-0.000 282 006 34 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 34(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
-0.000 282 006 34(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:

|-0.000 282 006 34| = 0.000 282 006 34


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: 0.000 282 006 34.

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) 0.000 282 006 34 × 2 = 0 + 0.000 564 012 68;
  • 2) 0.000 564 012 68 × 2 = 0 + 0.001 128 025 36;
  • 3) 0.001 128 025 36 × 2 = 0 + 0.002 256 050 72;
  • 4) 0.002 256 050 72 × 2 = 0 + 0.004 512 101 44;
  • 5) 0.004 512 101 44 × 2 = 0 + 0.009 024 202 88;
  • 6) 0.009 024 202 88 × 2 = 0 + 0.018 048 405 76;
  • 7) 0.018 048 405 76 × 2 = 0 + 0.036 096 811 52;
  • 8) 0.036 096 811 52 × 2 = 0 + 0.072 193 623 04;
  • 9) 0.072 193 623 04 × 2 = 0 + 0.144 387 246 08;
  • 10) 0.144 387 246 08 × 2 = 0 + 0.288 774 492 16;
  • 11) 0.288 774 492 16 × 2 = 0 + 0.577 548 984 32;
  • 12) 0.577 548 984 32 × 2 = 1 + 0.155 097 968 64;
  • 13) 0.155 097 968 64 × 2 = 0 + 0.310 195 937 28;
  • 14) 0.310 195 937 28 × 2 = 0 + 0.620 391 874 56;
  • 15) 0.620 391 874 56 × 2 = 1 + 0.240 783 749 12;
  • 16) 0.240 783 749 12 × 2 = 0 + 0.481 567 498 24;
  • 17) 0.481 567 498 24 × 2 = 0 + 0.963 134 996 48;
  • 18) 0.963 134 996 48 × 2 = 1 + 0.926 269 992 96;
  • 19) 0.926 269 992 96 × 2 = 1 + 0.852 539 985 92;
  • 20) 0.852 539 985 92 × 2 = 1 + 0.705 079 971 84;
  • 21) 0.705 079 971 84 × 2 = 1 + 0.410 159 943 68;
  • 22) 0.410 159 943 68 × 2 = 0 + 0.820 319 887 36;
  • 23) 0.820 319 887 36 × 2 = 1 + 0.640 639 774 72;
  • 24) 0.640 639 774 72 × 2 = 1 + 0.281 279 549 44;
  • 25) 0.281 279 549 44 × 2 = 0 + 0.562 559 098 88;
  • 26) 0.562 559 098 88 × 2 = 1 + 0.125 118 197 76;
  • 27) 0.125 118 197 76 × 2 = 0 + 0.250 236 395 52;
  • 28) 0.250 236 395 52 × 2 = 0 + 0.500 472 791 04;
  • 29) 0.500 472 791 04 × 2 = 1 + 0.000 945 582 08;
  • 30) 0.000 945 582 08 × 2 = 0 + 0.001 891 164 16;
  • 31) 0.001 891 164 16 × 2 = 0 + 0.003 782 328 32;
  • 32) 0.003 782 328 32 × 2 = 0 + 0.007 564 656 64;
  • 33) 0.007 564 656 64 × 2 = 0 + 0.015 129 313 28;
  • 34) 0.015 129 313 28 × 2 = 0 + 0.030 258 626 56;
  • 35) 0.030 258 626 56 × 2 = 0 + 0.060 517 253 12;
  • 36) 0.060 517 253 12 × 2 = 0 + 0.121 034 506 24;
  • 37) 0.121 034 506 24 × 2 = 0 + 0.242 069 012 48;
  • 38) 0.242 069 012 48 × 2 = 0 + 0.484 138 024 96;
  • 39) 0.484 138 024 96 × 2 = 0 + 0.968 276 049 92;
  • 40) 0.968 276 049 92 × 2 = 1 + 0.936 552 099 84;
  • 41) 0.936 552 099 84 × 2 = 1 + 0.873 104 199 68;
  • 42) 0.873 104 199 68 × 2 = 1 + 0.746 208 399 36;
  • 43) 0.746 208 399 36 × 2 = 1 + 0.492 416 798 72;
  • 44) 0.492 416 798 72 × 2 = 0 + 0.984 833 597 44;
  • 45) 0.984 833 597 44 × 2 = 1 + 0.969 667 194 88;
  • 46) 0.969 667 194 88 × 2 = 1 + 0.939 334 389 76;
  • 47) 0.939 334 389 76 × 2 = 1 + 0.878 668 779 52;
  • 48) 0.878 668 779 52 × 2 = 1 + 0.757 337 559 04;
  • 49) 0.757 337 559 04 × 2 = 1 + 0.514 675 118 08;
  • 50) 0.514 675 118 08 × 2 = 1 + 0.029 350 236 16;
  • 51) 0.029 350 236 16 × 2 = 0 + 0.058 700 472 32;
  • 52) 0.058 700 472 32 × 2 = 0 + 0.117 400 944 64;
  • 53) 0.117 400 944 64 × 2 = 0 + 0.234 801 889 28;
  • 54) 0.234 801 889 28 × 2 = 0 + 0.469 603 778 56;
  • 55) 0.469 603 778 56 × 2 = 0 + 0.939 207 557 12;
  • 56) 0.939 207 557 12 × 2 = 1 + 0.878 415 114 24;
  • 57) 0.878 415 114 24 × 2 = 1 + 0.756 830 228 48;
  • 58) 0.756 830 228 48 × 2 = 1 + 0.513 660 456 96;
  • 59) 0.513 660 456 96 × 2 = 1 + 0.027 320 913 92;
  • 60) 0.027 320 913 92 × 2 = 0 + 0.054 641 827 84;
  • 61) 0.054 641 827 84 × 2 = 0 + 0.109 283 655 68;
  • 62) 0.109 283 655 68 × 2 = 0 + 0.218 567 311 36;
  • 63) 0.218 567 311 36 × 2 = 0 + 0.437 134 622 72;
  • 64) 0.437 134 622 72 × 2 = 0 + 0.874 269 245 44;

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:


0.000 282 006 34(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 0001 1110 1111 1100 0001 1110 0000(2)

6. Positive number before normalization:

0.000 282 006 34(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 0001 1110 1111 1100 0001 1110 0000(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:


0.000 282 006 34(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 0001 1110 1111 1100 0001 1110 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0000 0001 1110 1111 1100 0001 1110 0000(2) × 20 =

1.0010 0111 1011 0100 1000 0000 0001 1110 1111 1100 0001 1110 0000(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 0111 1011 0100 1000 0000 0001 1110 1111 1100 0001 1110 0000


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 0111 1011 0100 1000 0000 0001 1110 1111 1100 0001 1110 0000 =

0010 0111 1011 0100 1000 0000 0001 1110 1111 1100 0001 1110 0000


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 0111 1011 0100 1000 0000 0001 1110 1111 1100 0001 1110 0000


Decimal number -0.000 282 006 34 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0000 0001 1110 1111 1100 0001 1110 0000

Share

» Convert decimal -0.000 282 006 13 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