-1 036.699 999 999 999 818 101 038 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -1 036.699 999 999 999 818 101 038(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
-1 036.699 999 999 999 818 101 038(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:

|-1 036.699 999 999 999 818 101 038| = 1 036.699 999 999 999 818 101 038


2. First, convert to binary (in base 2) the integer part: 1 036.
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;
  • 1 036 ÷ 2 = 518 + 0;
  • 518 ÷ 2 = 259 + 0;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 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;

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.

1 036(10) =

100 0000 1100(2)


4. Convert to binary (base 2) the fractional part: 0.699 999 999 999 818 101 038.

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) 0.699 999 999 999 818 101 038 × 2 = 1 + 0.399 999 999 999 636 202 076;
  • 2) 0.399 999 999 999 636 202 076 × 2 = 0 + 0.799 999 999 999 272 404 152;
  • 3) 0.799 999 999 999 272 404 152 × 2 = 1 + 0.599 999 999 998 544 808 304;
  • 4) 0.599 999 999 998 544 808 304 × 2 = 1 + 0.199 999 999 997 089 616 608;
  • 5) 0.199 999 999 997 089 616 608 × 2 = 0 + 0.399 999 999 994 179 233 216;
  • 6) 0.399 999 999 994 179 233 216 × 2 = 0 + 0.799 999 999 988 358 466 432;
  • 7) 0.799 999 999 988 358 466 432 × 2 = 1 + 0.599 999 999 976 716 932 864;
  • 8) 0.599 999 999 976 716 932 864 × 2 = 1 + 0.199 999 999 953 433 865 728;
  • 9) 0.199 999 999 953 433 865 728 × 2 = 0 + 0.399 999 999 906 867 731 456;
  • 10) 0.399 999 999 906 867 731 456 × 2 = 0 + 0.799 999 999 813 735 462 912;
  • 11) 0.799 999 999 813 735 462 912 × 2 = 1 + 0.599 999 999 627 470 925 824;
  • 12) 0.599 999 999 627 470 925 824 × 2 = 1 + 0.199 999 999 254 941 851 648;
  • 13) 0.199 999 999 254 941 851 648 × 2 = 0 + 0.399 999 998 509 883 703 296;
  • 14) 0.399 999 998 509 883 703 296 × 2 = 0 + 0.799 999 997 019 767 406 592;
  • 15) 0.799 999 997 019 767 406 592 × 2 = 1 + 0.599 999 994 039 534 813 184;
  • 16) 0.599 999 994 039 534 813 184 × 2 = 1 + 0.199 999 988 079 069 626 368;
  • 17) 0.199 999 988 079 069 626 368 × 2 = 0 + 0.399 999 976 158 139 252 736;
  • 18) 0.399 999 976 158 139 252 736 × 2 = 0 + 0.799 999 952 316 278 505 472;
  • 19) 0.799 999 952 316 278 505 472 × 2 = 1 + 0.599 999 904 632 557 010 944;
  • 20) 0.599 999 904 632 557 010 944 × 2 = 1 + 0.199 999 809 265 114 021 888;
  • 21) 0.199 999 809 265 114 021 888 × 2 = 0 + 0.399 999 618 530 228 043 776;
  • 22) 0.399 999 618 530 228 043 776 × 2 = 0 + 0.799 999 237 060 456 087 552;
  • 23) 0.799 999 237 060 456 087 552 × 2 = 1 + 0.599 998 474 120 912 175 104;
  • 24) 0.599 998 474 120 912 175 104 × 2 = 1 + 0.199 996 948 241 824 350 208;
  • 25) 0.199 996 948 241 824 350 208 × 2 = 0 + 0.399 993 896 483 648 700 416;
  • 26) 0.399 993 896 483 648 700 416 × 2 = 0 + 0.799 987 792 967 297 400 832;
  • 27) 0.799 987 792 967 297 400 832 × 2 = 1 + 0.599 975 585 934 594 801 664;
  • 28) 0.599 975 585 934 594 801 664 × 2 = 1 + 0.199 951 171 869 189 603 328;
  • 29) 0.199 951 171 869 189 603 328 × 2 = 0 + 0.399 902 343 738 379 206 656;
  • 30) 0.399 902 343 738 379 206 656 × 2 = 0 + 0.799 804 687 476 758 413 312;
  • 31) 0.799 804 687 476 758 413 312 × 2 = 1 + 0.599 609 374 953 516 826 624;
  • 32) 0.599 609 374 953 516 826 624 × 2 = 1 + 0.199 218 749 907 033 653 248;
  • 33) 0.199 218 749 907 033 653 248 × 2 = 0 + 0.398 437 499 814 067 306 496;
  • 34) 0.398 437 499 814 067 306 496 × 2 = 0 + 0.796 874 999 628 134 612 992;
  • 35) 0.796 874 999 628 134 612 992 × 2 = 1 + 0.593 749 999 256 269 225 984;
  • 36) 0.593 749 999 256 269 225 984 × 2 = 1 + 0.187 499 998 512 538 451 968;
  • 37) 0.187 499 998 512 538 451 968 × 2 = 0 + 0.374 999 997 025 076 903 936;
  • 38) 0.374 999 997 025 076 903 936 × 2 = 0 + 0.749 999 994 050 153 807 872;
  • 39) 0.749 999 994 050 153 807 872 × 2 = 1 + 0.499 999 988 100 307 615 744;
  • 40) 0.499 999 988 100 307 615 744 × 2 = 0 + 0.999 999 976 200 615 231 488;
  • 41) 0.999 999 976 200 615 231 488 × 2 = 1 + 0.999 999 952 401 230 462 976;
  • 42) 0.999 999 952 401 230 462 976 × 2 = 1 + 0.999 999 904 802 460 925 952;
  • 43) 0.999 999 904 802 460 925 952 × 2 = 1 + 0.999 999 809 604 921 851 904;
  • 44) 0.999 999 809 604 921 851 904 × 2 = 1 + 0.999 999 619 209 843 703 808;
  • 45) 0.999 999 619 209 843 703 808 × 2 = 1 + 0.999 999 238 419 687 407 616;
  • 46) 0.999 999 238 419 687 407 616 × 2 = 1 + 0.999 998 476 839 374 815 232;
  • 47) 0.999 998 476 839 374 815 232 × 2 = 1 + 0.999 996 953 678 749 630 464;
  • 48) 0.999 996 953 678 749 630 464 × 2 = 1 + 0.999 993 907 357 499 260 928;
  • 49) 0.999 993 907 357 499 260 928 × 2 = 1 + 0.999 987 814 714 998 521 856;
  • 50) 0.999 987 814 714 998 521 856 × 2 = 1 + 0.999 975 629 429 997 043 712;
  • 51) 0.999 975 629 429 997 043 712 × 2 = 1 + 0.999 951 258 859 994 087 424;
  • 52) 0.999 951 258 859 994 087 424 × 2 = 1 + 0.999 902 517 719 988 174 848;
  • 53) 0.999 902 517 719 988 174 848 × 2 = 1 + 0.999 805 035 439 976 349 696;

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:


0.699 999 999 999 818 101 038(10) =

0.1011 0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 1111 1111 1(2)

6. Positive number before normalization:

1 036.699 999 999 999 818 101 038(10) =

100 0000 1100.1011 0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 1111 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the left, so that only one non zero digit remains to the left of it:


1 036.699 999 999 999 818 101 038(10) =

100 0000 1100.1011 0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 1111 1111 1(2) =

100 0000 1100.1011 0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 1111 1111 1(2) × 20 =

1.0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1111 111(2) × 210


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): 10

Mantissa (not normalized):
1.0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1111 111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

10 + 2(11-1) - 1 =

(10 + 1 023)(10) =

1 033(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;
  • 1 033 ÷ 2 = 516 + 1;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 129 ÷ 2 = 64 + 1;
  • 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) =

1033(10) =

100 0000 1001(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. 0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 111 1111 1111 =

0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 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 1001

Mantissa (52 bits) =
0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011


Decimal number -1 036.699 999 999 999 818 101 038 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 1001 - 0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 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