-2.211 829 052 383 358 303 67 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 303 67(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
-2.211 829 052 383 358 303 67(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:

|-2.211 829 052 383 358 303 67| = 2.211 829 052 383 358 303 67


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: 0.211 829 052 383 358 303 67.

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) 0.211 829 052 383 358 303 67 × 2 = 0 + 0.423 658 104 766 716 607 34;
  • 2) 0.423 658 104 766 716 607 34 × 2 = 0 + 0.847 316 209 533 433 214 68;
  • 3) 0.847 316 209 533 433 214 68 × 2 = 1 + 0.694 632 419 066 866 429 36;
  • 4) 0.694 632 419 066 866 429 36 × 2 = 1 + 0.389 264 838 133 732 858 72;
  • 5) 0.389 264 838 133 732 858 72 × 2 = 0 + 0.778 529 676 267 465 717 44;
  • 6) 0.778 529 676 267 465 717 44 × 2 = 1 + 0.557 059 352 534 931 434 88;
  • 7) 0.557 059 352 534 931 434 88 × 2 = 1 + 0.114 118 705 069 862 869 76;
  • 8) 0.114 118 705 069 862 869 76 × 2 = 0 + 0.228 237 410 139 725 739 52;
  • 9) 0.228 237 410 139 725 739 52 × 2 = 0 + 0.456 474 820 279 451 479 04;
  • 10) 0.456 474 820 279 451 479 04 × 2 = 0 + 0.912 949 640 558 902 958 08;
  • 11) 0.912 949 640 558 902 958 08 × 2 = 1 + 0.825 899 281 117 805 916 16;
  • 12) 0.825 899 281 117 805 916 16 × 2 = 1 + 0.651 798 562 235 611 832 32;
  • 13) 0.651 798 562 235 611 832 32 × 2 = 1 + 0.303 597 124 471 223 664 64;
  • 14) 0.303 597 124 471 223 664 64 × 2 = 0 + 0.607 194 248 942 447 329 28;
  • 15) 0.607 194 248 942 447 329 28 × 2 = 1 + 0.214 388 497 884 894 658 56;
  • 16) 0.214 388 497 884 894 658 56 × 2 = 0 + 0.428 776 995 769 789 317 12;
  • 17) 0.428 776 995 769 789 317 12 × 2 = 0 + 0.857 553 991 539 578 634 24;
  • 18) 0.857 553 991 539 578 634 24 × 2 = 1 + 0.715 107 983 079 157 268 48;
  • 19) 0.715 107 983 079 157 268 48 × 2 = 1 + 0.430 215 966 158 314 536 96;
  • 20) 0.430 215 966 158 314 536 96 × 2 = 0 + 0.860 431 932 316 629 073 92;
  • 21) 0.860 431 932 316 629 073 92 × 2 = 1 + 0.720 863 864 633 258 147 84;
  • 22) 0.720 863 864 633 258 147 84 × 2 = 1 + 0.441 727 729 266 516 295 68;
  • 23) 0.441 727 729 266 516 295 68 × 2 = 0 + 0.883 455 458 533 032 591 36;
  • 24) 0.883 455 458 533 032 591 36 × 2 = 1 + 0.766 910 917 066 065 182 72;
  • 25) 0.766 910 917 066 065 182 72 × 2 = 1 + 0.533 821 834 132 130 365 44;
  • 26) 0.533 821 834 132 130 365 44 × 2 = 1 + 0.067 643 668 264 260 730 88;
  • 27) 0.067 643 668 264 260 730 88 × 2 = 0 + 0.135 287 336 528 521 461 76;
  • 28) 0.135 287 336 528 521 461 76 × 2 = 0 + 0.270 574 673 057 042 923 52;
  • 29) 0.270 574 673 057 042 923 52 × 2 = 0 + 0.541 149 346 114 085 847 04;
  • 30) 0.541 149 346 114 085 847 04 × 2 = 1 + 0.082 298 692 228 171 694 08;
  • 31) 0.082 298 692 228 171 694 08 × 2 = 0 + 0.164 597 384 456 343 388 16;
  • 32) 0.164 597 384 456 343 388 16 × 2 = 0 + 0.329 194 768 912 686 776 32;
  • 33) 0.329 194 768 912 686 776 32 × 2 = 0 + 0.658 389 537 825 373 552 64;
  • 34) 0.658 389 537 825 373 552 64 × 2 = 1 + 0.316 779 075 650 747 105 28;
  • 35) 0.316 779 075 650 747 105 28 × 2 = 0 + 0.633 558 151 301 494 210 56;
  • 36) 0.633 558 151 301 494 210 56 × 2 = 1 + 0.267 116 302 602 988 421 12;
  • 37) 0.267 116 302 602 988 421 12 × 2 = 0 + 0.534 232 605 205 976 842 24;
  • 38) 0.534 232 605 205 976 842 24 × 2 = 1 + 0.068 465 210 411 953 684 48;
  • 39) 0.068 465 210 411 953 684 48 × 2 = 0 + 0.136 930 420 823 907 368 96;
  • 40) 0.136 930 420 823 907 368 96 × 2 = 0 + 0.273 860 841 647 814 737 92;
  • 41) 0.273 860 841 647 814 737 92 × 2 = 0 + 0.547 721 683 295 629 475 84;
  • 42) 0.547 721 683 295 629 475 84 × 2 = 1 + 0.095 443 366 591 258 951 68;
  • 43) 0.095 443 366 591 258 951 68 × 2 = 0 + 0.190 886 733 182 517 903 36;
  • 44) 0.190 886 733 182 517 903 36 × 2 = 0 + 0.381 773 466 365 035 806 72;
  • 45) 0.381 773 466 365 035 806 72 × 2 = 0 + 0.763 546 932 730 071 613 44;
  • 46) 0.763 546 932 730 071 613 44 × 2 = 1 + 0.527 093 865 460 143 226 88;
  • 47) 0.527 093 865 460 143 226 88 × 2 = 1 + 0.054 187 730 920 286 453 76;
  • 48) 0.054 187 730 920 286 453 76 × 2 = 0 + 0.108 375 461 840 572 907 52;
  • 49) 0.108 375 461 840 572 907 52 × 2 = 0 + 0.216 750 923 681 145 815 04;
  • 50) 0.216 750 923 681 145 815 04 × 2 = 0 + 0.433 501 847 362 291 630 08;
  • 51) 0.433 501 847 362 291 630 08 × 2 = 0 + 0.867 003 694 724 583 260 16;
  • 52) 0.867 003 694 724 583 260 16 × 2 = 1 + 0.734 007 389 449 166 520 32;
  • 53) 0.734 007 389 449 166 520 32 × 2 = 1 + 0.468 014 778 898 333 040 64;

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:


0.211 829 052 383 358 303 67(10) =

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

6. Positive number before normalization:

2.211 829 052 383 358 303 67(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:


2.211 829 052 383 358 303 67(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 -2.211 829 052 383 358 303 67 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

» Convert decimal -2.211 829 052 383 358 304 4 to 64 bit double precision IEEE 754 binary floating point representation standard

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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