-0.016 738 891 601 562 493 87 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 493 87(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
-0.016 738 891 601 562 493 87(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:

|-0.016 738 891 601 562 493 87| = 0.016 738 891 601 562 493 87


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: 0.016 738 891 601 562 493 87.

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) 0.016 738 891 601 562 493 87 × 2 = 0 + 0.033 477 783 203 124 987 74;
  • 2) 0.033 477 783 203 124 987 74 × 2 = 0 + 0.066 955 566 406 249 975 48;
  • 3) 0.066 955 566 406 249 975 48 × 2 = 0 + 0.133 911 132 812 499 950 96;
  • 4) 0.133 911 132 812 499 950 96 × 2 = 0 + 0.267 822 265 624 999 901 92;
  • 5) 0.267 822 265 624 999 901 92 × 2 = 0 + 0.535 644 531 249 999 803 84;
  • 6) 0.535 644 531 249 999 803 84 × 2 = 1 + 0.071 289 062 499 999 607 68;
  • 7) 0.071 289 062 499 999 607 68 × 2 = 0 + 0.142 578 124 999 999 215 36;
  • 8) 0.142 578 124 999 999 215 36 × 2 = 0 + 0.285 156 249 999 998 430 72;
  • 9) 0.285 156 249 999 998 430 72 × 2 = 0 + 0.570 312 499 999 996 861 44;
  • 10) 0.570 312 499 999 996 861 44 × 2 = 1 + 0.140 624 999 999 993 722 88;
  • 11) 0.140 624 999 999 993 722 88 × 2 = 0 + 0.281 249 999 999 987 445 76;
  • 12) 0.281 249 999 999 987 445 76 × 2 = 0 + 0.562 499 999 999 974 891 52;
  • 13) 0.562 499 999 999 974 891 52 × 2 = 1 + 0.124 999 999 999 949 783 04;
  • 14) 0.124 999 999 999 949 783 04 × 2 = 0 + 0.249 999 999 999 899 566 08;
  • 15) 0.249 999 999 999 899 566 08 × 2 = 0 + 0.499 999 999 999 799 132 16;
  • 16) 0.499 999 999 999 799 132 16 × 2 = 0 + 0.999 999 999 999 598 264 32;
  • 17) 0.999 999 999 999 598 264 32 × 2 = 1 + 0.999 999 999 999 196 528 64;
  • 18) 0.999 999 999 999 196 528 64 × 2 = 1 + 0.999 999 999 998 393 057 28;
  • 19) 0.999 999 999 998 393 057 28 × 2 = 1 + 0.999 999 999 996 786 114 56;
  • 20) 0.999 999 999 996 786 114 56 × 2 = 1 + 0.999 999 999 993 572 229 12;
  • 21) 0.999 999 999 993 572 229 12 × 2 = 1 + 0.999 999 999 987 144 458 24;
  • 22) 0.999 999 999 987 144 458 24 × 2 = 1 + 0.999 999 999 974 288 916 48;
  • 23) 0.999 999 999 974 288 916 48 × 2 = 1 + 0.999 999 999 948 577 832 96;
  • 24) 0.999 999 999 948 577 832 96 × 2 = 1 + 0.999 999 999 897 155 665 92;
  • 25) 0.999 999 999 897 155 665 92 × 2 = 1 + 0.999 999 999 794 311 331 84;
  • 26) 0.999 999 999 794 311 331 84 × 2 = 1 + 0.999 999 999 588 622 663 68;
  • 27) 0.999 999 999 588 622 663 68 × 2 = 1 + 0.999 999 999 177 245 327 36;
  • 28) 0.999 999 999 177 245 327 36 × 2 = 1 + 0.999 999 998 354 490 654 72;
  • 29) 0.999 999 998 354 490 654 72 × 2 = 1 + 0.999 999 996 708 981 309 44;
  • 30) 0.999 999 996 708 981 309 44 × 2 = 1 + 0.999 999 993 417 962 618 88;
  • 31) 0.999 999 993 417 962 618 88 × 2 = 1 + 0.999 999 986 835 925 237 76;
  • 32) 0.999 999 986 835 925 237 76 × 2 = 1 + 0.999 999 973 671 850 475 52;
  • 33) 0.999 999 973 671 850 475 52 × 2 = 1 + 0.999 999 947 343 700 951 04;
  • 34) 0.999 999 947 343 700 951 04 × 2 = 1 + 0.999 999 894 687 401 902 08;
  • 35) 0.999 999 894 687 401 902 08 × 2 = 1 + 0.999 999 789 374 803 804 16;
  • 36) 0.999 999 789 374 803 804 16 × 2 = 1 + 0.999 999 578 749 607 608 32;
  • 37) 0.999 999 578 749 607 608 32 × 2 = 1 + 0.999 999 157 499 215 216 64;
  • 38) 0.999 999 157 499 215 216 64 × 2 = 1 + 0.999 998 314 998 430 433 28;
  • 39) 0.999 998 314 998 430 433 28 × 2 = 1 + 0.999 996 629 996 860 866 56;
  • 40) 0.999 996 629 996 860 866 56 × 2 = 1 + 0.999 993 259 993 721 733 12;
  • 41) 0.999 993 259 993 721 733 12 × 2 = 1 + 0.999 986 519 987 443 466 24;
  • 42) 0.999 986 519 987 443 466 24 × 2 = 1 + 0.999 973 039 974 886 932 48;
  • 43) 0.999 973 039 974 886 932 48 × 2 = 1 + 0.999 946 079 949 773 864 96;
  • 44) 0.999 946 079 949 773 864 96 × 2 = 1 + 0.999 892 159 899 547 729 92;
  • 45) 0.999 892 159 899 547 729 92 × 2 = 1 + 0.999 784 319 799 095 459 84;
  • 46) 0.999 784 319 799 095 459 84 × 2 = 1 + 0.999 568 639 598 190 919 68;
  • 47) 0.999 568 639 598 190 919 68 × 2 = 1 + 0.999 137 279 196 381 839 36;
  • 48) 0.999 137 279 196 381 839 36 × 2 = 1 + 0.998 274 558 392 763 678 72;
  • 49) 0.998 274 558 392 763 678 72 × 2 = 1 + 0.996 549 116 785 527 357 44;
  • 50) 0.996 549 116 785 527 357 44 × 2 = 1 + 0.993 098 233 571 054 714 88;
  • 51) 0.993 098 233 571 054 714 88 × 2 = 1 + 0.986 196 467 142 109 429 76;
  • 52) 0.986 196 467 142 109 429 76 × 2 = 1 + 0.972 392 934 284 218 859 52;
  • 53) 0.972 392 934 284 218 859 52 × 2 = 1 + 0.944 785 868 568 437 719 04;
  • 54) 0.944 785 868 568 437 719 04 × 2 = 1 + 0.889 571 737 136 875 438 08;
  • 55) 0.889 571 737 136 875 438 08 × 2 = 1 + 0.779 143 474 273 750 876 16;
  • 56) 0.779 143 474 273 750 876 16 × 2 = 1 + 0.558 286 948 547 501 752 32;
  • 57) 0.558 286 948 547 501 752 32 × 2 = 1 + 0.116 573 897 095 003 504 64;
  • 58) 0.116 573 897 095 003 504 64 × 2 = 0 + 0.233 147 794 190 007 009 28;

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:


0.016 738 891 601 562 493 87(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 493 87(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 493 87(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 493 87 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 1110

Share

» Convert decimal -0.016 738 891 601 562 494 42 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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