-1 036.699 999 999 999 818 68 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -1 036.699 999 999 999 818 68(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 68(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 68| = 1 036.699 999 999 999 818 68


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 68.

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 68 × 2 = 1 + 0.399 999 999 999 637 36;
  • 2) 0.399 999 999 999 637 36 × 2 = 0 + 0.799 999 999 999 274 72;
  • 3) 0.799 999 999 999 274 72 × 2 = 1 + 0.599 999 999 998 549 44;
  • 4) 0.599 999 999 998 549 44 × 2 = 1 + 0.199 999 999 997 098 88;
  • 5) 0.199 999 999 997 098 88 × 2 = 0 + 0.399 999 999 994 197 76;
  • 6) 0.399 999 999 994 197 76 × 2 = 0 + 0.799 999 999 988 395 52;
  • 7) 0.799 999 999 988 395 52 × 2 = 1 + 0.599 999 999 976 791 04;
  • 8) 0.599 999 999 976 791 04 × 2 = 1 + 0.199 999 999 953 582 08;
  • 9) 0.199 999 999 953 582 08 × 2 = 0 + 0.399 999 999 907 164 16;
  • 10) 0.399 999 999 907 164 16 × 2 = 0 + 0.799 999 999 814 328 32;
  • 11) 0.799 999 999 814 328 32 × 2 = 1 + 0.599 999 999 628 656 64;
  • 12) 0.599 999 999 628 656 64 × 2 = 1 + 0.199 999 999 257 313 28;
  • 13) 0.199 999 999 257 313 28 × 2 = 0 + 0.399 999 998 514 626 56;
  • 14) 0.399 999 998 514 626 56 × 2 = 0 + 0.799 999 997 029 253 12;
  • 15) 0.799 999 997 029 253 12 × 2 = 1 + 0.599 999 994 058 506 24;
  • 16) 0.599 999 994 058 506 24 × 2 = 1 + 0.199 999 988 117 012 48;
  • 17) 0.199 999 988 117 012 48 × 2 = 0 + 0.399 999 976 234 024 96;
  • 18) 0.399 999 976 234 024 96 × 2 = 0 + 0.799 999 952 468 049 92;
  • 19) 0.799 999 952 468 049 92 × 2 = 1 + 0.599 999 904 936 099 84;
  • 20) 0.599 999 904 936 099 84 × 2 = 1 + 0.199 999 809 872 199 68;
  • 21) 0.199 999 809 872 199 68 × 2 = 0 + 0.399 999 619 744 399 36;
  • 22) 0.399 999 619 744 399 36 × 2 = 0 + 0.799 999 239 488 798 72;
  • 23) 0.799 999 239 488 798 72 × 2 = 1 + 0.599 998 478 977 597 44;
  • 24) 0.599 998 478 977 597 44 × 2 = 1 + 0.199 996 957 955 194 88;
  • 25) 0.199 996 957 955 194 88 × 2 = 0 + 0.399 993 915 910 389 76;
  • 26) 0.399 993 915 910 389 76 × 2 = 0 + 0.799 987 831 820 779 52;
  • 27) 0.799 987 831 820 779 52 × 2 = 1 + 0.599 975 663 641 559 04;
  • 28) 0.599 975 663 641 559 04 × 2 = 1 + 0.199 951 327 283 118 08;
  • 29) 0.199 951 327 283 118 08 × 2 = 0 + 0.399 902 654 566 236 16;
  • 30) 0.399 902 654 566 236 16 × 2 = 0 + 0.799 805 309 132 472 32;
  • 31) 0.799 805 309 132 472 32 × 2 = 1 + 0.599 610 618 264 944 64;
  • 32) 0.599 610 618 264 944 64 × 2 = 1 + 0.199 221 236 529 889 28;
  • 33) 0.199 221 236 529 889 28 × 2 = 0 + 0.398 442 473 059 778 56;
  • 34) 0.398 442 473 059 778 56 × 2 = 0 + 0.796 884 946 119 557 12;
  • 35) 0.796 884 946 119 557 12 × 2 = 1 + 0.593 769 892 239 114 24;
  • 36) 0.593 769 892 239 114 24 × 2 = 1 + 0.187 539 784 478 228 48;
  • 37) 0.187 539 784 478 228 48 × 2 = 0 + 0.375 079 568 956 456 96;
  • 38) 0.375 079 568 956 456 96 × 2 = 0 + 0.750 159 137 912 913 92;
  • 39) 0.750 159 137 912 913 92 × 2 = 1 + 0.500 318 275 825 827 84;
  • 40) 0.500 318 275 825 827 84 × 2 = 1 + 0.000 636 551 651 655 68;
  • 41) 0.000 636 551 651 655 68 × 2 = 0 + 0.001 273 103 303 311 36;
  • 42) 0.001 273 103 303 311 36 × 2 = 0 + 0.002 546 206 606 622 72;
  • 43) 0.002 546 206 606 622 72 × 2 = 0 + 0.005 092 413 213 245 44;
  • 44) 0.005 092 413 213 245 44 × 2 = 0 + 0.010 184 826 426 490 88;
  • 45) 0.010 184 826 426 490 88 × 2 = 0 + 0.020 369 652 852 981 76;
  • 46) 0.020 369 652 852 981 76 × 2 = 0 + 0.040 739 305 705 963 52;
  • 47) 0.040 739 305 705 963 52 × 2 = 0 + 0.081 478 611 411 927 04;
  • 48) 0.081 478 611 411 927 04 × 2 = 0 + 0.162 957 222 823 854 08;
  • 49) 0.162 957 222 823 854 08 × 2 = 0 + 0.325 914 445 647 708 16;
  • 50) 0.325 914 445 647 708 16 × 2 = 0 + 0.651 828 891 295 416 32;
  • 51) 0.651 828 891 295 416 32 × 2 = 1 + 0.303 657 782 590 832 64;
  • 52) 0.303 657 782 590 832 64 × 2 = 0 + 0.607 315 565 181 665 28;
  • 53) 0.607 315 565 181 665 28 × 2 = 1 + 0.214 631 130 363 330 56;

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 68(10) =

0.1011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0000 0000 0010 1(2)

6. Positive number before normalization:

1 036.699 999 999 999 818 68(10) =

100 0000 1100.1011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0000 0000 0010 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 68(10) =

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

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

1.0000 0011 0010 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0000 101(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 1100 0000 0000 101


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 1100 000 0000 0101 =

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


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 1100


Decimal number -1 036.699 999 999 999 818 68 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 1100

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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