-2.211 829 052 383 358 300 119 548 661 699 571 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 300 119 548 661 699 571(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
-2.211 829 052 383 358 300 119 548 661 699 571(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:

|-2.211 829 052 383 358 300 119 548 661 699 571| = 2.211 829 052 383 358 300 119 548 661 699 571


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: 0.211 829 052 383 358 300 119 548 661 699 571.

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) 0.211 829 052 383 358 300 119 548 661 699 571 × 2 = 0 + 0.423 658 104 766 716 600 239 097 323 399 142;
  • 2) 0.423 658 104 766 716 600 239 097 323 399 142 × 2 = 0 + 0.847 316 209 533 433 200 478 194 646 798 284;
  • 3) 0.847 316 209 533 433 200 478 194 646 798 284 × 2 = 1 + 0.694 632 419 066 866 400 956 389 293 596 568;
  • 4) 0.694 632 419 066 866 400 956 389 293 596 568 × 2 = 1 + 0.389 264 838 133 732 801 912 778 587 193 136;
  • 5) 0.389 264 838 133 732 801 912 778 587 193 136 × 2 = 0 + 0.778 529 676 267 465 603 825 557 174 386 272;
  • 6) 0.778 529 676 267 465 603 825 557 174 386 272 × 2 = 1 + 0.557 059 352 534 931 207 651 114 348 772 544;
  • 7) 0.557 059 352 534 931 207 651 114 348 772 544 × 2 = 1 + 0.114 118 705 069 862 415 302 228 697 545 088;
  • 8) 0.114 118 705 069 862 415 302 228 697 545 088 × 2 = 0 + 0.228 237 410 139 724 830 604 457 395 090 176;
  • 9) 0.228 237 410 139 724 830 604 457 395 090 176 × 2 = 0 + 0.456 474 820 279 449 661 208 914 790 180 352;
  • 10) 0.456 474 820 279 449 661 208 914 790 180 352 × 2 = 0 + 0.912 949 640 558 899 322 417 829 580 360 704;
  • 11) 0.912 949 640 558 899 322 417 829 580 360 704 × 2 = 1 + 0.825 899 281 117 798 644 835 659 160 721 408;
  • 12) 0.825 899 281 117 798 644 835 659 160 721 408 × 2 = 1 + 0.651 798 562 235 597 289 671 318 321 442 816;
  • 13) 0.651 798 562 235 597 289 671 318 321 442 816 × 2 = 1 + 0.303 597 124 471 194 579 342 636 642 885 632;
  • 14) 0.303 597 124 471 194 579 342 636 642 885 632 × 2 = 0 + 0.607 194 248 942 389 158 685 273 285 771 264;
  • 15) 0.607 194 248 942 389 158 685 273 285 771 264 × 2 = 1 + 0.214 388 497 884 778 317 370 546 571 542 528;
  • 16) 0.214 388 497 884 778 317 370 546 571 542 528 × 2 = 0 + 0.428 776 995 769 556 634 741 093 143 085 056;
  • 17) 0.428 776 995 769 556 634 741 093 143 085 056 × 2 = 0 + 0.857 553 991 539 113 269 482 186 286 170 112;
  • 18) 0.857 553 991 539 113 269 482 186 286 170 112 × 2 = 1 + 0.715 107 983 078 226 538 964 372 572 340 224;
  • 19) 0.715 107 983 078 226 538 964 372 572 340 224 × 2 = 1 + 0.430 215 966 156 453 077 928 745 144 680 448;
  • 20) 0.430 215 966 156 453 077 928 745 144 680 448 × 2 = 0 + 0.860 431 932 312 906 155 857 490 289 360 896;
  • 21) 0.860 431 932 312 906 155 857 490 289 360 896 × 2 = 1 + 0.720 863 864 625 812 311 714 980 578 721 792;
  • 22) 0.720 863 864 625 812 311 714 980 578 721 792 × 2 = 1 + 0.441 727 729 251 624 623 429 961 157 443 584;
  • 23) 0.441 727 729 251 624 623 429 961 157 443 584 × 2 = 0 + 0.883 455 458 503 249 246 859 922 314 887 168;
  • 24) 0.883 455 458 503 249 246 859 922 314 887 168 × 2 = 1 + 0.766 910 917 006 498 493 719 844 629 774 336;
  • 25) 0.766 910 917 006 498 493 719 844 629 774 336 × 2 = 1 + 0.533 821 834 012 996 987 439 689 259 548 672;
  • 26) 0.533 821 834 012 996 987 439 689 259 548 672 × 2 = 1 + 0.067 643 668 025 993 974 879 378 519 097 344;
  • 27) 0.067 643 668 025 993 974 879 378 519 097 344 × 2 = 0 + 0.135 287 336 051 987 949 758 757 038 194 688;
  • 28) 0.135 287 336 051 987 949 758 757 038 194 688 × 2 = 0 + 0.270 574 672 103 975 899 517 514 076 389 376;
  • 29) 0.270 574 672 103 975 899 517 514 076 389 376 × 2 = 0 + 0.541 149 344 207 951 799 035 028 152 778 752;
  • 30) 0.541 149 344 207 951 799 035 028 152 778 752 × 2 = 1 + 0.082 298 688 415 903 598 070 056 305 557 504;
  • 31) 0.082 298 688 415 903 598 070 056 305 557 504 × 2 = 0 + 0.164 597 376 831 807 196 140 112 611 115 008;
  • 32) 0.164 597 376 831 807 196 140 112 611 115 008 × 2 = 0 + 0.329 194 753 663 614 392 280 225 222 230 016;
  • 33) 0.329 194 753 663 614 392 280 225 222 230 016 × 2 = 0 + 0.658 389 507 327 228 784 560 450 444 460 032;
  • 34) 0.658 389 507 327 228 784 560 450 444 460 032 × 2 = 1 + 0.316 779 014 654 457 569 120 900 888 920 064;
  • 35) 0.316 779 014 654 457 569 120 900 888 920 064 × 2 = 0 + 0.633 558 029 308 915 138 241 801 777 840 128;
  • 36) 0.633 558 029 308 915 138 241 801 777 840 128 × 2 = 1 + 0.267 116 058 617 830 276 483 603 555 680 256;
  • 37) 0.267 116 058 617 830 276 483 603 555 680 256 × 2 = 0 + 0.534 232 117 235 660 552 967 207 111 360 512;
  • 38) 0.534 232 117 235 660 552 967 207 111 360 512 × 2 = 1 + 0.068 464 234 471 321 105 934 414 222 721 024;
  • 39) 0.068 464 234 471 321 105 934 414 222 721 024 × 2 = 0 + 0.136 928 468 942 642 211 868 828 445 442 048;
  • 40) 0.136 928 468 942 642 211 868 828 445 442 048 × 2 = 0 + 0.273 856 937 885 284 423 737 656 890 884 096;
  • 41) 0.273 856 937 885 284 423 737 656 890 884 096 × 2 = 0 + 0.547 713 875 770 568 847 475 313 781 768 192;
  • 42) 0.547 713 875 770 568 847 475 313 781 768 192 × 2 = 1 + 0.095 427 751 541 137 694 950 627 563 536 384;
  • 43) 0.095 427 751 541 137 694 950 627 563 536 384 × 2 = 0 + 0.190 855 503 082 275 389 901 255 127 072 768;
  • 44) 0.190 855 503 082 275 389 901 255 127 072 768 × 2 = 0 + 0.381 711 006 164 550 779 802 510 254 145 536;
  • 45) 0.381 711 006 164 550 779 802 510 254 145 536 × 2 = 0 + 0.763 422 012 329 101 559 605 020 508 291 072;
  • 46) 0.763 422 012 329 101 559 605 020 508 291 072 × 2 = 1 + 0.526 844 024 658 203 119 210 041 016 582 144;
  • 47) 0.526 844 024 658 203 119 210 041 016 582 144 × 2 = 1 + 0.053 688 049 316 406 238 420 082 033 164 288;
  • 48) 0.053 688 049 316 406 238 420 082 033 164 288 × 2 = 0 + 0.107 376 098 632 812 476 840 164 066 328 576;
  • 49) 0.107 376 098 632 812 476 840 164 066 328 576 × 2 = 0 + 0.214 752 197 265 624 953 680 328 132 657 152;
  • 50) 0.214 752 197 265 624 953 680 328 132 657 152 × 2 = 0 + 0.429 504 394 531 249 907 360 656 265 314 304;
  • 51) 0.429 504 394 531 249 907 360 656 265 314 304 × 2 = 0 + 0.859 008 789 062 499 814 721 312 530 628 608;
  • 52) 0.859 008 789 062 499 814 721 312 530 628 608 × 2 = 1 + 0.718 017 578 124 999 629 442 625 061 257 216;
  • 53) 0.718 017 578 124 999 629 442 625 061 257 216 × 2 = 1 + 0.436 035 156 249 999 258 885 250 122 514 432;

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:


0.211 829 052 383 358 300 119 548 661 699 571(10) =

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

6. Positive number before normalization:

2.211 829 052 383 358 300 119 548 661 699 571(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:


2.211 829 052 383 358 300 119 548 661 699 571(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 -2.211 829 052 383 358 300 119 548 661 699 571 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