-2.211 829 052 383 358 300 119 548 662 51 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 300 119 548 662 51(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
-2.211 829 052 383 358 300 119 548 662 51(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:

|-2.211 829 052 383 358 300 119 548 662 51| = 2.211 829 052 383 358 300 119 548 662 51


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: 0.211 829 052 383 358 300 119 548 662 51.

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) 0.211 829 052 383 358 300 119 548 662 51 × 2 = 0 + 0.423 658 104 766 716 600 239 097 325 02;
  • 2) 0.423 658 104 766 716 600 239 097 325 02 × 2 = 0 + 0.847 316 209 533 433 200 478 194 650 04;
  • 3) 0.847 316 209 533 433 200 478 194 650 04 × 2 = 1 + 0.694 632 419 066 866 400 956 389 300 08;
  • 4) 0.694 632 419 066 866 400 956 389 300 08 × 2 = 1 + 0.389 264 838 133 732 801 912 778 600 16;
  • 5) 0.389 264 838 133 732 801 912 778 600 16 × 2 = 0 + 0.778 529 676 267 465 603 825 557 200 32;
  • 6) 0.778 529 676 267 465 603 825 557 200 32 × 2 = 1 + 0.557 059 352 534 931 207 651 114 400 64;
  • 7) 0.557 059 352 534 931 207 651 114 400 64 × 2 = 1 + 0.114 118 705 069 862 415 302 228 801 28;
  • 8) 0.114 118 705 069 862 415 302 228 801 28 × 2 = 0 + 0.228 237 410 139 724 830 604 457 602 56;
  • 9) 0.228 237 410 139 724 830 604 457 602 56 × 2 = 0 + 0.456 474 820 279 449 661 208 915 205 12;
  • 10) 0.456 474 820 279 449 661 208 915 205 12 × 2 = 0 + 0.912 949 640 558 899 322 417 830 410 24;
  • 11) 0.912 949 640 558 899 322 417 830 410 24 × 2 = 1 + 0.825 899 281 117 798 644 835 660 820 48;
  • 12) 0.825 899 281 117 798 644 835 660 820 48 × 2 = 1 + 0.651 798 562 235 597 289 671 321 640 96;
  • 13) 0.651 798 562 235 597 289 671 321 640 96 × 2 = 1 + 0.303 597 124 471 194 579 342 643 281 92;
  • 14) 0.303 597 124 471 194 579 342 643 281 92 × 2 = 0 + 0.607 194 248 942 389 158 685 286 563 84;
  • 15) 0.607 194 248 942 389 158 685 286 563 84 × 2 = 1 + 0.214 388 497 884 778 317 370 573 127 68;
  • 16) 0.214 388 497 884 778 317 370 573 127 68 × 2 = 0 + 0.428 776 995 769 556 634 741 146 255 36;
  • 17) 0.428 776 995 769 556 634 741 146 255 36 × 2 = 0 + 0.857 553 991 539 113 269 482 292 510 72;
  • 18) 0.857 553 991 539 113 269 482 292 510 72 × 2 = 1 + 0.715 107 983 078 226 538 964 585 021 44;
  • 19) 0.715 107 983 078 226 538 964 585 021 44 × 2 = 1 + 0.430 215 966 156 453 077 929 170 042 88;
  • 20) 0.430 215 966 156 453 077 929 170 042 88 × 2 = 0 + 0.860 431 932 312 906 155 858 340 085 76;
  • 21) 0.860 431 932 312 906 155 858 340 085 76 × 2 = 1 + 0.720 863 864 625 812 311 716 680 171 52;
  • 22) 0.720 863 864 625 812 311 716 680 171 52 × 2 = 1 + 0.441 727 729 251 624 623 433 360 343 04;
  • 23) 0.441 727 729 251 624 623 433 360 343 04 × 2 = 0 + 0.883 455 458 503 249 246 866 720 686 08;
  • 24) 0.883 455 458 503 249 246 866 720 686 08 × 2 = 1 + 0.766 910 917 006 498 493 733 441 372 16;
  • 25) 0.766 910 917 006 498 493 733 441 372 16 × 2 = 1 + 0.533 821 834 012 996 987 466 882 744 32;
  • 26) 0.533 821 834 012 996 987 466 882 744 32 × 2 = 1 + 0.067 643 668 025 993 974 933 765 488 64;
  • 27) 0.067 643 668 025 993 974 933 765 488 64 × 2 = 0 + 0.135 287 336 051 987 949 867 530 977 28;
  • 28) 0.135 287 336 051 987 949 867 530 977 28 × 2 = 0 + 0.270 574 672 103 975 899 735 061 954 56;
  • 29) 0.270 574 672 103 975 899 735 061 954 56 × 2 = 0 + 0.541 149 344 207 951 799 470 123 909 12;
  • 30) 0.541 149 344 207 951 799 470 123 909 12 × 2 = 1 + 0.082 298 688 415 903 598 940 247 818 24;
  • 31) 0.082 298 688 415 903 598 940 247 818 24 × 2 = 0 + 0.164 597 376 831 807 197 880 495 636 48;
  • 32) 0.164 597 376 831 807 197 880 495 636 48 × 2 = 0 + 0.329 194 753 663 614 395 760 991 272 96;
  • 33) 0.329 194 753 663 614 395 760 991 272 96 × 2 = 0 + 0.658 389 507 327 228 791 521 982 545 92;
  • 34) 0.658 389 507 327 228 791 521 982 545 92 × 2 = 1 + 0.316 779 014 654 457 583 043 965 091 84;
  • 35) 0.316 779 014 654 457 583 043 965 091 84 × 2 = 0 + 0.633 558 029 308 915 166 087 930 183 68;
  • 36) 0.633 558 029 308 915 166 087 930 183 68 × 2 = 1 + 0.267 116 058 617 830 332 175 860 367 36;
  • 37) 0.267 116 058 617 830 332 175 860 367 36 × 2 = 0 + 0.534 232 117 235 660 664 351 720 734 72;
  • 38) 0.534 232 117 235 660 664 351 720 734 72 × 2 = 1 + 0.068 464 234 471 321 328 703 441 469 44;
  • 39) 0.068 464 234 471 321 328 703 441 469 44 × 2 = 0 + 0.136 928 468 942 642 657 406 882 938 88;
  • 40) 0.136 928 468 942 642 657 406 882 938 88 × 2 = 0 + 0.273 856 937 885 285 314 813 765 877 76;
  • 41) 0.273 856 937 885 285 314 813 765 877 76 × 2 = 0 + 0.547 713 875 770 570 629 627 531 755 52;
  • 42) 0.547 713 875 770 570 629 627 531 755 52 × 2 = 1 + 0.095 427 751 541 141 259 255 063 511 04;
  • 43) 0.095 427 751 541 141 259 255 063 511 04 × 2 = 0 + 0.190 855 503 082 282 518 510 127 022 08;
  • 44) 0.190 855 503 082 282 518 510 127 022 08 × 2 = 0 + 0.381 711 006 164 565 037 020 254 044 16;
  • 45) 0.381 711 006 164 565 037 020 254 044 16 × 2 = 0 + 0.763 422 012 329 130 074 040 508 088 32;
  • 46) 0.763 422 012 329 130 074 040 508 088 32 × 2 = 1 + 0.526 844 024 658 260 148 081 016 176 64;
  • 47) 0.526 844 024 658 260 148 081 016 176 64 × 2 = 1 + 0.053 688 049 316 520 296 162 032 353 28;
  • 48) 0.053 688 049 316 520 296 162 032 353 28 × 2 = 0 + 0.107 376 098 633 040 592 324 064 706 56;
  • 49) 0.107 376 098 633 040 592 324 064 706 56 × 2 = 0 + 0.214 752 197 266 081 184 648 129 413 12;
  • 50) 0.214 752 197 266 081 184 648 129 413 12 × 2 = 0 + 0.429 504 394 532 162 369 296 258 826 24;
  • 51) 0.429 504 394 532 162 369 296 258 826 24 × 2 = 0 + 0.859 008 789 064 324 738 592 517 652 48;
  • 52) 0.859 008 789 064 324 738 592 517 652 48 × 2 = 1 + 0.718 017 578 128 649 477 185 035 304 96;
  • 53) 0.718 017 578 128 649 477 185 035 304 96 × 2 = 1 + 0.436 035 156 257 298 954 370 070 609 92;

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:


0.211 829 052 383 358 300 119 548 662 51(10) =

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

6. Positive number before normalization:

2.211 829 052 383 358 300 119 548 662 51(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:


2.211 829 052 383 358 300 119 548 662 51(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 -2.211 829 052 383 358 300 119 548 662 51 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