-2.211 829 052 383 358 300 119 548 661 699 653 276 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 300 119 548 661 699 653 276 2(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
-2.211 829 052 383 358 300 119 548 661 699 653 276 2(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:

|-2.211 829 052 383 358 300 119 548 661 699 653 276 2| = 2.211 829 052 383 358 300 119 548 661 699 653 276 2


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: 0.211 829 052 383 358 300 119 548 661 699 653 276 2.

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) 0.211 829 052 383 358 300 119 548 661 699 653 276 2 × 2 = 0 + 0.423 658 104 766 716 600 239 097 323 399 306 552 4;
  • 2) 0.423 658 104 766 716 600 239 097 323 399 306 552 4 × 2 = 0 + 0.847 316 209 533 433 200 478 194 646 798 613 104 8;
  • 3) 0.847 316 209 533 433 200 478 194 646 798 613 104 8 × 2 = 1 + 0.694 632 419 066 866 400 956 389 293 597 226 209 6;
  • 4) 0.694 632 419 066 866 400 956 389 293 597 226 209 6 × 2 = 1 + 0.389 264 838 133 732 801 912 778 587 194 452 419 2;
  • 5) 0.389 264 838 133 732 801 912 778 587 194 452 419 2 × 2 = 0 + 0.778 529 676 267 465 603 825 557 174 388 904 838 4;
  • 6) 0.778 529 676 267 465 603 825 557 174 388 904 838 4 × 2 = 1 + 0.557 059 352 534 931 207 651 114 348 777 809 676 8;
  • 7) 0.557 059 352 534 931 207 651 114 348 777 809 676 8 × 2 = 1 + 0.114 118 705 069 862 415 302 228 697 555 619 353 6;
  • 8) 0.114 118 705 069 862 415 302 228 697 555 619 353 6 × 2 = 0 + 0.228 237 410 139 724 830 604 457 395 111 238 707 2;
  • 9) 0.228 237 410 139 724 830 604 457 395 111 238 707 2 × 2 = 0 + 0.456 474 820 279 449 661 208 914 790 222 477 414 4;
  • 10) 0.456 474 820 279 449 661 208 914 790 222 477 414 4 × 2 = 0 + 0.912 949 640 558 899 322 417 829 580 444 954 828 8;
  • 11) 0.912 949 640 558 899 322 417 829 580 444 954 828 8 × 2 = 1 + 0.825 899 281 117 798 644 835 659 160 889 909 657 6;
  • 12) 0.825 899 281 117 798 644 835 659 160 889 909 657 6 × 2 = 1 + 0.651 798 562 235 597 289 671 318 321 779 819 315 2;
  • 13) 0.651 798 562 235 597 289 671 318 321 779 819 315 2 × 2 = 1 + 0.303 597 124 471 194 579 342 636 643 559 638 630 4;
  • 14) 0.303 597 124 471 194 579 342 636 643 559 638 630 4 × 2 = 0 + 0.607 194 248 942 389 158 685 273 287 119 277 260 8;
  • 15) 0.607 194 248 942 389 158 685 273 287 119 277 260 8 × 2 = 1 + 0.214 388 497 884 778 317 370 546 574 238 554 521 6;
  • 16) 0.214 388 497 884 778 317 370 546 574 238 554 521 6 × 2 = 0 + 0.428 776 995 769 556 634 741 093 148 477 109 043 2;
  • 17) 0.428 776 995 769 556 634 741 093 148 477 109 043 2 × 2 = 0 + 0.857 553 991 539 113 269 482 186 296 954 218 086 4;
  • 18) 0.857 553 991 539 113 269 482 186 296 954 218 086 4 × 2 = 1 + 0.715 107 983 078 226 538 964 372 593 908 436 172 8;
  • 19) 0.715 107 983 078 226 538 964 372 593 908 436 172 8 × 2 = 1 + 0.430 215 966 156 453 077 928 745 187 816 872 345 6;
  • 20) 0.430 215 966 156 453 077 928 745 187 816 872 345 6 × 2 = 0 + 0.860 431 932 312 906 155 857 490 375 633 744 691 2;
  • 21) 0.860 431 932 312 906 155 857 490 375 633 744 691 2 × 2 = 1 + 0.720 863 864 625 812 311 714 980 751 267 489 382 4;
  • 22) 0.720 863 864 625 812 311 714 980 751 267 489 382 4 × 2 = 1 + 0.441 727 729 251 624 623 429 961 502 534 978 764 8;
  • 23) 0.441 727 729 251 624 623 429 961 502 534 978 764 8 × 2 = 0 + 0.883 455 458 503 249 246 859 923 005 069 957 529 6;
  • 24) 0.883 455 458 503 249 246 859 923 005 069 957 529 6 × 2 = 1 + 0.766 910 917 006 498 493 719 846 010 139 915 059 2;
  • 25) 0.766 910 917 006 498 493 719 846 010 139 915 059 2 × 2 = 1 + 0.533 821 834 012 996 987 439 692 020 279 830 118 4;
  • 26) 0.533 821 834 012 996 987 439 692 020 279 830 118 4 × 2 = 1 + 0.067 643 668 025 993 974 879 384 040 559 660 236 8;
  • 27) 0.067 643 668 025 993 974 879 384 040 559 660 236 8 × 2 = 0 + 0.135 287 336 051 987 949 758 768 081 119 320 473 6;
  • 28) 0.135 287 336 051 987 949 758 768 081 119 320 473 6 × 2 = 0 + 0.270 574 672 103 975 899 517 536 162 238 640 947 2;
  • 29) 0.270 574 672 103 975 899 517 536 162 238 640 947 2 × 2 = 0 + 0.541 149 344 207 951 799 035 072 324 477 281 894 4;
  • 30) 0.541 149 344 207 951 799 035 072 324 477 281 894 4 × 2 = 1 + 0.082 298 688 415 903 598 070 144 648 954 563 788 8;
  • 31) 0.082 298 688 415 903 598 070 144 648 954 563 788 8 × 2 = 0 + 0.164 597 376 831 807 196 140 289 297 909 127 577 6;
  • 32) 0.164 597 376 831 807 196 140 289 297 909 127 577 6 × 2 = 0 + 0.329 194 753 663 614 392 280 578 595 818 255 155 2;
  • 33) 0.329 194 753 663 614 392 280 578 595 818 255 155 2 × 2 = 0 + 0.658 389 507 327 228 784 561 157 191 636 510 310 4;
  • 34) 0.658 389 507 327 228 784 561 157 191 636 510 310 4 × 2 = 1 + 0.316 779 014 654 457 569 122 314 383 273 020 620 8;
  • 35) 0.316 779 014 654 457 569 122 314 383 273 020 620 8 × 2 = 0 + 0.633 558 029 308 915 138 244 628 766 546 041 241 6;
  • 36) 0.633 558 029 308 915 138 244 628 766 546 041 241 6 × 2 = 1 + 0.267 116 058 617 830 276 489 257 533 092 082 483 2;
  • 37) 0.267 116 058 617 830 276 489 257 533 092 082 483 2 × 2 = 0 + 0.534 232 117 235 660 552 978 515 066 184 164 966 4;
  • 38) 0.534 232 117 235 660 552 978 515 066 184 164 966 4 × 2 = 1 + 0.068 464 234 471 321 105 957 030 132 368 329 932 8;
  • 39) 0.068 464 234 471 321 105 957 030 132 368 329 932 8 × 2 = 0 + 0.136 928 468 942 642 211 914 060 264 736 659 865 6;
  • 40) 0.136 928 468 942 642 211 914 060 264 736 659 865 6 × 2 = 0 + 0.273 856 937 885 284 423 828 120 529 473 319 731 2;
  • 41) 0.273 856 937 885 284 423 828 120 529 473 319 731 2 × 2 = 0 + 0.547 713 875 770 568 847 656 241 058 946 639 462 4;
  • 42) 0.547 713 875 770 568 847 656 241 058 946 639 462 4 × 2 = 1 + 0.095 427 751 541 137 695 312 482 117 893 278 924 8;
  • 43) 0.095 427 751 541 137 695 312 482 117 893 278 924 8 × 2 = 0 + 0.190 855 503 082 275 390 624 964 235 786 557 849 6;
  • 44) 0.190 855 503 082 275 390 624 964 235 786 557 849 6 × 2 = 0 + 0.381 711 006 164 550 781 249 928 471 573 115 699 2;
  • 45) 0.381 711 006 164 550 781 249 928 471 573 115 699 2 × 2 = 0 + 0.763 422 012 329 101 562 499 856 943 146 231 398 4;
  • 46) 0.763 422 012 329 101 562 499 856 943 146 231 398 4 × 2 = 1 + 0.526 844 024 658 203 124 999 713 886 292 462 796 8;
  • 47) 0.526 844 024 658 203 124 999 713 886 292 462 796 8 × 2 = 1 + 0.053 688 049 316 406 249 999 427 772 584 925 593 6;
  • 48) 0.053 688 049 316 406 249 999 427 772 584 925 593 6 × 2 = 0 + 0.107 376 098 632 812 499 998 855 545 169 851 187 2;
  • 49) 0.107 376 098 632 812 499 998 855 545 169 851 187 2 × 2 = 0 + 0.214 752 197 265 624 999 997 711 090 339 702 374 4;
  • 50) 0.214 752 197 265 624 999 997 711 090 339 702 374 4 × 2 = 0 + 0.429 504 394 531 249 999 995 422 180 679 404 748 8;
  • 51) 0.429 504 394 531 249 999 995 422 180 679 404 748 8 × 2 = 0 + 0.859 008 789 062 499 999 990 844 361 358 809 497 6;
  • 52) 0.859 008 789 062 499 999 990 844 361 358 809 497 6 × 2 = 1 + 0.718 017 578 124 999 999 981 688 722 717 618 995 2;
  • 53) 0.718 017 578 124 999 999 981 688 722 717 618 995 2 × 2 = 1 + 0.436 035 156 249 999 999 963 377 445 435 237 990 4;

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:


0.211 829 052 383 358 300 119 548 661 699 653 276 2(10) =

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

6. Positive number before normalization:

2.211 829 052 383 358 300 119 548 661 699 653 276 2(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:


2.211 829 052 383 358 300 119 548 661 699 653 276 2(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 -2.211 829 052 383 358 300 119 548 661 699 653 276 2 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