-69.814 999 999 999 997 726 263 193 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -69.814 999 999 999 997 726 263 193(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
-69.814 999 999 999 997 726 263 193(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:

|-69.814 999 999 999 997 726 263 193| = 69.814 999 999 999 997 726 263 193


2. First, convert to binary (in base 2) the integer part: 69.
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;
  • 69 ÷ 2 = 34 + 1;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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.

69(10) =

100 0101(2)


4. Convert to binary (base 2) the fractional part: 0.814 999 999 999 997 726 263 193.

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) 0.814 999 999 999 997 726 263 193 × 2 = 1 + 0.629 999 999 999 995 452 526 386;
  • 2) 0.629 999 999 999 995 452 526 386 × 2 = 1 + 0.259 999 999 999 990 905 052 772;
  • 3) 0.259 999 999 999 990 905 052 772 × 2 = 0 + 0.519 999 999 999 981 810 105 544;
  • 4) 0.519 999 999 999 981 810 105 544 × 2 = 1 + 0.039 999 999 999 963 620 211 088;
  • 5) 0.039 999 999 999 963 620 211 088 × 2 = 0 + 0.079 999 999 999 927 240 422 176;
  • 6) 0.079 999 999 999 927 240 422 176 × 2 = 0 + 0.159 999 999 999 854 480 844 352;
  • 7) 0.159 999 999 999 854 480 844 352 × 2 = 0 + 0.319 999 999 999 708 961 688 704;
  • 8) 0.319 999 999 999 708 961 688 704 × 2 = 0 + 0.639 999 999 999 417 923 377 408;
  • 9) 0.639 999 999 999 417 923 377 408 × 2 = 1 + 0.279 999 999 998 835 846 754 816;
  • 10) 0.279 999 999 998 835 846 754 816 × 2 = 0 + 0.559 999 999 997 671 693 509 632;
  • 11) 0.559 999 999 997 671 693 509 632 × 2 = 1 + 0.119 999 999 995 343 387 019 264;
  • 12) 0.119 999 999 995 343 387 019 264 × 2 = 0 + 0.239 999 999 990 686 774 038 528;
  • 13) 0.239 999 999 990 686 774 038 528 × 2 = 0 + 0.479 999 999 981 373 548 077 056;
  • 14) 0.479 999 999 981 373 548 077 056 × 2 = 0 + 0.959 999 999 962 747 096 154 112;
  • 15) 0.959 999 999 962 747 096 154 112 × 2 = 1 + 0.919 999 999 925 494 192 308 224;
  • 16) 0.919 999 999 925 494 192 308 224 × 2 = 1 + 0.839 999 999 850 988 384 616 448;
  • 17) 0.839 999 999 850 988 384 616 448 × 2 = 1 + 0.679 999 999 701 976 769 232 896;
  • 18) 0.679 999 999 701 976 769 232 896 × 2 = 1 + 0.359 999 999 403 953 538 465 792;
  • 19) 0.359 999 999 403 953 538 465 792 × 2 = 0 + 0.719 999 998 807 907 076 931 584;
  • 20) 0.719 999 998 807 907 076 931 584 × 2 = 1 + 0.439 999 997 615 814 153 863 168;
  • 21) 0.439 999 997 615 814 153 863 168 × 2 = 0 + 0.879 999 995 231 628 307 726 336;
  • 22) 0.879 999 995 231 628 307 726 336 × 2 = 1 + 0.759 999 990 463 256 615 452 672;
  • 23) 0.759 999 990 463 256 615 452 672 × 2 = 1 + 0.519 999 980 926 513 230 905 344;
  • 24) 0.519 999 980 926 513 230 905 344 × 2 = 1 + 0.039 999 961 853 026 461 810 688;
  • 25) 0.039 999 961 853 026 461 810 688 × 2 = 0 + 0.079 999 923 706 052 923 621 376;
  • 26) 0.079 999 923 706 052 923 621 376 × 2 = 0 + 0.159 999 847 412 105 847 242 752;
  • 27) 0.159 999 847 412 105 847 242 752 × 2 = 0 + 0.319 999 694 824 211 694 485 504;
  • 28) 0.319 999 694 824 211 694 485 504 × 2 = 0 + 0.639 999 389 648 423 388 971 008;
  • 29) 0.639 999 389 648 423 388 971 008 × 2 = 1 + 0.279 998 779 296 846 777 942 016;
  • 30) 0.279 998 779 296 846 777 942 016 × 2 = 0 + 0.559 997 558 593 693 555 884 032;
  • 31) 0.559 997 558 593 693 555 884 032 × 2 = 1 + 0.119 995 117 187 387 111 768 064;
  • 32) 0.119 995 117 187 387 111 768 064 × 2 = 0 + 0.239 990 234 374 774 223 536 128;
  • 33) 0.239 990 234 374 774 223 536 128 × 2 = 0 + 0.479 980 468 749 548 447 072 256;
  • 34) 0.479 980 468 749 548 447 072 256 × 2 = 0 + 0.959 960 937 499 096 894 144 512;
  • 35) 0.959 960 937 499 096 894 144 512 × 2 = 1 + 0.919 921 874 998 193 788 289 024;
  • 36) 0.919 921 874 998 193 788 289 024 × 2 = 1 + 0.839 843 749 996 387 576 578 048;
  • 37) 0.839 843 749 996 387 576 578 048 × 2 = 1 + 0.679 687 499 992 775 153 156 096;
  • 38) 0.679 687 499 992 775 153 156 096 × 2 = 1 + 0.359 374 999 985 550 306 312 192;
  • 39) 0.359 374 999 985 550 306 312 192 × 2 = 0 + 0.718 749 999 971 100 612 624 384;
  • 40) 0.718 749 999 971 100 612 624 384 × 2 = 1 + 0.437 499 999 942 201 225 248 768;
  • 41) 0.437 499 999 942 201 225 248 768 × 2 = 0 + 0.874 999 999 884 402 450 497 536;
  • 42) 0.874 999 999 884 402 450 497 536 × 2 = 1 + 0.749 999 999 768 804 900 995 072;
  • 43) 0.749 999 999 768 804 900 995 072 × 2 = 1 + 0.499 999 999 537 609 801 990 144;
  • 44) 0.499 999 999 537 609 801 990 144 × 2 = 0 + 0.999 999 999 075 219 603 980 288;
  • 45) 0.999 999 999 075 219 603 980 288 × 2 = 1 + 0.999 999 998 150 439 207 960 576;
  • 46) 0.999 999 998 150 439 207 960 576 × 2 = 1 + 0.999 999 996 300 878 415 921 152;
  • 47) 0.999 999 996 300 878 415 921 152 × 2 = 1 + 0.999 999 992 601 756 831 842 304;
  • 48) 0.999 999 992 601 756 831 842 304 × 2 = 1 + 0.999 999 985 203 513 663 684 608;
  • 49) 0.999 999 985 203 513 663 684 608 × 2 = 1 + 0.999 999 970 407 027 327 369 216;
  • 50) 0.999 999 970 407 027 327 369 216 × 2 = 1 + 0.999 999 940 814 054 654 738 432;
  • 51) 0.999 999 940 814 054 654 738 432 × 2 = 1 + 0.999 999 881 628 109 309 476 864;
  • 52) 0.999 999 881 628 109 309 476 864 × 2 = 1 + 0.999 999 763 256 218 618 953 728;
  • 53) 0.999 999 763 256 218 618 953 728 × 2 = 1 + 0.999 999 526 512 437 237 907 456;

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:


0.814 999 999 999 997 726 263 193(10) =

0.1101 0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1111 1111 1(2)

6. Positive number before normalization:

69.814 999 999 999 997 726 263 193(10) =

100 0101.1101 0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1111 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


69.814 999 999 999 997 726 263 193(10) =

100 0101.1101 0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1111 1111 1(2) =

100 0101.1101 0000 1010 0011 1101 0111 0000 1010 0011 1101 0110 1111 1111 1(2) × 20 =

1.0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1011 1111 111(2) × 26


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): 6

Mantissa (not normalized):
1.0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1011 1111 111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

6 + 2(11-1) - 1 =

(6 + 1 023)(10) =

1 029(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;
  • 1 029 ÷ 2 = 514 + 1;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 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) =

1029(10) =

100 0000 0101(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 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1011 111 1111 =

0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1011


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 0101

Mantissa (52 bits) =
0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1011


Decimal number -69.814 999 999 999 997 726 263 193 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0101 - 0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1011

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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