-0.000 280 58 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 280 58(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
-0.000 280 58(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:

|-0.000 280 58| = 0.000 280 58


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: 0.000 280 58.

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) 0.000 280 58 × 2 = 0 + 0.000 561 16;
  • 2) 0.000 561 16 × 2 = 0 + 0.001 122 32;
  • 3) 0.001 122 32 × 2 = 0 + 0.002 244 64;
  • 4) 0.002 244 64 × 2 = 0 + 0.004 489 28;
  • 5) 0.004 489 28 × 2 = 0 + 0.008 978 56;
  • 6) 0.008 978 56 × 2 = 0 + 0.017 957 12;
  • 7) 0.017 957 12 × 2 = 0 + 0.035 914 24;
  • 8) 0.035 914 24 × 2 = 0 + 0.071 828 48;
  • 9) 0.071 828 48 × 2 = 0 + 0.143 656 96;
  • 10) 0.143 656 96 × 2 = 0 + 0.287 313 92;
  • 11) 0.287 313 92 × 2 = 0 + 0.574 627 84;
  • 12) 0.574 627 84 × 2 = 1 + 0.149 255 68;
  • 13) 0.149 255 68 × 2 = 0 + 0.298 511 36;
  • 14) 0.298 511 36 × 2 = 0 + 0.597 022 72;
  • 15) 0.597 022 72 × 2 = 1 + 0.194 045 44;
  • 16) 0.194 045 44 × 2 = 0 + 0.388 090 88;
  • 17) 0.388 090 88 × 2 = 0 + 0.776 181 76;
  • 18) 0.776 181 76 × 2 = 1 + 0.552 363 52;
  • 19) 0.552 363 52 × 2 = 1 + 0.104 727 04;
  • 20) 0.104 727 04 × 2 = 0 + 0.209 454 08;
  • 21) 0.209 454 08 × 2 = 0 + 0.418 908 16;
  • 22) 0.418 908 16 × 2 = 0 + 0.837 816 32;
  • 23) 0.837 816 32 × 2 = 1 + 0.675 632 64;
  • 24) 0.675 632 64 × 2 = 1 + 0.351 265 28;
  • 25) 0.351 265 28 × 2 = 0 + 0.702 530 56;
  • 26) 0.702 530 56 × 2 = 1 + 0.405 061 12;
  • 27) 0.405 061 12 × 2 = 0 + 0.810 122 24;
  • 28) 0.810 122 24 × 2 = 1 + 0.620 244 48;
  • 29) 0.620 244 48 × 2 = 1 + 0.240 488 96;
  • 30) 0.240 488 96 × 2 = 0 + 0.480 977 92;
  • 31) 0.480 977 92 × 2 = 0 + 0.961 955 84;
  • 32) 0.961 955 84 × 2 = 1 + 0.923 911 68;
  • 33) 0.923 911 68 × 2 = 1 + 0.847 823 36;
  • 34) 0.847 823 36 × 2 = 1 + 0.695 646 72;
  • 35) 0.695 646 72 × 2 = 1 + 0.391 293 44;
  • 36) 0.391 293 44 × 2 = 0 + 0.782 586 88;
  • 37) 0.782 586 88 × 2 = 1 + 0.565 173 76;
  • 38) 0.565 173 76 × 2 = 1 + 0.130 347 52;
  • 39) 0.130 347 52 × 2 = 0 + 0.260 695 04;
  • 40) 0.260 695 04 × 2 = 0 + 0.521 390 08;
  • 41) 0.521 390 08 × 2 = 1 + 0.042 780 16;
  • 42) 0.042 780 16 × 2 = 0 + 0.085 560 32;
  • 43) 0.085 560 32 × 2 = 0 + 0.171 120 64;
  • 44) 0.171 120 64 × 2 = 0 + 0.342 241 28;
  • 45) 0.342 241 28 × 2 = 0 + 0.684 482 56;
  • 46) 0.684 482 56 × 2 = 1 + 0.368 965 12;
  • 47) 0.368 965 12 × 2 = 0 + 0.737 930 24;
  • 48) 0.737 930 24 × 2 = 1 + 0.475 860 48;
  • 49) 0.475 860 48 × 2 = 0 + 0.951 720 96;
  • 50) 0.951 720 96 × 2 = 1 + 0.903 441 92;
  • 51) 0.903 441 92 × 2 = 1 + 0.806 883 84;
  • 52) 0.806 883 84 × 2 = 1 + 0.613 767 68;
  • 53) 0.613 767 68 × 2 = 1 + 0.227 535 36;
  • 54) 0.227 535 36 × 2 = 0 + 0.455 070 72;
  • 55) 0.455 070 72 × 2 = 0 + 0.910 141 44;
  • 56) 0.910 141 44 × 2 = 1 + 0.820 282 88;
  • 57) 0.820 282 88 × 2 = 1 + 0.640 565 76;
  • 58) 0.640 565 76 × 2 = 1 + 0.281 131 52;
  • 59) 0.281 131 52 × 2 = 0 + 0.562 263 04;
  • 60) 0.562 263 04 × 2 = 1 + 0.124 526 08;
  • 61) 0.124 526 08 × 2 = 0 + 0.249 052 16;
  • 62) 0.249 052 16 × 2 = 0 + 0.498 104 32;
  • 63) 0.498 104 32 × 2 = 0 + 0.996 208 64;
  • 64) 0.996 208 64 × 2 = 1 + 0.992 417 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:


0.000 280 58(10) =

0.0000 0000 0001 0010 0110 0011 0101 1001 1110 1100 1000 0101 0111 1001 1101 0001(2)

6. Positive number before normalization:

0.000 280 58(10) =

0.0000 0000 0001 0010 0110 0011 0101 1001 1110 1100 1000 0101 0111 1001 1101 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 280 58(10) =

0.0000 0000 0001 0010 0110 0011 0101 1001 1110 1100 1000 0101 0111 1001 1101 0001(2) =

0.0000 0000 0001 0010 0110 0011 0101 1001 1110 1100 1000 0101 0111 1001 1101 0001(2) × 20 =

1.0010 0110 0011 0101 1001 1110 1100 1000 0101 0111 1001 1101 0001(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0110 0011 0101 1001 1110 1100 1000 0101 0111 1001 1101 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0110 0011 0101 1001 1110 1100 1000 0101 0111 1001 1101 0001 =

0010 0110 0011 0101 1001 1110 1100 1000 0101 0111 1001 1101 0001


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 0011

Mantissa (52 bits) =
0010 0110 0011 0101 1001 1110 1100 1000 0101 0111 1001 1101 0001


Decimal number -0.000 280 58 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0110 0011 0101 1001 1110 1100 1000 0101 0111 1001 1101 0001

Share

» Convert decimal -0.000 281 37 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 Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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