-1 036.699 999 999 999 818 100 949 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -1 036.699 999 999 999 818 100 949(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
-1 036.699 999 999 999 818 100 949(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:

|-1 036.699 999 999 999 818 100 949| = 1 036.699 999 999 999 818 100 949


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: 0.699 999 999 999 818 100 949.

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) 0.699 999 999 999 818 100 949 × 2 = 1 + 0.399 999 999 999 636 201 898;
  • 2) 0.399 999 999 999 636 201 898 × 2 = 0 + 0.799 999 999 999 272 403 796;
  • 3) 0.799 999 999 999 272 403 796 × 2 = 1 + 0.599 999 999 998 544 807 592;
  • 4) 0.599 999 999 998 544 807 592 × 2 = 1 + 0.199 999 999 997 089 615 184;
  • 5) 0.199 999 999 997 089 615 184 × 2 = 0 + 0.399 999 999 994 179 230 368;
  • 6) 0.399 999 999 994 179 230 368 × 2 = 0 + 0.799 999 999 988 358 460 736;
  • 7) 0.799 999 999 988 358 460 736 × 2 = 1 + 0.599 999 999 976 716 921 472;
  • 8) 0.599 999 999 976 716 921 472 × 2 = 1 + 0.199 999 999 953 433 842 944;
  • 9) 0.199 999 999 953 433 842 944 × 2 = 0 + 0.399 999 999 906 867 685 888;
  • 10) 0.399 999 999 906 867 685 888 × 2 = 0 + 0.799 999 999 813 735 371 776;
  • 11) 0.799 999 999 813 735 371 776 × 2 = 1 + 0.599 999 999 627 470 743 552;
  • 12) 0.599 999 999 627 470 743 552 × 2 = 1 + 0.199 999 999 254 941 487 104;
  • 13) 0.199 999 999 254 941 487 104 × 2 = 0 + 0.399 999 998 509 882 974 208;
  • 14) 0.399 999 998 509 882 974 208 × 2 = 0 + 0.799 999 997 019 765 948 416;
  • 15) 0.799 999 997 019 765 948 416 × 2 = 1 + 0.599 999 994 039 531 896 832;
  • 16) 0.599 999 994 039 531 896 832 × 2 = 1 + 0.199 999 988 079 063 793 664;
  • 17) 0.199 999 988 079 063 793 664 × 2 = 0 + 0.399 999 976 158 127 587 328;
  • 18) 0.399 999 976 158 127 587 328 × 2 = 0 + 0.799 999 952 316 255 174 656;
  • 19) 0.799 999 952 316 255 174 656 × 2 = 1 + 0.599 999 904 632 510 349 312;
  • 20) 0.599 999 904 632 510 349 312 × 2 = 1 + 0.199 999 809 265 020 698 624;
  • 21) 0.199 999 809 265 020 698 624 × 2 = 0 + 0.399 999 618 530 041 397 248;
  • 22) 0.399 999 618 530 041 397 248 × 2 = 0 + 0.799 999 237 060 082 794 496;
  • 23) 0.799 999 237 060 082 794 496 × 2 = 1 + 0.599 998 474 120 165 588 992;
  • 24) 0.599 998 474 120 165 588 992 × 2 = 1 + 0.199 996 948 240 331 177 984;
  • 25) 0.199 996 948 240 331 177 984 × 2 = 0 + 0.399 993 896 480 662 355 968;
  • 26) 0.399 993 896 480 662 355 968 × 2 = 0 + 0.799 987 792 961 324 711 936;
  • 27) 0.799 987 792 961 324 711 936 × 2 = 1 + 0.599 975 585 922 649 423 872;
  • 28) 0.599 975 585 922 649 423 872 × 2 = 1 + 0.199 951 171 845 298 847 744;
  • 29) 0.199 951 171 845 298 847 744 × 2 = 0 + 0.399 902 343 690 597 695 488;
  • 30) 0.399 902 343 690 597 695 488 × 2 = 0 + 0.799 804 687 381 195 390 976;
  • 31) 0.799 804 687 381 195 390 976 × 2 = 1 + 0.599 609 374 762 390 781 952;
  • 32) 0.599 609 374 762 390 781 952 × 2 = 1 + 0.199 218 749 524 781 563 904;
  • 33) 0.199 218 749 524 781 563 904 × 2 = 0 + 0.398 437 499 049 563 127 808;
  • 34) 0.398 437 499 049 563 127 808 × 2 = 0 + 0.796 874 998 099 126 255 616;
  • 35) 0.796 874 998 099 126 255 616 × 2 = 1 + 0.593 749 996 198 252 511 232;
  • 36) 0.593 749 996 198 252 511 232 × 2 = 1 + 0.187 499 992 396 505 022 464;
  • 37) 0.187 499 992 396 505 022 464 × 2 = 0 + 0.374 999 984 793 010 044 928;
  • 38) 0.374 999 984 793 010 044 928 × 2 = 0 + 0.749 999 969 586 020 089 856;
  • 39) 0.749 999 969 586 020 089 856 × 2 = 1 + 0.499 999 939 172 040 179 712;
  • 40) 0.499 999 939 172 040 179 712 × 2 = 0 + 0.999 999 878 344 080 359 424;
  • 41) 0.999 999 878 344 080 359 424 × 2 = 1 + 0.999 999 756 688 160 718 848;
  • 42) 0.999 999 756 688 160 718 848 × 2 = 1 + 0.999 999 513 376 321 437 696;
  • 43) 0.999 999 513 376 321 437 696 × 2 = 1 + 0.999 999 026 752 642 875 392;
  • 44) 0.999 999 026 752 642 875 392 × 2 = 1 + 0.999 998 053 505 285 750 784;
  • 45) 0.999 998 053 505 285 750 784 × 2 = 1 + 0.999 996 107 010 571 501 568;
  • 46) 0.999 996 107 010 571 501 568 × 2 = 1 + 0.999 992 214 021 143 003 136;
  • 47) 0.999 992 214 021 143 003 136 × 2 = 1 + 0.999 984 428 042 286 006 272;
  • 48) 0.999 984 428 042 286 006 272 × 2 = 1 + 0.999 968 856 084 572 012 544;
  • 49) 0.999 968 856 084 572 012 544 × 2 = 1 + 0.999 937 712 169 144 025 088;
  • 50) 0.999 937 712 169 144 025 088 × 2 = 1 + 0.999 875 424 338 288 050 176;
  • 51) 0.999 875 424 338 288 050 176 × 2 = 1 + 0.999 750 848 676 576 100 352;
  • 52) 0.999 750 848 676 576 100 352 × 2 = 1 + 0.999 501 697 353 152 200 704;
  • 53) 0.999 501 697 353 152 200 704 × 2 = 1 + 0.999 003 394 706 304 401 408;

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:


0.699 999 999 999 818 100 949(10) =

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

6. Positive number before normalization:

1 036.699 999 999 999 818 100 949(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:


1 036.699 999 999 999 818 100 949(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 -1 036.699 999 999 999 818 100 949 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