-2.211 829 052 383 358 300 119 548 661 699 653 282 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 300 119 548 661 699 653 282 3(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
-2.211 829 052 383 358 300 119 548 661 699 653 282 3(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:

|-2.211 829 052 383 358 300 119 548 661 699 653 282 3| = 2.211 829 052 383 358 300 119 548 661 699 653 282 3


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: 0.211 829 052 383 358 300 119 548 661 699 653 282 3.

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) 0.211 829 052 383 358 300 119 548 661 699 653 282 3 × 2 = 0 + 0.423 658 104 766 716 600 239 097 323 399 306 564 6;
  • 2) 0.423 658 104 766 716 600 239 097 323 399 306 564 6 × 2 = 0 + 0.847 316 209 533 433 200 478 194 646 798 613 129 2;
  • 3) 0.847 316 209 533 433 200 478 194 646 798 613 129 2 × 2 = 1 + 0.694 632 419 066 866 400 956 389 293 597 226 258 4;
  • 4) 0.694 632 419 066 866 400 956 389 293 597 226 258 4 × 2 = 1 + 0.389 264 838 133 732 801 912 778 587 194 452 516 8;
  • 5) 0.389 264 838 133 732 801 912 778 587 194 452 516 8 × 2 = 0 + 0.778 529 676 267 465 603 825 557 174 388 905 033 6;
  • 6) 0.778 529 676 267 465 603 825 557 174 388 905 033 6 × 2 = 1 + 0.557 059 352 534 931 207 651 114 348 777 810 067 2;
  • 7) 0.557 059 352 534 931 207 651 114 348 777 810 067 2 × 2 = 1 + 0.114 118 705 069 862 415 302 228 697 555 620 134 4;
  • 8) 0.114 118 705 069 862 415 302 228 697 555 620 134 4 × 2 = 0 + 0.228 237 410 139 724 830 604 457 395 111 240 268 8;
  • 9) 0.228 237 410 139 724 830 604 457 395 111 240 268 8 × 2 = 0 + 0.456 474 820 279 449 661 208 914 790 222 480 537 6;
  • 10) 0.456 474 820 279 449 661 208 914 790 222 480 537 6 × 2 = 0 + 0.912 949 640 558 899 322 417 829 580 444 961 075 2;
  • 11) 0.912 949 640 558 899 322 417 829 580 444 961 075 2 × 2 = 1 + 0.825 899 281 117 798 644 835 659 160 889 922 150 4;
  • 12) 0.825 899 281 117 798 644 835 659 160 889 922 150 4 × 2 = 1 + 0.651 798 562 235 597 289 671 318 321 779 844 300 8;
  • 13) 0.651 798 562 235 597 289 671 318 321 779 844 300 8 × 2 = 1 + 0.303 597 124 471 194 579 342 636 643 559 688 601 6;
  • 14) 0.303 597 124 471 194 579 342 636 643 559 688 601 6 × 2 = 0 + 0.607 194 248 942 389 158 685 273 287 119 377 203 2;
  • 15) 0.607 194 248 942 389 158 685 273 287 119 377 203 2 × 2 = 1 + 0.214 388 497 884 778 317 370 546 574 238 754 406 4;
  • 16) 0.214 388 497 884 778 317 370 546 574 238 754 406 4 × 2 = 0 + 0.428 776 995 769 556 634 741 093 148 477 508 812 8;
  • 17) 0.428 776 995 769 556 634 741 093 148 477 508 812 8 × 2 = 0 + 0.857 553 991 539 113 269 482 186 296 955 017 625 6;
  • 18) 0.857 553 991 539 113 269 482 186 296 955 017 625 6 × 2 = 1 + 0.715 107 983 078 226 538 964 372 593 910 035 251 2;
  • 19) 0.715 107 983 078 226 538 964 372 593 910 035 251 2 × 2 = 1 + 0.430 215 966 156 453 077 928 745 187 820 070 502 4;
  • 20) 0.430 215 966 156 453 077 928 745 187 820 070 502 4 × 2 = 0 + 0.860 431 932 312 906 155 857 490 375 640 141 004 8;
  • 21) 0.860 431 932 312 906 155 857 490 375 640 141 004 8 × 2 = 1 + 0.720 863 864 625 812 311 714 980 751 280 282 009 6;
  • 22) 0.720 863 864 625 812 311 714 980 751 280 282 009 6 × 2 = 1 + 0.441 727 729 251 624 623 429 961 502 560 564 019 2;
  • 23) 0.441 727 729 251 624 623 429 961 502 560 564 019 2 × 2 = 0 + 0.883 455 458 503 249 246 859 923 005 121 128 038 4;
  • 24) 0.883 455 458 503 249 246 859 923 005 121 128 038 4 × 2 = 1 + 0.766 910 917 006 498 493 719 846 010 242 256 076 8;
  • 25) 0.766 910 917 006 498 493 719 846 010 242 256 076 8 × 2 = 1 + 0.533 821 834 012 996 987 439 692 020 484 512 153 6;
  • 26) 0.533 821 834 012 996 987 439 692 020 484 512 153 6 × 2 = 1 + 0.067 643 668 025 993 974 879 384 040 969 024 307 2;
  • 27) 0.067 643 668 025 993 974 879 384 040 969 024 307 2 × 2 = 0 + 0.135 287 336 051 987 949 758 768 081 938 048 614 4;
  • 28) 0.135 287 336 051 987 949 758 768 081 938 048 614 4 × 2 = 0 + 0.270 574 672 103 975 899 517 536 163 876 097 228 8;
  • 29) 0.270 574 672 103 975 899 517 536 163 876 097 228 8 × 2 = 0 + 0.541 149 344 207 951 799 035 072 327 752 194 457 6;
  • 30) 0.541 149 344 207 951 799 035 072 327 752 194 457 6 × 2 = 1 + 0.082 298 688 415 903 598 070 144 655 504 388 915 2;
  • 31) 0.082 298 688 415 903 598 070 144 655 504 388 915 2 × 2 = 0 + 0.164 597 376 831 807 196 140 289 311 008 777 830 4;
  • 32) 0.164 597 376 831 807 196 140 289 311 008 777 830 4 × 2 = 0 + 0.329 194 753 663 614 392 280 578 622 017 555 660 8;
  • 33) 0.329 194 753 663 614 392 280 578 622 017 555 660 8 × 2 = 0 + 0.658 389 507 327 228 784 561 157 244 035 111 321 6;
  • 34) 0.658 389 507 327 228 784 561 157 244 035 111 321 6 × 2 = 1 + 0.316 779 014 654 457 569 122 314 488 070 222 643 2;
  • 35) 0.316 779 014 654 457 569 122 314 488 070 222 643 2 × 2 = 0 + 0.633 558 029 308 915 138 244 628 976 140 445 286 4;
  • 36) 0.633 558 029 308 915 138 244 628 976 140 445 286 4 × 2 = 1 + 0.267 116 058 617 830 276 489 257 952 280 890 572 8;
  • 37) 0.267 116 058 617 830 276 489 257 952 280 890 572 8 × 2 = 0 + 0.534 232 117 235 660 552 978 515 904 561 781 145 6;
  • 38) 0.534 232 117 235 660 552 978 515 904 561 781 145 6 × 2 = 1 + 0.068 464 234 471 321 105 957 031 809 123 562 291 2;
  • 39) 0.068 464 234 471 321 105 957 031 809 123 562 291 2 × 2 = 0 + 0.136 928 468 942 642 211 914 063 618 247 124 582 4;
  • 40) 0.136 928 468 942 642 211 914 063 618 247 124 582 4 × 2 = 0 + 0.273 856 937 885 284 423 828 127 236 494 249 164 8;
  • 41) 0.273 856 937 885 284 423 828 127 236 494 249 164 8 × 2 = 0 + 0.547 713 875 770 568 847 656 254 472 988 498 329 6;
  • 42) 0.547 713 875 770 568 847 656 254 472 988 498 329 6 × 2 = 1 + 0.095 427 751 541 137 695 312 508 945 976 996 659 2;
  • 43) 0.095 427 751 541 137 695 312 508 945 976 996 659 2 × 2 = 0 + 0.190 855 503 082 275 390 625 017 891 953 993 318 4;
  • 44) 0.190 855 503 082 275 390 625 017 891 953 993 318 4 × 2 = 0 + 0.381 711 006 164 550 781 250 035 783 907 986 636 8;
  • 45) 0.381 711 006 164 550 781 250 035 783 907 986 636 8 × 2 = 0 + 0.763 422 012 329 101 562 500 071 567 815 973 273 6;
  • 46) 0.763 422 012 329 101 562 500 071 567 815 973 273 6 × 2 = 1 + 0.526 844 024 658 203 125 000 143 135 631 946 547 2;
  • 47) 0.526 844 024 658 203 125 000 143 135 631 946 547 2 × 2 = 1 + 0.053 688 049 316 406 250 000 286 271 263 893 094 4;
  • 48) 0.053 688 049 316 406 250 000 286 271 263 893 094 4 × 2 = 0 + 0.107 376 098 632 812 500 000 572 542 527 786 188 8;
  • 49) 0.107 376 098 632 812 500 000 572 542 527 786 188 8 × 2 = 0 + 0.214 752 197 265 625 000 001 145 085 055 572 377 6;
  • 50) 0.214 752 197 265 625 000 001 145 085 055 572 377 6 × 2 = 0 + 0.429 504 394 531 250 000 002 290 170 111 144 755 2;
  • 51) 0.429 504 394 531 250 000 002 290 170 111 144 755 2 × 2 = 0 + 0.859 008 789 062 500 000 004 580 340 222 289 510 4;
  • 52) 0.859 008 789 062 500 000 004 580 340 222 289 510 4 × 2 = 1 + 0.718 017 578 125 000 000 009 160 680 444 579 020 8;
  • 53) 0.718 017 578 125 000 000 009 160 680 444 579 020 8 × 2 = 1 + 0.436 035 156 250 000 000 018 321 360 889 158 041 6;

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:


0.211 829 052 383 358 300 119 548 661 699 653 282 3(10) =

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

6. Positive number before normalization:

2.211 829 052 383 358 300 119 548 661 699 653 282 3(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:


2.211 829 052 383 358 300 119 548 661 699 653 282 3(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 -2.211 829 052 383 358 300 119 548 661 699 653 282 3 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