-1 036.699 999 999 999 818 101 059 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -1 036.699 999 999 999 818 101 059 1(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
-1 036.699 999 999 999 818 101 059 1(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:

|-1 036.699 999 999 999 818 101 059 1| = 1 036.699 999 999 999 818 101 059 1


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: 0.699 999 999 999 818 101 059 1.

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) 0.699 999 999 999 818 101 059 1 × 2 = 1 + 0.399 999 999 999 636 202 118 2;
  • 2) 0.399 999 999 999 636 202 118 2 × 2 = 0 + 0.799 999 999 999 272 404 236 4;
  • 3) 0.799 999 999 999 272 404 236 4 × 2 = 1 + 0.599 999 999 998 544 808 472 8;
  • 4) 0.599 999 999 998 544 808 472 8 × 2 = 1 + 0.199 999 999 997 089 616 945 6;
  • 5) 0.199 999 999 997 089 616 945 6 × 2 = 0 + 0.399 999 999 994 179 233 891 2;
  • 6) 0.399 999 999 994 179 233 891 2 × 2 = 0 + 0.799 999 999 988 358 467 782 4;
  • 7) 0.799 999 999 988 358 467 782 4 × 2 = 1 + 0.599 999 999 976 716 935 564 8;
  • 8) 0.599 999 999 976 716 935 564 8 × 2 = 1 + 0.199 999 999 953 433 871 129 6;
  • 9) 0.199 999 999 953 433 871 129 6 × 2 = 0 + 0.399 999 999 906 867 742 259 2;
  • 10) 0.399 999 999 906 867 742 259 2 × 2 = 0 + 0.799 999 999 813 735 484 518 4;
  • 11) 0.799 999 999 813 735 484 518 4 × 2 = 1 + 0.599 999 999 627 470 969 036 8;
  • 12) 0.599 999 999 627 470 969 036 8 × 2 = 1 + 0.199 999 999 254 941 938 073 6;
  • 13) 0.199 999 999 254 941 938 073 6 × 2 = 0 + 0.399 999 998 509 883 876 147 2;
  • 14) 0.399 999 998 509 883 876 147 2 × 2 = 0 + 0.799 999 997 019 767 752 294 4;
  • 15) 0.799 999 997 019 767 752 294 4 × 2 = 1 + 0.599 999 994 039 535 504 588 8;
  • 16) 0.599 999 994 039 535 504 588 8 × 2 = 1 + 0.199 999 988 079 071 009 177 6;
  • 17) 0.199 999 988 079 071 009 177 6 × 2 = 0 + 0.399 999 976 158 142 018 355 2;
  • 18) 0.399 999 976 158 142 018 355 2 × 2 = 0 + 0.799 999 952 316 284 036 710 4;
  • 19) 0.799 999 952 316 284 036 710 4 × 2 = 1 + 0.599 999 904 632 568 073 420 8;
  • 20) 0.599 999 904 632 568 073 420 8 × 2 = 1 + 0.199 999 809 265 136 146 841 6;
  • 21) 0.199 999 809 265 136 146 841 6 × 2 = 0 + 0.399 999 618 530 272 293 683 2;
  • 22) 0.399 999 618 530 272 293 683 2 × 2 = 0 + 0.799 999 237 060 544 587 366 4;
  • 23) 0.799 999 237 060 544 587 366 4 × 2 = 1 + 0.599 998 474 121 089 174 732 8;
  • 24) 0.599 998 474 121 089 174 732 8 × 2 = 1 + 0.199 996 948 242 178 349 465 6;
  • 25) 0.199 996 948 242 178 349 465 6 × 2 = 0 + 0.399 993 896 484 356 698 931 2;
  • 26) 0.399 993 896 484 356 698 931 2 × 2 = 0 + 0.799 987 792 968 713 397 862 4;
  • 27) 0.799 987 792 968 713 397 862 4 × 2 = 1 + 0.599 975 585 937 426 795 724 8;
  • 28) 0.599 975 585 937 426 795 724 8 × 2 = 1 + 0.199 951 171 874 853 591 449 6;
  • 29) 0.199 951 171 874 853 591 449 6 × 2 = 0 + 0.399 902 343 749 707 182 899 2;
  • 30) 0.399 902 343 749 707 182 899 2 × 2 = 0 + 0.799 804 687 499 414 365 798 4;
  • 31) 0.799 804 687 499 414 365 798 4 × 2 = 1 + 0.599 609 374 998 828 731 596 8;
  • 32) 0.599 609 374 998 828 731 596 8 × 2 = 1 + 0.199 218 749 997 657 463 193 6;
  • 33) 0.199 218 749 997 657 463 193 6 × 2 = 0 + 0.398 437 499 995 314 926 387 2;
  • 34) 0.398 437 499 995 314 926 387 2 × 2 = 0 + 0.796 874 999 990 629 852 774 4;
  • 35) 0.796 874 999 990 629 852 774 4 × 2 = 1 + 0.593 749 999 981 259 705 548 8;
  • 36) 0.593 749 999 981 259 705 548 8 × 2 = 1 + 0.187 499 999 962 519 411 097 6;
  • 37) 0.187 499 999 962 519 411 097 6 × 2 = 0 + 0.374 999 999 925 038 822 195 2;
  • 38) 0.374 999 999 925 038 822 195 2 × 2 = 0 + 0.749 999 999 850 077 644 390 4;
  • 39) 0.749 999 999 850 077 644 390 4 × 2 = 1 + 0.499 999 999 700 155 288 780 8;
  • 40) 0.499 999 999 700 155 288 780 8 × 2 = 0 + 0.999 999 999 400 310 577 561 6;
  • 41) 0.999 999 999 400 310 577 561 6 × 2 = 1 + 0.999 999 998 800 621 155 123 2;
  • 42) 0.999 999 998 800 621 155 123 2 × 2 = 1 + 0.999 999 997 601 242 310 246 4;
  • 43) 0.999 999 997 601 242 310 246 4 × 2 = 1 + 0.999 999 995 202 484 620 492 8;
  • 44) 0.999 999 995 202 484 620 492 8 × 2 = 1 + 0.999 999 990 404 969 240 985 6;
  • 45) 0.999 999 990 404 969 240 985 6 × 2 = 1 + 0.999 999 980 809 938 481 971 2;
  • 46) 0.999 999 980 809 938 481 971 2 × 2 = 1 + 0.999 999 961 619 876 963 942 4;
  • 47) 0.999 999 961 619 876 963 942 4 × 2 = 1 + 0.999 999 923 239 753 927 884 8;
  • 48) 0.999 999 923 239 753 927 884 8 × 2 = 1 + 0.999 999 846 479 507 855 769 6;
  • 49) 0.999 999 846 479 507 855 769 6 × 2 = 1 + 0.999 999 692 959 015 711 539 2;
  • 50) 0.999 999 692 959 015 711 539 2 × 2 = 1 + 0.999 999 385 918 031 423 078 4;
  • 51) 0.999 999 385 918 031 423 078 4 × 2 = 1 + 0.999 998 771 836 062 846 156 8;
  • 52) 0.999 998 771 836 062 846 156 8 × 2 = 1 + 0.999 997 543 672 125 692 313 6;
  • 53) 0.999 997 543 672 125 692 313 6 × 2 = 1 + 0.999 995 087 344 251 384 627 2;

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:


0.699 999 999 999 818 101 059 1(10) =

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

6. Positive number before normalization:

1 036.699 999 999 999 818 101 059 1(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:


1 036.699 999 999 999 818 101 059 1(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 -1 036.699 999 999 999 818 101 059 1 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