-1 036.699 999 999 999 818 101 059 341 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -1 036.699 999 999 999 818 101 059 341(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
-1 036.699 999 999 999 818 101 059 341(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:

|-1 036.699 999 999 999 818 101 059 341| = 1 036.699 999 999 999 818 101 059 341


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: 0.699 999 999 999 818 101 059 341.

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) 0.699 999 999 999 818 101 059 341 × 2 = 1 + 0.399 999 999 999 636 202 118 682;
  • 2) 0.399 999 999 999 636 202 118 682 × 2 = 0 + 0.799 999 999 999 272 404 237 364;
  • 3) 0.799 999 999 999 272 404 237 364 × 2 = 1 + 0.599 999 999 998 544 808 474 728;
  • 4) 0.599 999 999 998 544 808 474 728 × 2 = 1 + 0.199 999 999 997 089 616 949 456;
  • 5) 0.199 999 999 997 089 616 949 456 × 2 = 0 + 0.399 999 999 994 179 233 898 912;
  • 6) 0.399 999 999 994 179 233 898 912 × 2 = 0 + 0.799 999 999 988 358 467 797 824;
  • 7) 0.799 999 999 988 358 467 797 824 × 2 = 1 + 0.599 999 999 976 716 935 595 648;
  • 8) 0.599 999 999 976 716 935 595 648 × 2 = 1 + 0.199 999 999 953 433 871 191 296;
  • 9) 0.199 999 999 953 433 871 191 296 × 2 = 0 + 0.399 999 999 906 867 742 382 592;
  • 10) 0.399 999 999 906 867 742 382 592 × 2 = 0 + 0.799 999 999 813 735 484 765 184;
  • 11) 0.799 999 999 813 735 484 765 184 × 2 = 1 + 0.599 999 999 627 470 969 530 368;
  • 12) 0.599 999 999 627 470 969 530 368 × 2 = 1 + 0.199 999 999 254 941 939 060 736;
  • 13) 0.199 999 999 254 941 939 060 736 × 2 = 0 + 0.399 999 998 509 883 878 121 472;
  • 14) 0.399 999 998 509 883 878 121 472 × 2 = 0 + 0.799 999 997 019 767 756 242 944;
  • 15) 0.799 999 997 019 767 756 242 944 × 2 = 1 + 0.599 999 994 039 535 512 485 888;
  • 16) 0.599 999 994 039 535 512 485 888 × 2 = 1 + 0.199 999 988 079 071 024 971 776;
  • 17) 0.199 999 988 079 071 024 971 776 × 2 = 0 + 0.399 999 976 158 142 049 943 552;
  • 18) 0.399 999 976 158 142 049 943 552 × 2 = 0 + 0.799 999 952 316 284 099 887 104;
  • 19) 0.799 999 952 316 284 099 887 104 × 2 = 1 + 0.599 999 904 632 568 199 774 208;
  • 20) 0.599 999 904 632 568 199 774 208 × 2 = 1 + 0.199 999 809 265 136 399 548 416;
  • 21) 0.199 999 809 265 136 399 548 416 × 2 = 0 + 0.399 999 618 530 272 799 096 832;
  • 22) 0.399 999 618 530 272 799 096 832 × 2 = 0 + 0.799 999 237 060 545 598 193 664;
  • 23) 0.799 999 237 060 545 598 193 664 × 2 = 1 + 0.599 998 474 121 091 196 387 328;
  • 24) 0.599 998 474 121 091 196 387 328 × 2 = 1 + 0.199 996 948 242 182 392 774 656;
  • 25) 0.199 996 948 242 182 392 774 656 × 2 = 0 + 0.399 993 896 484 364 785 549 312;
  • 26) 0.399 993 896 484 364 785 549 312 × 2 = 0 + 0.799 987 792 968 729 571 098 624;
  • 27) 0.799 987 792 968 729 571 098 624 × 2 = 1 + 0.599 975 585 937 459 142 197 248;
  • 28) 0.599 975 585 937 459 142 197 248 × 2 = 1 + 0.199 951 171 874 918 284 394 496;
  • 29) 0.199 951 171 874 918 284 394 496 × 2 = 0 + 0.399 902 343 749 836 568 788 992;
  • 30) 0.399 902 343 749 836 568 788 992 × 2 = 0 + 0.799 804 687 499 673 137 577 984;
  • 31) 0.799 804 687 499 673 137 577 984 × 2 = 1 + 0.599 609 374 999 346 275 155 968;
  • 32) 0.599 609 374 999 346 275 155 968 × 2 = 1 + 0.199 218 749 998 692 550 311 936;
  • 33) 0.199 218 749 998 692 550 311 936 × 2 = 0 + 0.398 437 499 997 385 100 623 872;
  • 34) 0.398 437 499 997 385 100 623 872 × 2 = 0 + 0.796 874 999 994 770 201 247 744;
  • 35) 0.796 874 999 994 770 201 247 744 × 2 = 1 + 0.593 749 999 989 540 402 495 488;
  • 36) 0.593 749 999 989 540 402 495 488 × 2 = 1 + 0.187 499 999 979 080 804 990 976;
  • 37) 0.187 499 999 979 080 804 990 976 × 2 = 0 + 0.374 999 999 958 161 609 981 952;
  • 38) 0.374 999 999 958 161 609 981 952 × 2 = 0 + 0.749 999 999 916 323 219 963 904;
  • 39) 0.749 999 999 916 323 219 963 904 × 2 = 1 + 0.499 999 999 832 646 439 927 808;
  • 40) 0.499 999 999 832 646 439 927 808 × 2 = 0 + 0.999 999 999 665 292 879 855 616;
  • 41) 0.999 999 999 665 292 879 855 616 × 2 = 1 + 0.999 999 999 330 585 759 711 232;
  • 42) 0.999 999 999 330 585 759 711 232 × 2 = 1 + 0.999 999 998 661 171 519 422 464;
  • 43) 0.999 999 998 661 171 519 422 464 × 2 = 1 + 0.999 999 997 322 343 038 844 928;
  • 44) 0.999 999 997 322 343 038 844 928 × 2 = 1 + 0.999 999 994 644 686 077 689 856;
  • 45) 0.999 999 994 644 686 077 689 856 × 2 = 1 + 0.999 999 989 289 372 155 379 712;
  • 46) 0.999 999 989 289 372 155 379 712 × 2 = 1 + 0.999 999 978 578 744 310 759 424;
  • 47) 0.999 999 978 578 744 310 759 424 × 2 = 1 + 0.999 999 957 157 488 621 518 848;
  • 48) 0.999 999 957 157 488 621 518 848 × 2 = 1 + 0.999 999 914 314 977 243 037 696;
  • 49) 0.999 999 914 314 977 243 037 696 × 2 = 1 + 0.999 999 828 629 954 486 075 392;
  • 50) 0.999 999 828 629 954 486 075 392 × 2 = 1 + 0.999 999 657 259 908 972 150 784;
  • 51) 0.999 999 657 259 908 972 150 784 × 2 = 1 + 0.999 999 314 519 817 944 301 568;
  • 52) 0.999 999 314 519 817 944 301 568 × 2 = 1 + 0.999 998 629 039 635 888 603 136;
  • 53) 0.999 998 629 039 635 888 603 136 × 2 = 1 + 0.999 997 258 079 271 777 206 272;

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:


0.699 999 999 999 818 101 059 341(10) =

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

6. Positive number before normalization:

1 036.699 999 999 999 818 101 059 341(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:


1 036.699 999 999 999 818 101 059 341(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 -1 036.699 999 999 999 818 101 059 341 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