-0.000 282 021 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 021 4(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
-0.000 282 021 4(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:

|-0.000 282 021 4| = 0.000 282 021 4


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: 0.000 282 021 4.

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) 0.000 282 021 4 × 2 = 0 + 0.000 564 042 8;
  • 2) 0.000 564 042 8 × 2 = 0 + 0.001 128 085 6;
  • 3) 0.001 128 085 6 × 2 = 0 + 0.002 256 171 2;
  • 4) 0.002 256 171 2 × 2 = 0 + 0.004 512 342 4;
  • 5) 0.004 512 342 4 × 2 = 0 + 0.009 024 684 8;
  • 6) 0.009 024 684 8 × 2 = 0 + 0.018 049 369 6;
  • 7) 0.018 049 369 6 × 2 = 0 + 0.036 098 739 2;
  • 8) 0.036 098 739 2 × 2 = 0 + 0.072 197 478 4;
  • 9) 0.072 197 478 4 × 2 = 0 + 0.144 394 956 8;
  • 10) 0.144 394 956 8 × 2 = 0 + 0.288 789 913 6;
  • 11) 0.288 789 913 6 × 2 = 0 + 0.577 579 827 2;
  • 12) 0.577 579 827 2 × 2 = 1 + 0.155 159 654 4;
  • 13) 0.155 159 654 4 × 2 = 0 + 0.310 319 308 8;
  • 14) 0.310 319 308 8 × 2 = 0 + 0.620 638 617 6;
  • 15) 0.620 638 617 6 × 2 = 1 + 0.241 277 235 2;
  • 16) 0.241 277 235 2 × 2 = 0 + 0.482 554 470 4;
  • 17) 0.482 554 470 4 × 2 = 0 + 0.965 108 940 8;
  • 18) 0.965 108 940 8 × 2 = 1 + 0.930 217 881 6;
  • 19) 0.930 217 881 6 × 2 = 1 + 0.860 435 763 2;
  • 20) 0.860 435 763 2 × 2 = 1 + 0.720 871 526 4;
  • 21) 0.720 871 526 4 × 2 = 1 + 0.441 743 052 8;
  • 22) 0.441 743 052 8 × 2 = 0 + 0.883 486 105 6;
  • 23) 0.883 486 105 6 × 2 = 1 + 0.766 972 211 2;
  • 24) 0.766 972 211 2 × 2 = 1 + 0.533 944 422 4;
  • 25) 0.533 944 422 4 × 2 = 1 + 0.067 888 844 8;
  • 26) 0.067 888 844 8 × 2 = 0 + 0.135 777 689 6;
  • 27) 0.135 777 689 6 × 2 = 0 + 0.271 555 379 2;
  • 28) 0.271 555 379 2 × 2 = 0 + 0.543 110 758 4;
  • 29) 0.543 110 758 4 × 2 = 1 + 0.086 221 516 8;
  • 30) 0.086 221 516 8 × 2 = 0 + 0.172 443 033 6;
  • 31) 0.172 443 033 6 × 2 = 0 + 0.344 886 067 2;
  • 32) 0.344 886 067 2 × 2 = 0 + 0.689 772 134 4;
  • 33) 0.689 772 134 4 × 2 = 1 + 0.379 544 268 8;
  • 34) 0.379 544 268 8 × 2 = 0 + 0.759 088 537 6;
  • 35) 0.759 088 537 6 × 2 = 1 + 0.518 177 075 2;
  • 36) 0.518 177 075 2 × 2 = 1 + 0.036 354 150 4;
  • 37) 0.036 354 150 4 × 2 = 0 + 0.072 708 300 8;
  • 38) 0.072 708 300 8 × 2 = 0 + 0.145 416 601 6;
  • 39) 0.145 416 601 6 × 2 = 0 + 0.290 833 203 2;
  • 40) 0.290 833 203 2 × 2 = 0 + 0.581 666 406 4;
  • 41) 0.581 666 406 4 × 2 = 1 + 0.163 332 812 8;
  • 42) 0.163 332 812 8 × 2 = 0 + 0.326 665 625 6;
  • 43) 0.326 665 625 6 × 2 = 0 + 0.653 331 251 2;
  • 44) 0.653 331 251 2 × 2 = 1 + 0.306 662 502 4;
  • 45) 0.306 662 502 4 × 2 = 0 + 0.613 325 004 8;
  • 46) 0.613 325 004 8 × 2 = 1 + 0.226 650 009 6;
  • 47) 0.226 650 009 6 × 2 = 0 + 0.453 300 019 2;
  • 48) 0.453 300 019 2 × 2 = 0 + 0.906 600 038 4;
  • 49) 0.906 600 038 4 × 2 = 1 + 0.813 200 076 8;
  • 50) 0.813 200 076 8 × 2 = 1 + 0.626 400 153 6;
  • 51) 0.626 400 153 6 × 2 = 1 + 0.252 800 307 2;
  • 52) 0.252 800 307 2 × 2 = 0 + 0.505 600 614 4;
  • 53) 0.505 600 614 4 × 2 = 1 + 0.011 201 228 8;
  • 54) 0.011 201 228 8 × 2 = 0 + 0.022 402 457 6;
  • 55) 0.022 402 457 6 × 2 = 0 + 0.044 804 915 2;
  • 56) 0.044 804 915 2 × 2 = 0 + 0.089 609 830 4;
  • 57) 0.089 609 830 4 × 2 = 0 + 0.179 219 660 8;
  • 58) 0.179 219 660 8 × 2 = 0 + 0.358 439 321 6;
  • 59) 0.358 439 321 6 × 2 = 0 + 0.716 878 643 2;
  • 60) 0.716 878 643 2 × 2 = 1 + 0.433 757 286 4;
  • 61) 0.433 757 286 4 × 2 = 0 + 0.867 514 572 8;
  • 62) 0.867 514 572 8 × 2 = 1 + 0.735 029 145 6;
  • 63) 0.735 029 145 6 × 2 = 1 + 0.470 058 291 2;
  • 64) 0.470 058 291 2 × 2 = 0 + 0.940 116 582 4;

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:


0.000 282 021 4(10) =

0.0000 0000 0001 0010 0111 1011 1000 1000 1011 0000 1001 0100 1110 1000 0001 0110(2)

6. Positive number before normalization:

0.000 282 021 4(10) =

0.0000 0000 0001 0010 0111 1011 1000 1000 1011 0000 1001 0100 1110 1000 0001 0110(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:


0.000 282 021 4(10) =

0.0000 0000 0001 0010 0111 1011 1000 1000 1011 0000 1001 0100 1110 1000 0001 0110(2) =

0.0000 0000 0001 0010 0111 1011 1000 1000 1011 0000 1001 0100 1110 1000 0001 0110(2) × 20 =

1.0010 0111 1011 1000 1000 1011 0000 1001 0100 1110 1000 0001 0110(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 1000 1000 1011 0000 1001 0100 1110 1000 0001 0110


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 1000 1000 1011 0000 1001 0100 1110 1000 0001 0110 =

0010 0111 1011 1000 1000 1011 0000 1001 0100 1110 1000 0001 0110


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 1000 1000 1011 0000 1001 0100 1110 1000 0001 0110


Decimal number -0.000 282 021 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1000 1000 1011 0000 1001 0100 1110 1000 0001 0110

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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