-0.016 738 891 601 562 496 530 88 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 88(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
-0.016 738 891 601 562 496 530 88(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:

|-0.016 738 891 601 562 496 530 88| = 0.016 738 891 601 562 496 530 88


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 496 530 88.

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) 0.016 738 891 601 562 496 530 88 × 2 = 0 + 0.033 477 783 203 124 993 061 76;
  • 2) 0.033 477 783 203 124 993 061 76 × 2 = 0 + 0.066 955 566 406 249 986 123 52;
  • 3) 0.066 955 566 406 249 986 123 52 × 2 = 0 + 0.133 911 132 812 499 972 247 04;
  • 4) 0.133 911 132 812 499 972 247 04 × 2 = 0 + 0.267 822 265 624 999 944 494 08;
  • 5) 0.267 822 265 624 999 944 494 08 × 2 = 0 + 0.535 644 531 249 999 888 988 16;
  • 6) 0.535 644 531 249 999 888 988 16 × 2 = 1 + 0.071 289 062 499 999 777 976 32;
  • 7) 0.071 289 062 499 999 777 976 32 × 2 = 0 + 0.142 578 124 999 999 555 952 64;
  • 8) 0.142 578 124 999 999 555 952 64 × 2 = 0 + 0.285 156 249 999 999 111 905 28;
  • 9) 0.285 156 249 999 999 111 905 28 × 2 = 0 + 0.570 312 499 999 998 223 810 56;
  • 10) 0.570 312 499 999 998 223 810 56 × 2 = 1 + 0.140 624 999 999 996 447 621 12;
  • 11) 0.140 624 999 999 996 447 621 12 × 2 = 0 + 0.281 249 999 999 992 895 242 24;
  • 12) 0.281 249 999 999 992 895 242 24 × 2 = 0 + 0.562 499 999 999 985 790 484 48;
  • 13) 0.562 499 999 999 985 790 484 48 × 2 = 1 + 0.124 999 999 999 971 580 968 96;
  • 14) 0.124 999 999 999 971 580 968 96 × 2 = 0 + 0.249 999 999 999 943 161 937 92;
  • 15) 0.249 999 999 999 943 161 937 92 × 2 = 0 + 0.499 999 999 999 886 323 875 84;
  • 16) 0.499 999 999 999 886 323 875 84 × 2 = 0 + 0.999 999 999 999 772 647 751 68;
  • 17) 0.999 999 999 999 772 647 751 68 × 2 = 1 + 0.999 999 999 999 545 295 503 36;
  • 18) 0.999 999 999 999 545 295 503 36 × 2 = 1 + 0.999 999 999 999 090 591 006 72;
  • 19) 0.999 999 999 999 090 591 006 72 × 2 = 1 + 0.999 999 999 998 181 182 013 44;
  • 20) 0.999 999 999 998 181 182 013 44 × 2 = 1 + 0.999 999 999 996 362 364 026 88;
  • 21) 0.999 999 999 996 362 364 026 88 × 2 = 1 + 0.999 999 999 992 724 728 053 76;
  • 22) 0.999 999 999 992 724 728 053 76 × 2 = 1 + 0.999 999 999 985 449 456 107 52;
  • 23) 0.999 999 999 985 449 456 107 52 × 2 = 1 + 0.999 999 999 970 898 912 215 04;
  • 24) 0.999 999 999 970 898 912 215 04 × 2 = 1 + 0.999 999 999 941 797 824 430 08;
  • 25) 0.999 999 999 941 797 824 430 08 × 2 = 1 + 0.999 999 999 883 595 648 860 16;
  • 26) 0.999 999 999 883 595 648 860 16 × 2 = 1 + 0.999 999 999 767 191 297 720 32;
  • 27) 0.999 999 999 767 191 297 720 32 × 2 = 1 + 0.999 999 999 534 382 595 440 64;
  • 28) 0.999 999 999 534 382 595 440 64 × 2 = 1 + 0.999 999 999 068 765 190 881 28;
  • 29) 0.999 999 999 068 765 190 881 28 × 2 = 1 + 0.999 999 998 137 530 381 762 56;
  • 30) 0.999 999 998 137 530 381 762 56 × 2 = 1 + 0.999 999 996 275 060 763 525 12;
  • 31) 0.999 999 996 275 060 763 525 12 × 2 = 1 + 0.999 999 992 550 121 527 050 24;
  • 32) 0.999 999 992 550 121 527 050 24 × 2 = 1 + 0.999 999 985 100 243 054 100 48;
  • 33) 0.999 999 985 100 243 054 100 48 × 2 = 1 + 0.999 999 970 200 486 108 200 96;
  • 34) 0.999 999 970 200 486 108 200 96 × 2 = 1 + 0.999 999 940 400 972 216 401 92;
  • 35) 0.999 999 940 400 972 216 401 92 × 2 = 1 + 0.999 999 880 801 944 432 803 84;
  • 36) 0.999 999 880 801 944 432 803 84 × 2 = 1 + 0.999 999 761 603 888 865 607 68;
  • 37) 0.999 999 761 603 888 865 607 68 × 2 = 1 + 0.999 999 523 207 777 731 215 36;
  • 38) 0.999 999 523 207 777 731 215 36 × 2 = 1 + 0.999 999 046 415 555 462 430 72;
  • 39) 0.999 999 046 415 555 462 430 72 × 2 = 1 + 0.999 998 092 831 110 924 861 44;
  • 40) 0.999 998 092 831 110 924 861 44 × 2 = 1 + 0.999 996 185 662 221 849 722 88;
  • 41) 0.999 996 185 662 221 849 722 88 × 2 = 1 + 0.999 992 371 324 443 699 445 76;
  • 42) 0.999 992 371 324 443 699 445 76 × 2 = 1 + 0.999 984 742 648 887 398 891 52;
  • 43) 0.999 984 742 648 887 398 891 52 × 2 = 1 + 0.999 969 485 297 774 797 783 04;
  • 44) 0.999 969 485 297 774 797 783 04 × 2 = 1 + 0.999 938 970 595 549 595 566 08;
  • 45) 0.999 938 970 595 549 595 566 08 × 2 = 1 + 0.999 877 941 191 099 191 132 16;
  • 46) 0.999 877 941 191 099 191 132 16 × 2 = 1 + 0.999 755 882 382 198 382 264 32;
  • 47) 0.999 755 882 382 198 382 264 32 × 2 = 1 + 0.999 511 764 764 396 764 528 64;
  • 48) 0.999 511 764 764 396 764 528 64 × 2 = 1 + 0.999 023 529 528 793 529 057 28;
  • 49) 0.999 023 529 528 793 529 057 28 × 2 = 1 + 0.998 047 059 057 587 058 114 56;
  • 50) 0.998 047 059 057 587 058 114 56 × 2 = 1 + 0.996 094 118 115 174 116 229 12;
  • 51) 0.996 094 118 115 174 116 229 12 × 2 = 1 + 0.992 188 236 230 348 232 458 24;
  • 52) 0.992 188 236 230 348 232 458 24 × 2 = 1 + 0.984 376 472 460 696 464 916 48;
  • 53) 0.984 376 472 460 696 464 916 48 × 2 = 1 + 0.968 752 944 921 392 929 832 96;
  • 54) 0.968 752 944 921 392 929 832 96 × 2 = 1 + 0.937 505 889 842 785 859 665 92;
  • 55) 0.937 505 889 842 785 859 665 92 × 2 = 1 + 0.875 011 779 685 571 719 331 84;
  • 56) 0.875 011 779 685 571 719 331 84 × 2 = 1 + 0.750 023 559 371 143 438 663 68;
  • 57) 0.750 023 559 371 143 438 663 68 × 2 = 1 + 0.500 047 118 742 286 877 327 36;
  • 58) 0.500 047 118 742 286 877 327 36 × 2 = 1 + 0.000 094 237 484 573 754 654 72;

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:


0.016 738 891 601 562 496 530 88(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 88(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 530 88(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1017(10) =

011 1111 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number -0.016 738 891 601 562 496 530 88 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

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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