-0.016 738 891 601 562 493 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 493 3(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 3(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 3| = 0.016 738 891 601 562 493 3


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 3.

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 3 × 2 = 0 + 0.033 477 783 203 124 986 6;
  • 2) 0.033 477 783 203 124 986 6 × 2 = 0 + 0.066 955 566 406 249 973 2;
  • 3) 0.066 955 566 406 249 973 2 × 2 = 0 + 0.133 911 132 812 499 946 4;
  • 4) 0.133 911 132 812 499 946 4 × 2 = 0 + 0.267 822 265 624 999 892 8;
  • 5) 0.267 822 265 624 999 892 8 × 2 = 0 + 0.535 644 531 249 999 785 6;
  • 6) 0.535 644 531 249 999 785 6 × 2 = 1 + 0.071 289 062 499 999 571 2;
  • 7) 0.071 289 062 499 999 571 2 × 2 = 0 + 0.142 578 124 999 999 142 4;
  • 8) 0.142 578 124 999 999 142 4 × 2 = 0 + 0.285 156 249 999 998 284 8;
  • 9) 0.285 156 249 999 998 284 8 × 2 = 0 + 0.570 312 499 999 996 569 6;
  • 10) 0.570 312 499 999 996 569 6 × 2 = 1 + 0.140 624 999 999 993 139 2;
  • 11) 0.140 624 999 999 993 139 2 × 2 = 0 + 0.281 249 999 999 986 278 4;
  • 12) 0.281 249 999 999 986 278 4 × 2 = 0 + 0.562 499 999 999 972 556 8;
  • 13) 0.562 499 999 999 972 556 8 × 2 = 1 + 0.124 999 999 999 945 113 6;
  • 14) 0.124 999 999 999 945 113 6 × 2 = 0 + 0.249 999 999 999 890 227 2;
  • 15) 0.249 999 999 999 890 227 2 × 2 = 0 + 0.499 999 999 999 780 454 4;
  • 16) 0.499 999 999 999 780 454 4 × 2 = 0 + 0.999 999 999 999 560 908 8;
  • 17) 0.999 999 999 999 560 908 8 × 2 = 1 + 0.999 999 999 999 121 817 6;
  • 18) 0.999 999 999 999 121 817 6 × 2 = 1 + 0.999 999 999 998 243 635 2;
  • 19) 0.999 999 999 998 243 635 2 × 2 = 1 + 0.999 999 999 996 487 270 4;
  • 20) 0.999 999 999 996 487 270 4 × 2 = 1 + 0.999 999 999 992 974 540 8;
  • 21) 0.999 999 999 992 974 540 8 × 2 = 1 + 0.999 999 999 985 949 081 6;
  • 22) 0.999 999 999 985 949 081 6 × 2 = 1 + 0.999 999 999 971 898 163 2;
  • 23) 0.999 999 999 971 898 163 2 × 2 = 1 + 0.999 999 999 943 796 326 4;
  • 24) 0.999 999 999 943 796 326 4 × 2 = 1 + 0.999 999 999 887 592 652 8;
  • 25) 0.999 999 999 887 592 652 8 × 2 = 1 + 0.999 999 999 775 185 305 6;
  • 26) 0.999 999 999 775 185 305 6 × 2 = 1 + 0.999 999 999 550 370 611 2;
  • 27) 0.999 999 999 550 370 611 2 × 2 = 1 + 0.999 999 999 100 741 222 4;
  • 28) 0.999 999 999 100 741 222 4 × 2 = 1 + 0.999 999 998 201 482 444 8;
  • 29) 0.999 999 998 201 482 444 8 × 2 = 1 + 0.999 999 996 402 964 889 6;
  • 30) 0.999 999 996 402 964 889 6 × 2 = 1 + 0.999 999 992 805 929 779 2;
  • 31) 0.999 999 992 805 929 779 2 × 2 = 1 + 0.999 999 985 611 859 558 4;
  • 32) 0.999 999 985 611 859 558 4 × 2 = 1 + 0.999 999 971 223 719 116 8;
  • 33) 0.999 999 971 223 719 116 8 × 2 = 1 + 0.999 999 942 447 438 233 6;
  • 34) 0.999 999 942 447 438 233 6 × 2 = 1 + 0.999 999 884 894 876 467 2;
  • 35) 0.999 999 884 894 876 467 2 × 2 = 1 + 0.999 999 769 789 752 934 4;
  • 36) 0.999 999 769 789 752 934 4 × 2 = 1 + 0.999 999 539 579 505 868 8;
  • 37) 0.999 999 539 579 505 868 8 × 2 = 1 + 0.999 999 079 159 011 737 6;
  • 38) 0.999 999 079 159 011 737 6 × 2 = 1 + 0.999 998 158 318 023 475 2;
  • 39) 0.999 998 158 318 023 475 2 × 2 = 1 + 0.999 996 316 636 046 950 4;
  • 40) 0.999 996 316 636 046 950 4 × 2 = 1 + 0.999 992 633 272 093 900 8;
  • 41) 0.999 992 633 272 093 900 8 × 2 = 1 + 0.999 985 266 544 187 801 6;
  • 42) 0.999 985 266 544 187 801 6 × 2 = 1 + 0.999 970 533 088 375 603 2;
  • 43) 0.999 970 533 088 375 603 2 × 2 = 1 + 0.999 941 066 176 751 206 4;
  • 44) 0.999 941 066 176 751 206 4 × 2 = 1 + 0.999 882 132 353 502 412 8;
  • 45) 0.999 882 132 353 502 412 8 × 2 = 1 + 0.999 764 264 707 004 825 6;
  • 46) 0.999 764 264 707 004 825 6 × 2 = 1 + 0.999 528 529 414 009 651 2;
  • 47) 0.999 528 529 414 009 651 2 × 2 = 1 + 0.999 057 058 828 019 302 4;
  • 48) 0.999 057 058 828 019 302 4 × 2 = 1 + 0.998 114 117 656 038 604 8;
  • 49) 0.998 114 117 656 038 604 8 × 2 = 1 + 0.996 228 235 312 077 209 6;
  • 50) 0.996 228 235 312 077 209 6 × 2 = 1 + 0.992 456 470 624 154 419 2;
  • 51) 0.992 456 470 624 154 419 2 × 2 = 1 + 0.984 912 941 248 308 838 4;
  • 52) 0.984 912 941 248 308 838 4 × 2 = 1 + 0.969 825 882 496 617 676 8;
  • 53) 0.969 825 882 496 617 676 8 × 2 = 1 + 0.939 651 764 993 235 353 6;
  • 54) 0.939 651 764 993 235 353 6 × 2 = 1 + 0.879 303 529 986 470 707 2;
  • 55) 0.879 303 529 986 470 707 2 × 2 = 1 + 0.758 607 059 972 941 414 4;
  • 56) 0.758 607 059 972 941 414 4 × 2 = 1 + 0.517 214 119 945 882 828 8;
  • 57) 0.517 214 119 945 882 828 8 × 2 = 1 + 0.034 428 239 891 765 657 6;
  • 58) 0.034 428 239 891 765 657 6 × 2 = 0 + 0.068 856 479 783 531 315 2;

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 3(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 3(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 3(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 3 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 -0.016 738 891 601 562 495 1 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: Pull vs. Push Leadership: The Invisible Power. NotebookLM YouTube Video of 11.31 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