-2.211 829 052 383 358 300 119 548 661 699 761 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 300 119 548 661 699 761(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
-2.211 829 052 383 358 300 119 548 661 699 761(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:

|-2.211 829 052 383 358 300 119 548 661 699 761| = 2.211 829 052 383 358 300 119 548 661 699 761


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: 0.211 829 052 383 358 300 119 548 661 699 761.

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) 0.211 829 052 383 358 300 119 548 661 699 761 × 2 = 0 + 0.423 658 104 766 716 600 239 097 323 399 522;
  • 2) 0.423 658 104 766 716 600 239 097 323 399 522 × 2 = 0 + 0.847 316 209 533 433 200 478 194 646 799 044;
  • 3) 0.847 316 209 533 433 200 478 194 646 799 044 × 2 = 1 + 0.694 632 419 066 866 400 956 389 293 598 088;
  • 4) 0.694 632 419 066 866 400 956 389 293 598 088 × 2 = 1 + 0.389 264 838 133 732 801 912 778 587 196 176;
  • 5) 0.389 264 838 133 732 801 912 778 587 196 176 × 2 = 0 + 0.778 529 676 267 465 603 825 557 174 392 352;
  • 6) 0.778 529 676 267 465 603 825 557 174 392 352 × 2 = 1 + 0.557 059 352 534 931 207 651 114 348 784 704;
  • 7) 0.557 059 352 534 931 207 651 114 348 784 704 × 2 = 1 + 0.114 118 705 069 862 415 302 228 697 569 408;
  • 8) 0.114 118 705 069 862 415 302 228 697 569 408 × 2 = 0 + 0.228 237 410 139 724 830 604 457 395 138 816;
  • 9) 0.228 237 410 139 724 830 604 457 395 138 816 × 2 = 0 + 0.456 474 820 279 449 661 208 914 790 277 632;
  • 10) 0.456 474 820 279 449 661 208 914 790 277 632 × 2 = 0 + 0.912 949 640 558 899 322 417 829 580 555 264;
  • 11) 0.912 949 640 558 899 322 417 829 580 555 264 × 2 = 1 + 0.825 899 281 117 798 644 835 659 161 110 528;
  • 12) 0.825 899 281 117 798 644 835 659 161 110 528 × 2 = 1 + 0.651 798 562 235 597 289 671 318 322 221 056;
  • 13) 0.651 798 562 235 597 289 671 318 322 221 056 × 2 = 1 + 0.303 597 124 471 194 579 342 636 644 442 112;
  • 14) 0.303 597 124 471 194 579 342 636 644 442 112 × 2 = 0 + 0.607 194 248 942 389 158 685 273 288 884 224;
  • 15) 0.607 194 248 942 389 158 685 273 288 884 224 × 2 = 1 + 0.214 388 497 884 778 317 370 546 577 768 448;
  • 16) 0.214 388 497 884 778 317 370 546 577 768 448 × 2 = 0 + 0.428 776 995 769 556 634 741 093 155 536 896;
  • 17) 0.428 776 995 769 556 634 741 093 155 536 896 × 2 = 0 + 0.857 553 991 539 113 269 482 186 311 073 792;
  • 18) 0.857 553 991 539 113 269 482 186 311 073 792 × 2 = 1 + 0.715 107 983 078 226 538 964 372 622 147 584;
  • 19) 0.715 107 983 078 226 538 964 372 622 147 584 × 2 = 1 + 0.430 215 966 156 453 077 928 745 244 295 168;
  • 20) 0.430 215 966 156 453 077 928 745 244 295 168 × 2 = 0 + 0.860 431 932 312 906 155 857 490 488 590 336;
  • 21) 0.860 431 932 312 906 155 857 490 488 590 336 × 2 = 1 + 0.720 863 864 625 812 311 714 980 977 180 672;
  • 22) 0.720 863 864 625 812 311 714 980 977 180 672 × 2 = 1 + 0.441 727 729 251 624 623 429 961 954 361 344;
  • 23) 0.441 727 729 251 624 623 429 961 954 361 344 × 2 = 0 + 0.883 455 458 503 249 246 859 923 908 722 688;
  • 24) 0.883 455 458 503 249 246 859 923 908 722 688 × 2 = 1 + 0.766 910 917 006 498 493 719 847 817 445 376;
  • 25) 0.766 910 917 006 498 493 719 847 817 445 376 × 2 = 1 + 0.533 821 834 012 996 987 439 695 634 890 752;
  • 26) 0.533 821 834 012 996 987 439 695 634 890 752 × 2 = 1 + 0.067 643 668 025 993 974 879 391 269 781 504;
  • 27) 0.067 643 668 025 993 974 879 391 269 781 504 × 2 = 0 + 0.135 287 336 051 987 949 758 782 539 563 008;
  • 28) 0.135 287 336 051 987 949 758 782 539 563 008 × 2 = 0 + 0.270 574 672 103 975 899 517 565 079 126 016;
  • 29) 0.270 574 672 103 975 899 517 565 079 126 016 × 2 = 0 + 0.541 149 344 207 951 799 035 130 158 252 032;
  • 30) 0.541 149 344 207 951 799 035 130 158 252 032 × 2 = 1 + 0.082 298 688 415 903 598 070 260 316 504 064;
  • 31) 0.082 298 688 415 903 598 070 260 316 504 064 × 2 = 0 + 0.164 597 376 831 807 196 140 520 633 008 128;
  • 32) 0.164 597 376 831 807 196 140 520 633 008 128 × 2 = 0 + 0.329 194 753 663 614 392 281 041 266 016 256;
  • 33) 0.329 194 753 663 614 392 281 041 266 016 256 × 2 = 0 + 0.658 389 507 327 228 784 562 082 532 032 512;
  • 34) 0.658 389 507 327 228 784 562 082 532 032 512 × 2 = 1 + 0.316 779 014 654 457 569 124 165 064 065 024;
  • 35) 0.316 779 014 654 457 569 124 165 064 065 024 × 2 = 0 + 0.633 558 029 308 915 138 248 330 128 130 048;
  • 36) 0.633 558 029 308 915 138 248 330 128 130 048 × 2 = 1 + 0.267 116 058 617 830 276 496 660 256 260 096;
  • 37) 0.267 116 058 617 830 276 496 660 256 260 096 × 2 = 0 + 0.534 232 117 235 660 552 993 320 512 520 192;
  • 38) 0.534 232 117 235 660 552 993 320 512 520 192 × 2 = 1 + 0.068 464 234 471 321 105 986 641 025 040 384;
  • 39) 0.068 464 234 471 321 105 986 641 025 040 384 × 2 = 0 + 0.136 928 468 942 642 211 973 282 050 080 768;
  • 40) 0.136 928 468 942 642 211 973 282 050 080 768 × 2 = 0 + 0.273 856 937 885 284 423 946 564 100 161 536;
  • 41) 0.273 856 937 885 284 423 946 564 100 161 536 × 2 = 0 + 0.547 713 875 770 568 847 893 128 200 323 072;
  • 42) 0.547 713 875 770 568 847 893 128 200 323 072 × 2 = 1 + 0.095 427 751 541 137 695 786 256 400 646 144;
  • 43) 0.095 427 751 541 137 695 786 256 400 646 144 × 2 = 0 + 0.190 855 503 082 275 391 572 512 801 292 288;
  • 44) 0.190 855 503 082 275 391 572 512 801 292 288 × 2 = 0 + 0.381 711 006 164 550 783 145 025 602 584 576;
  • 45) 0.381 711 006 164 550 783 145 025 602 584 576 × 2 = 0 + 0.763 422 012 329 101 566 290 051 205 169 152;
  • 46) 0.763 422 012 329 101 566 290 051 205 169 152 × 2 = 1 + 0.526 844 024 658 203 132 580 102 410 338 304;
  • 47) 0.526 844 024 658 203 132 580 102 410 338 304 × 2 = 1 + 0.053 688 049 316 406 265 160 204 820 676 608;
  • 48) 0.053 688 049 316 406 265 160 204 820 676 608 × 2 = 0 + 0.107 376 098 632 812 530 320 409 641 353 216;
  • 49) 0.107 376 098 632 812 530 320 409 641 353 216 × 2 = 0 + 0.214 752 197 265 625 060 640 819 282 706 432;
  • 50) 0.214 752 197 265 625 060 640 819 282 706 432 × 2 = 0 + 0.429 504 394 531 250 121 281 638 565 412 864;
  • 51) 0.429 504 394 531 250 121 281 638 565 412 864 × 2 = 0 + 0.859 008 789 062 500 242 563 277 130 825 728;
  • 52) 0.859 008 789 062 500 242 563 277 130 825 728 × 2 = 1 + 0.718 017 578 125 000 485 126 554 261 651 456;
  • 53) 0.718 017 578 125 000 485 126 554 261 651 456 × 2 = 1 + 0.436 035 156 250 000 970 253 108 523 302 912;

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:


0.211 829 052 383 358 300 119 548 661 699 761(10) =

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

6. Positive number before normalization:

2.211 829 052 383 358 300 119 548 661 699 761(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:


2.211 829 052 383 358 300 119 548 661 699 761(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 -2.211 829 052 383 358 300 119 548 661 699 761 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