-2.211 829 052 383 358 300 119 548 661 699 762 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 300 119 548 661 699 762(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
-2.211 829 052 383 358 300 119 548 661 699 762(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:

|-2.211 829 052 383 358 300 119 548 661 699 762| = 2.211 829 052 383 358 300 119 548 661 699 762


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: 0.211 829 052 383 358 300 119 548 661 699 762.

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) 0.211 829 052 383 358 300 119 548 661 699 762 × 2 = 0 + 0.423 658 104 766 716 600 239 097 323 399 524;
  • 2) 0.423 658 104 766 716 600 239 097 323 399 524 × 2 = 0 + 0.847 316 209 533 433 200 478 194 646 799 048;
  • 3) 0.847 316 209 533 433 200 478 194 646 799 048 × 2 = 1 + 0.694 632 419 066 866 400 956 389 293 598 096;
  • 4) 0.694 632 419 066 866 400 956 389 293 598 096 × 2 = 1 + 0.389 264 838 133 732 801 912 778 587 196 192;
  • 5) 0.389 264 838 133 732 801 912 778 587 196 192 × 2 = 0 + 0.778 529 676 267 465 603 825 557 174 392 384;
  • 6) 0.778 529 676 267 465 603 825 557 174 392 384 × 2 = 1 + 0.557 059 352 534 931 207 651 114 348 784 768;
  • 7) 0.557 059 352 534 931 207 651 114 348 784 768 × 2 = 1 + 0.114 118 705 069 862 415 302 228 697 569 536;
  • 8) 0.114 118 705 069 862 415 302 228 697 569 536 × 2 = 0 + 0.228 237 410 139 724 830 604 457 395 139 072;
  • 9) 0.228 237 410 139 724 830 604 457 395 139 072 × 2 = 0 + 0.456 474 820 279 449 661 208 914 790 278 144;
  • 10) 0.456 474 820 279 449 661 208 914 790 278 144 × 2 = 0 + 0.912 949 640 558 899 322 417 829 580 556 288;
  • 11) 0.912 949 640 558 899 322 417 829 580 556 288 × 2 = 1 + 0.825 899 281 117 798 644 835 659 161 112 576;
  • 12) 0.825 899 281 117 798 644 835 659 161 112 576 × 2 = 1 + 0.651 798 562 235 597 289 671 318 322 225 152;
  • 13) 0.651 798 562 235 597 289 671 318 322 225 152 × 2 = 1 + 0.303 597 124 471 194 579 342 636 644 450 304;
  • 14) 0.303 597 124 471 194 579 342 636 644 450 304 × 2 = 0 + 0.607 194 248 942 389 158 685 273 288 900 608;
  • 15) 0.607 194 248 942 389 158 685 273 288 900 608 × 2 = 1 + 0.214 388 497 884 778 317 370 546 577 801 216;
  • 16) 0.214 388 497 884 778 317 370 546 577 801 216 × 2 = 0 + 0.428 776 995 769 556 634 741 093 155 602 432;
  • 17) 0.428 776 995 769 556 634 741 093 155 602 432 × 2 = 0 + 0.857 553 991 539 113 269 482 186 311 204 864;
  • 18) 0.857 553 991 539 113 269 482 186 311 204 864 × 2 = 1 + 0.715 107 983 078 226 538 964 372 622 409 728;
  • 19) 0.715 107 983 078 226 538 964 372 622 409 728 × 2 = 1 + 0.430 215 966 156 453 077 928 745 244 819 456;
  • 20) 0.430 215 966 156 453 077 928 745 244 819 456 × 2 = 0 + 0.860 431 932 312 906 155 857 490 489 638 912;
  • 21) 0.860 431 932 312 906 155 857 490 489 638 912 × 2 = 1 + 0.720 863 864 625 812 311 714 980 979 277 824;
  • 22) 0.720 863 864 625 812 311 714 980 979 277 824 × 2 = 1 + 0.441 727 729 251 624 623 429 961 958 555 648;
  • 23) 0.441 727 729 251 624 623 429 961 958 555 648 × 2 = 0 + 0.883 455 458 503 249 246 859 923 917 111 296;
  • 24) 0.883 455 458 503 249 246 859 923 917 111 296 × 2 = 1 + 0.766 910 917 006 498 493 719 847 834 222 592;
  • 25) 0.766 910 917 006 498 493 719 847 834 222 592 × 2 = 1 + 0.533 821 834 012 996 987 439 695 668 445 184;
  • 26) 0.533 821 834 012 996 987 439 695 668 445 184 × 2 = 1 + 0.067 643 668 025 993 974 879 391 336 890 368;
  • 27) 0.067 643 668 025 993 974 879 391 336 890 368 × 2 = 0 + 0.135 287 336 051 987 949 758 782 673 780 736;
  • 28) 0.135 287 336 051 987 949 758 782 673 780 736 × 2 = 0 + 0.270 574 672 103 975 899 517 565 347 561 472;
  • 29) 0.270 574 672 103 975 899 517 565 347 561 472 × 2 = 0 + 0.541 149 344 207 951 799 035 130 695 122 944;
  • 30) 0.541 149 344 207 951 799 035 130 695 122 944 × 2 = 1 + 0.082 298 688 415 903 598 070 261 390 245 888;
  • 31) 0.082 298 688 415 903 598 070 261 390 245 888 × 2 = 0 + 0.164 597 376 831 807 196 140 522 780 491 776;
  • 32) 0.164 597 376 831 807 196 140 522 780 491 776 × 2 = 0 + 0.329 194 753 663 614 392 281 045 560 983 552;
  • 33) 0.329 194 753 663 614 392 281 045 560 983 552 × 2 = 0 + 0.658 389 507 327 228 784 562 091 121 967 104;
  • 34) 0.658 389 507 327 228 784 562 091 121 967 104 × 2 = 1 + 0.316 779 014 654 457 569 124 182 243 934 208;
  • 35) 0.316 779 014 654 457 569 124 182 243 934 208 × 2 = 0 + 0.633 558 029 308 915 138 248 364 487 868 416;
  • 36) 0.633 558 029 308 915 138 248 364 487 868 416 × 2 = 1 + 0.267 116 058 617 830 276 496 728 975 736 832;
  • 37) 0.267 116 058 617 830 276 496 728 975 736 832 × 2 = 0 + 0.534 232 117 235 660 552 993 457 951 473 664;
  • 38) 0.534 232 117 235 660 552 993 457 951 473 664 × 2 = 1 + 0.068 464 234 471 321 105 986 915 902 947 328;
  • 39) 0.068 464 234 471 321 105 986 915 902 947 328 × 2 = 0 + 0.136 928 468 942 642 211 973 831 805 894 656;
  • 40) 0.136 928 468 942 642 211 973 831 805 894 656 × 2 = 0 + 0.273 856 937 885 284 423 947 663 611 789 312;
  • 41) 0.273 856 937 885 284 423 947 663 611 789 312 × 2 = 0 + 0.547 713 875 770 568 847 895 327 223 578 624;
  • 42) 0.547 713 875 770 568 847 895 327 223 578 624 × 2 = 1 + 0.095 427 751 541 137 695 790 654 447 157 248;
  • 43) 0.095 427 751 541 137 695 790 654 447 157 248 × 2 = 0 + 0.190 855 503 082 275 391 581 308 894 314 496;
  • 44) 0.190 855 503 082 275 391 581 308 894 314 496 × 2 = 0 + 0.381 711 006 164 550 783 162 617 788 628 992;
  • 45) 0.381 711 006 164 550 783 162 617 788 628 992 × 2 = 0 + 0.763 422 012 329 101 566 325 235 577 257 984;
  • 46) 0.763 422 012 329 101 566 325 235 577 257 984 × 2 = 1 + 0.526 844 024 658 203 132 650 471 154 515 968;
  • 47) 0.526 844 024 658 203 132 650 471 154 515 968 × 2 = 1 + 0.053 688 049 316 406 265 300 942 309 031 936;
  • 48) 0.053 688 049 316 406 265 300 942 309 031 936 × 2 = 0 + 0.107 376 098 632 812 530 601 884 618 063 872;
  • 49) 0.107 376 098 632 812 530 601 884 618 063 872 × 2 = 0 + 0.214 752 197 265 625 061 203 769 236 127 744;
  • 50) 0.214 752 197 265 625 061 203 769 236 127 744 × 2 = 0 + 0.429 504 394 531 250 122 407 538 472 255 488;
  • 51) 0.429 504 394 531 250 122 407 538 472 255 488 × 2 = 0 + 0.859 008 789 062 500 244 815 076 944 510 976;
  • 52) 0.859 008 789 062 500 244 815 076 944 510 976 × 2 = 1 + 0.718 017 578 125 000 489 630 153 889 021 952;
  • 53) 0.718 017 578 125 000 489 630 153 889 021 952 × 2 = 1 + 0.436 035 156 250 000 979 260 307 778 043 904;

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:


0.211 829 052 383 358 300 119 548 661 699 762(10) =

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

6. Positive number before normalization:

2.211 829 052 383 358 300 119 548 661 699 762(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:


2.211 829 052 383 358 300 119 548 661 699 762(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 -2.211 829 052 383 358 300 119 548 661 699 762 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