-0.105 000 025 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.105 000 025 2(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
-0.105 000 025 2(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:

|-0.105 000 025 2| = 0.105 000 025 2


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: 0.105 000 025 2.

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) 0.105 000 025 2 × 2 = 0 + 0.210 000 050 4;
  • 2) 0.210 000 050 4 × 2 = 0 + 0.420 000 100 8;
  • 3) 0.420 000 100 8 × 2 = 0 + 0.840 000 201 6;
  • 4) 0.840 000 201 6 × 2 = 1 + 0.680 000 403 2;
  • 5) 0.680 000 403 2 × 2 = 1 + 0.360 000 806 4;
  • 6) 0.360 000 806 4 × 2 = 0 + 0.720 001 612 8;
  • 7) 0.720 001 612 8 × 2 = 1 + 0.440 003 225 6;
  • 8) 0.440 003 225 6 × 2 = 0 + 0.880 006 451 2;
  • 9) 0.880 006 451 2 × 2 = 1 + 0.760 012 902 4;
  • 10) 0.760 012 902 4 × 2 = 1 + 0.520 025 804 8;
  • 11) 0.520 025 804 8 × 2 = 1 + 0.040 051 609 6;
  • 12) 0.040 051 609 6 × 2 = 0 + 0.080 103 219 2;
  • 13) 0.080 103 219 2 × 2 = 0 + 0.160 206 438 4;
  • 14) 0.160 206 438 4 × 2 = 0 + 0.320 412 876 8;
  • 15) 0.320 412 876 8 × 2 = 0 + 0.640 825 753 6;
  • 16) 0.640 825 753 6 × 2 = 1 + 0.281 651 507 2;
  • 17) 0.281 651 507 2 × 2 = 0 + 0.563 303 014 4;
  • 18) 0.563 303 014 4 × 2 = 1 + 0.126 606 028 8;
  • 19) 0.126 606 028 8 × 2 = 0 + 0.253 212 057 6;
  • 20) 0.253 212 057 6 × 2 = 0 + 0.506 424 115 2;
  • 21) 0.506 424 115 2 × 2 = 1 + 0.012 848 230 4;
  • 22) 0.012 848 230 4 × 2 = 0 + 0.025 696 460 8;
  • 23) 0.025 696 460 8 × 2 = 0 + 0.051 392 921 6;
  • 24) 0.051 392 921 6 × 2 = 0 + 0.102 785 843 2;
  • 25) 0.102 785 843 2 × 2 = 0 + 0.205 571 686 4;
  • 26) 0.205 571 686 4 × 2 = 0 + 0.411 143 372 8;
  • 27) 0.411 143 372 8 × 2 = 0 + 0.822 286 745 6;
  • 28) 0.822 286 745 6 × 2 = 1 + 0.644 573 491 2;
  • 29) 0.644 573 491 2 × 2 = 1 + 0.289 146 982 4;
  • 30) 0.289 146 982 4 × 2 = 0 + 0.578 293 964 8;
  • 31) 0.578 293 964 8 × 2 = 1 + 0.156 587 929 6;
  • 32) 0.156 587 929 6 × 2 = 0 + 0.313 175 859 2;
  • 33) 0.313 175 859 2 × 2 = 0 + 0.626 351 718 4;
  • 34) 0.626 351 718 4 × 2 = 1 + 0.252 703 436 8;
  • 35) 0.252 703 436 8 × 2 = 0 + 0.505 406 873 6;
  • 36) 0.505 406 873 6 × 2 = 1 + 0.010 813 747 2;
  • 37) 0.010 813 747 2 × 2 = 0 + 0.021 627 494 4;
  • 38) 0.021 627 494 4 × 2 = 0 + 0.043 254 988 8;
  • 39) 0.043 254 988 8 × 2 = 0 + 0.086 509 977 6;
  • 40) 0.086 509 977 6 × 2 = 0 + 0.173 019 955 2;
  • 41) 0.173 019 955 2 × 2 = 0 + 0.346 039 910 4;
  • 42) 0.346 039 910 4 × 2 = 0 + 0.692 079 820 8;
  • 43) 0.692 079 820 8 × 2 = 1 + 0.384 159 641 6;
  • 44) 0.384 159 641 6 × 2 = 0 + 0.768 319 283 2;
  • 45) 0.768 319 283 2 × 2 = 1 + 0.536 638 566 4;
  • 46) 0.536 638 566 4 × 2 = 1 + 0.073 277 132 8;
  • 47) 0.073 277 132 8 × 2 = 0 + 0.146 554 265 6;
  • 48) 0.146 554 265 6 × 2 = 0 + 0.293 108 531 2;
  • 49) 0.293 108 531 2 × 2 = 0 + 0.586 217 062 4;
  • 50) 0.586 217 062 4 × 2 = 1 + 0.172 434 124 8;
  • 51) 0.172 434 124 8 × 2 = 0 + 0.344 868 249 6;
  • 52) 0.344 868 249 6 × 2 = 0 + 0.689 736 499 2;
  • 53) 0.689 736 499 2 × 2 = 1 + 0.379 472 998 4;
  • 54) 0.379 472 998 4 × 2 = 0 + 0.758 945 996 8;
  • 55) 0.758 945 996 8 × 2 = 1 + 0.517 891 993 6;
  • 56) 0.517 891 993 6 × 2 = 1 + 0.035 783 987 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:


0.105 000 025 2(10) =

0.0001 1010 1110 0001 0100 1000 0001 1010 0101 0000 0010 1100 0100 1011(2)

6. Positive number before normalization:

0.105 000 025 2(10) =

0.0001 1010 1110 0001 0100 1000 0001 1010 0101 0000 0010 1100 0100 1011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.105 000 025 2(10) =

0.0001 1010 1110 0001 0100 1000 0001 1010 0101 0000 0010 1100 0100 1011(2) =

0.0001 1010 1110 0001 0100 1000 0001 1010 0101 0000 0010 1100 0100 1011(2) × 20 =

1.1010 1110 0001 0100 1000 0001 1010 0101 0000 0010 1100 0100 1011(2) × 2-4


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):
1.1010 1110 0001 0100 1000 0001 1010 0101 0000 0010 1100 0100 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 1010 1110 0001 0100 1000 0001 1010 0101 0000 0010 1100 0100 1011 =

1010 1110 0001 0100 1000 0001 1010 0101 0000 0010 1100 0100 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) =
011 1111 1011

Mantissa (52 bits) =
1010 1110 0001 0100 1000 0001 1010 0101 0000 0010 1100 0100 1011


Decimal number -0.105 000 025 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1010 1110 0001 0100 1000 0001 1010 0101 0000 0010 1100 0100 1011

Share

» Convert decimal -0.105 000 032 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 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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