-0.016 738 891 601 562 498 23 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 498 23(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
-0.016 738 891 601 562 498 23(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:

|-0.016 738 891 601 562 498 23| = 0.016 738 891 601 562 498 23


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: 0.016 738 891 601 562 498 23.

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) 0.016 738 891 601 562 498 23 × 2 = 0 + 0.033 477 783 203 124 996 46;
  • 2) 0.033 477 783 203 124 996 46 × 2 = 0 + 0.066 955 566 406 249 992 92;
  • 3) 0.066 955 566 406 249 992 92 × 2 = 0 + 0.133 911 132 812 499 985 84;
  • 4) 0.133 911 132 812 499 985 84 × 2 = 0 + 0.267 822 265 624 999 971 68;
  • 5) 0.267 822 265 624 999 971 68 × 2 = 0 + 0.535 644 531 249 999 943 36;
  • 6) 0.535 644 531 249 999 943 36 × 2 = 1 + 0.071 289 062 499 999 886 72;
  • 7) 0.071 289 062 499 999 886 72 × 2 = 0 + 0.142 578 124 999 999 773 44;
  • 8) 0.142 578 124 999 999 773 44 × 2 = 0 + 0.285 156 249 999 999 546 88;
  • 9) 0.285 156 249 999 999 546 88 × 2 = 0 + 0.570 312 499 999 999 093 76;
  • 10) 0.570 312 499 999 999 093 76 × 2 = 1 + 0.140 624 999 999 998 187 52;
  • 11) 0.140 624 999 999 998 187 52 × 2 = 0 + 0.281 249 999 999 996 375 04;
  • 12) 0.281 249 999 999 996 375 04 × 2 = 0 + 0.562 499 999 999 992 750 08;
  • 13) 0.562 499 999 999 992 750 08 × 2 = 1 + 0.124 999 999 999 985 500 16;
  • 14) 0.124 999 999 999 985 500 16 × 2 = 0 + 0.249 999 999 999 971 000 32;
  • 15) 0.249 999 999 999 971 000 32 × 2 = 0 + 0.499 999 999 999 942 000 64;
  • 16) 0.499 999 999 999 942 000 64 × 2 = 0 + 0.999 999 999 999 884 001 28;
  • 17) 0.999 999 999 999 884 001 28 × 2 = 1 + 0.999 999 999 999 768 002 56;
  • 18) 0.999 999 999 999 768 002 56 × 2 = 1 + 0.999 999 999 999 536 005 12;
  • 19) 0.999 999 999 999 536 005 12 × 2 = 1 + 0.999 999 999 999 072 010 24;
  • 20) 0.999 999 999 999 072 010 24 × 2 = 1 + 0.999 999 999 998 144 020 48;
  • 21) 0.999 999 999 998 144 020 48 × 2 = 1 + 0.999 999 999 996 288 040 96;
  • 22) 0.999 999 999 996 288 040 96 × 2 = 1 + 0.999 999 999 992 576 081 92;
  • 23) 0.999 999 999 992 576 081 92 × 2 = 1 + 0.999 999 999 985 152 163 84;
  • 24) 0.999 999 999 985 152 163 84 × 2 = 1 + 0.999 999 999 970 304 327 68;
  • 25) 0.999 999 999 970 304 327 68 × 2 = 1 + 0.999 999 999 940 608 655 36;
  • 26) 0.999 999 999 940 608 655 36 × 2 = 1 + 0.999 999 999 881 217 310 72;
  • 27) 0.999 999 999 881 217 310 72 × 2 = 1 + 0.999 999 999 762 434 621 44;
  • 28) 0.999 999 999 762 434 621 44 × 2 = 1 + 0.999 999 999 524 869 242 88;
  • 29) 0.999 999 999 524 869 242 88 × 2 = 1 + 0.999 999 999 049 738 485 76;
  • 30) 0.999 999 999 049 738 485 76 × 2 = 1 + 0.999 999 998 099 476 971 52;
  • 31) 0.999 999 998 099 476 971 52 × 2 = 1 + 0.999 999 996 198 953 943 04;
  • 32) 0.999 999 996 198 953 943 04 × 2 = 1 + 0.999 999 992 397 907 886 08;
  • 33) 0.999 999 992 397 907 886 08 × 2 = 1 + 0.999 999 984 795 815 772 16;
  • 34) 0.999 999 984 795 815 772 16 × 2 = 1 + 0.999 999 969 591 631 544 32;
  • 35) 0.999 999 969 591 631 544 32 × 2 = 1 + 0.999 999 939 183 263 088 64;
  • 36) 0.999 999 939 183 263 088 64 × 2 = 1 + 0.999 999 878 366 526 177 28;
  • 37) 0.999 999 878 366 526 177 28 × 2 = 1 + 0.999 999 756 733 052 354 56;
  • 38) 0.999 999 756 733 052 354 56 × 2 = 1 + 0.999 999 513 466 104 709 12;
  • 39) 0.999 999 513 466 104 709 12 × 2 = 1 + 0.999 999 026 932 209 418 24;
  • 40) 0.999 999 026 932 209 418 24 × 2 = 1 + 0.999 998 053 864 418 836 48;
  • 41) 0.999 998 053 864 418 836 48 × 2 = 1 + 0.999 996 107 728 837 672 96;
  • 42) 0.999 996 107 728 837 672 96 × 2 = 1 + 0.999 992 215 457 675 345 92;
  • 43) 0.999 992 215 457 675 345 92 × 2 = 1 + 0.999 984 430 915 350 691 84;
  • 44) 0.999 984 430 915 350 691 84 × 2 = 1 + 0.999 968 861 830 701 383 68;
  • 45) 0.999 968 861 830 701 383 68 × 2 = 1 + 0.999 937 723 661 402 767 36;
  • 46) 0.999 937 723 661 402 767 36 × 2 = 1 + 0.999 875 447 322 805 534 72;
  • 47) 0.999 875 447 322 805 534 72 × 2 = 1 + 0.999 750 894 645 611 069 44;
  • 48) 0.999 750 894 645 611 069 44 × 2 = 1 + 0.999 501 789 291 222 138 88;
  • 49) 0.999 501 789 291 222 138 88 × 2 = 1 + 0.999 003 578 582 444 277 76;
  • 50) 0.999 003 578 582 444 277 76 × 2 = 1 + 0.998 007 157 164 888 555 52;
  • 51) 0.998 007 157 164 888 555 52 × 2 = 1 + 0.996 014 314 329 777 111 04;
  • 52) 0.996 014 314 329 777 111 04 × 2 = 1 + 0.992 028 628 659 554 222 08;
  • 53) 0.992 028 628 659 554 222 08 × 2 = 1 + 0.984 057 257 319 108 444 16;
  • 54) 0.984 057 257 319 108 444 16 × 2 = 1 + 0.968 114 514 638 216 888 32;
  • 55) 0.968 114 514 638 216 888 32 × 2 = 1 + 0.936 229 029 276 433 776 64;
  • 56) 0.936 229 029 276 433 776 64 × 2 = 1 + 0.872 458 058 552 867 553 28;
  • 57) 0.872 458 058 552 867 553 28 × 2 = 1 + 0.744 916 117 105 735 106 56;
  • 58) 0.744 916 117 105 735 106 56 × 2 = 1 + 0.489 832 234 211 470 213 12;

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:


0.016 738 891 601 562 498 23(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 498 23(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:


0.016 738 891 601 562 498 23(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 -0.016 738 891 601 562 498 23 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

Share

» Convert decimal -0.016 738 891 601 562 497 84 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 minutes

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