235 597 289 671 313 285 12;
  • 13) 0.651 798 562 235 597 289 671 313 285 12 × 2 = 1 + 0.303 597 124 471 194 579 342 626 570 24;
  • 14) 0.303 597 124 471 194 579 342 626 570 24 × 2 = 0 + 0.607 194 248 942 389 158 685 253 140 48;
  • 15) 0.607 194 248 942 389 158 685 253 140 48 × 2 = 1 + 0.214 388 497 884 778 317 370 506 280 96;
  • 16) 0.214 388 497 884 778 317 370 506 280 96 × 2 = 0 + 0.428 776 995 769 556 634 741 012 561 92;
  • 17) 0.428 776 995 769 556 634 741 012 561 92 × 2 = 0 + 0.857 553 991 539 113 269 482 025 123 84;
  • 18) 0.857 553 991 539 113 269 482 025 123 84 × 2 = 1 + 0.715 107 983 078 226 538 964 050 247 68;
  • 19) 0.715 107 983 078 226 538 964 050 247 68 × 2 = 1 + 0.430 215 966 156 453 077 928 100 495 36;
  • 20) 0.430 215 966 156 453 077 928 100 495 36 × 2 = 0 + 0.860 431 932 312 906 155 856 200 990 72;
  • 21) 0.860 431 932 312 906 155 856 200 990 72 × 2 = 1 + 0.720 863 864 625 812 311 712 401 981 44;
  • 22) 0.720 863 864 625 812 311 712 401 981 44 × 2 = 1 + 0.441 727 729 251 624 623 424 803 962 88;
  • 23) 0.441 727 729 251 624 623 424 803 962 88 × 2 = 0 + 0.883 455 458 503 249 246 849 607 925 76;
  • 24) 0.883 455 458 503 249 246 849 607 925 76 × 2 = 1 + 0.766 910 917 006 498 493 699 215 851 52;
  • 25) 0.766 910 917 006 498 493 699 215 851 52 × 2 = 1 + 0.533 821 834 012 996 987 398 431 703 04;
  • 26) 0.533 821 834 012 996 987 398 431 703 04 × 2 = 1 + 0.067 643 668 025 993 974 796 863 406 08;
  • 27) 0.067 643 668 025 993 974 796 863 406 08 × 2 = 0 + 0.135 287 336 051 987 949 593 726 812 16;
  • 28) 0.135 287 336 051 987 949 593 726 812 16 × 2 = 0 + 0.270 574 672 103 975 899 187 453 624 32;
  • 29) 0.270 574 672 103 975 899 187 453 624 32 × 2 = 0 + 0.541 149 344 207 951 798 374 907 248 64;
  • 30) 0.541 149 344 207 951 798 374 907 248 64 × 2 = 1 + 0.082 298 688 415 903 596 749 814 497 28;
  • 31) 0.082 298 688 415 903 596 749 814 497 28 × 2 = 0 + 0.164 597 376 831 807 193 499 628 994 56;
  • 32) 0.164 597 376 831 807 193 499 628 994 56 × 2 = 0 + 0.329 194 753 663 614 386 999 257 989 12;
  • 33) 0.329 194 753 663 614 386 999 257 989 12 × 2 = 0 + 0.658 389 507 327 228 773 998 515 978 24;
  • 34) 0.658 389 507 327 228 773 998 515 978 24 × 2 = 1 + 0.316 779 014 654 457 547 997 031 956 48;
  • 35) 0.316 779 014 654 457 547 997 031 956 48 × 2 = 0 + 0.633 558 029 308 915 095 994 063 912 96;
  • 36) 0.633 558 029 308 915 095 994 063 912 96 × 2 = 1 + 0.267 116 058 617 830 191 988 127 825 92;
  • 37) 0.267 116 058 617 830 191 988 127 825 92 × 2 = 0 + 0.534 232 117 235 660 383 976 255 651 84;
  • 38) 0.534 232 117 235 660 383 976 255 651 84 × 2 = 1 + 0.068 464 234 471 320 767 952 511 303 68;
  • 39) 0.068 464 234 471 320 767 952 511 303 68 × 2 = 0 + 0.136 928 468 942 641 535 905 022 607 36;
  • 40) 0.136 928 468 942 641 535 905 022 607 36 × 2 = 0 + 0.273 856 937 885 283 071 810 045 214 72;
  • 41) 0.273 856 937 885 283 071 810 045 214 72 × 2 = 0 + 0.547 713 875 770 566 143 620 090 429 44;
  • 42) 0.547 713 875 770 566 143 620 090 429 44 × 2 = 1 + 0.095 427 751 541 132 287 240 180 858 88;
  • 43) 0.095 427 751 541 132 287 240 180 858 88 × 2 = 0 + 0.190 855 503 082 264 574 480 361 717 76;
  • 44) 0.190 855 503 082 264 574 480 361 717 76 × 2 = 0 + 0.381 711 006 164 529 148 960 723 435 52;
  • 45) 0.381 711 006 164 529 148 960 723 435 52 × 2 = 0 + 0.763 422 012 329 058 297 921 446 871 04;
  • 46) 0.763 422 012 329 058 297 921 446 871 04 × 2 = 1 + 0.526 844 024 658 116 595 842 893 742 08;
  • 47) 0.526 844 024 658 116 595 842 893 742 08 × 2 = 1 + 0.053 688 049 316 233 191 685 787 484 16;
  • 48) 0.053 688 049 316 233 191 685 787 484 16 × 2 = 0 + 0.107 376 098 632 466 383 371 574 968 32;
  • 49) 0.107 376 098 632 466 383 371 574 968 32 × 2 = 0 + 0.214 752 197 264 932 766 743 149 936 64;
  • 50) 0.214 752 197 264 932 766 743 149 936 64 × 2 = 0 + 0.429 504 394 529 865 533 486 299 873 28;
  • 51) 0.429 504 394 529 865 533 486 299 873 28 × 2 = 0 + 0.859 008 789 059 731 066 972 599 746 56;
  • 52) 0.859 008 789 059 731 066 972 599 746 56 × 2 = 1 + 0.718 017 578 119 462 133 945 199 493 12;
  • 53) 0.718 017 578 119 462 133 945 199 493 12 × 2 = 1 + 0.436 035 156 238 924 267 890 398 986 24;

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).


5. 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:


0.211 829 052 383 358 300 119 548 660 47(10) =

0.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2)

6. Positive number before normalization:

2.211 829 052 383 358 300 119 548 660 47(10) =

10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.211 829 052 383 358 300 119 548 660 47(10) =

10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2) =

10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2) × 20 =

1.0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11(2) × 21


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): 1

Mantissa (not normalized):
1.0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(10)


10. 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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1024(10) =

100 0000 0000(2)


12. 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. 0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11 =

0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
100 0000 0000

Mantissa (52 bits) =
0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000


Decimal number -2.211 829 052 383 358 300 119 548 660 47 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000

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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