-0.000 282 034 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 034 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
-0.000 282 034 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:

|-0.000 282 034 3| = 0.000 282 034 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: 0.000 282 034 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) 0.000 282 034 3 × 2 = 0 + 0.000 564 068 6;
  • 2) 0.000 564 068 6 × 2 = 0 + 0.001 128 137 2;
  • 3) 0.001 128 137 2 × 2 = 0 + 0.002 256 274 4;
  • 4) 0.002 256 274 4 × 2 = 0 + 0.004 512 548 8;
  • 5) 0.004 512 548 8 × 2 = 0 + 0.009 025 097 6;
  • 6) 0.009 025 097 6 × 2 = 0 + 0.018 050 195 2;
  • 7) 0.018 050 195 2 × 2 = 0 + 0.036 100 390 4;
  • 8) 0.036 100 390 4 × 2 = 0 + 0.072 200 780 8;
  • 9) 0.072 200 780 8 × 2 = 0 + 0.144 401 561 6;
  • 10) 0.144 401 561 6 × 2 = 0 + 0.288 803 123 2;
  • 11) 0.288 803 123 2 × 2 = 0 + 0.577 606 246 4;
  • 12) 0.577 606 246 4 × 2 = 1 + 0.155 212 492 8;
  • 13) 0.155 212 492 8 × 2 = 0 + 0.310 424 985 6;
  • 14) 0.310 424 985 6 × 2 = 0 + 0.620 849 971 2;
  • 15) 0.620 849 971 2 × 2 = 1 + 0.241 699 942 4;
  • 16) 0.241 699 942 4 × 2 = 0 + 0.483 399 884 8;
  • 17) 0.483 399 884 8 × 2 = 0 + 0.966 799 769 6;
  • 18) 0.966 799 769 6 × 2 = 1 + 0.933 599 539 2;
  • 19) 0.933 599 539 2 × 2 = 1 + 0.867 199 078 4;
  • 20) 0.867 199 078 4 × 2 = 1 + 0.734 398 156 8;
  • 21) 0.734 398 156 8 × 2 = 1 + 0.468 796 313 6;
  • 22) 0.468 796 313 6 × 2 = 0 + 0.937 592 627 2;
  • 23) 0.937 592 627 2 × 2 = 1 + 0.875 185 254 4;
  • 24) 0.875 185 254 4 × 2 = 1 + 0.750 370 508 8;
  • 25) 0.750 370 508 8 × 2 = 1 + 0.500 741 017 6;
  • 26) 0.500 741 017 6 × 2 = 1 + 0.001 482 035 2;
  • 27) 0.001 482 035 2 × 2 = 0 + 0.002 964 070 4;
  • 28) 0.002 964 070 4 × 2 = 0 + 0.005 928 140 8;
  • 29) 0.005 928 140 8 × 2 = 0 + 0.011 856 281 6;
  • 30) 0.011 856 281 6 × 2 = 0 + 0.023 712 563 2;
  • 31) 0.023 712 563 2 × 2 = 0 + 0.047 425 126 4;
  • 32) 0.047 425 126 4 × 2 = 0 + 0.094 850 252 8;
  • 33) 0.094 850 252 8 × 2 = 0 + 0.189 700 505 6;
  • 34) 0.189 700 505 6 × 2 = 0 + 0.379 401 011 2;
  • 35) 0.379 401 011 2 × 2 = 0 + 0.758 802 022 4;
  • 36) 0.758 802 022 4 × 2 = 1 + 0.517 604 044 8;
  • 37) 0.517 604 044 8 × 2 = 1 + 0.035 208 089 6;
  • 38) 0.035 208 089 6 × 2 = 0 + 0.070 416 179 2;
  • 39) 0.070 416 179 2 × 2 = 0 + 0.140 832 358 4;
  • 40) 0.140 832 358 4 × 2 = 0 + 0.281 664 716 8;
  • 41) 0.281 664 716 8 × 2 = 0 + 0.563 329 433 6;
  • 42) 0.563 329 433 6 × 2 = 1 + 0.126 658 867 2;
  • 43) 0.126 658 867 2 × 2 = 0 + 0.253 317 734 4;
  • 44) 0.253 317 734 4 × 2 = 0 + 0.506 635 468 8;
  • 45) 0.506 635 468 8 × 2 = 1 + 0.013 270 937 6;
  • 46) 0.013 270 937 6 × 2 = 0 + 0.026 541 875 2;
  • 47) 0.026 541 875 2 × 2 = 0 + 0.053 083 750 4;
  • 48) 0.053 083 750 4 × 2 = 0 + 0.106 167 500 8;
  • 49) 0.106 167 500 8 × 2 = 0 + 0.212 335 001 6;
  • 50) 0.212 335 001 6 × 2 = 0 + 0.424 670 003 2;
  • 51) 0.424 670 003 2 × 2 = 0 + 0.849 340 006 4;
  • 52) 0.849 340 006 4 × 2 = 1 + 0.698 680 012 8;
  • 53) 0.698 680 012 8 × 2 = 1 + 0.397 360 025 6;
  • 54) 0.397 360 025 6 × 2 = 0 + 0.794 720 051 2;
  • 55) 0.794 720 051 2 × 2 = 1 + 0.589 440 102 4;
  • 56) 0.589 440 102 4 × 2 = 1 + 0.178 880 204 8;
  • 57) 0.178 880 204 8 × 2 = 0 + 0.357 760 409 6;
  • 58) 0.357 760 409 6 × 2 = 0 + 0.715 520 819 2;
  • 59) 0.715 520 819 2 × 2 = 1 + 0.431 041 638 4;
  • 60) 0.431 041 638 4 × 2 = 0 + 0.862 083 276 8;
  • 61) 0.862 083 276 8 × 2 = 1 + 0.724 166 553 6;
  • 62) 0.724 166 553 6 × 2 = 1 + 0.448 333 107 2;
  • 63) 0.448 333 107 2 × 2 = 0 + 0.896 666 214 4;
  • 64) 0.896 666 214 4 × 2 = 1 + 0.793 332 428 8;

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:


0.000 282 034 3(10) =

0.0000 0000 0001 0010 0111 1011 1100 0000 0001 1000 0100 1000 0001 1011 0010 1101(2)

6. Positive number before normalization:

0.000 282 034 3(10) =

0.0000 0000 0001 0010 0111 1011 1100 0000 0001 1000 0100 1000 0001 1011 0010 1101(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:


0.000 282 034 3(10) =

0.0000 0000 0001 0010 0111 1011 1100 0000 0001 1000 0100 1000 0001 1011 0010 1101(2) =

0.0000 0000 0001 0010 0111 1011 1100 0000 0001 1000 0100 1000 0001 1011 0010 1101(2) × 20 =

1.0010 0111 1011 1100 0000 0001 1000 0100 1000 0001 1011 0010 1101(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 1100 0000 0001 1000 0100 1000 0001 1011 0010 1101


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 1100 0000 0001 1000 0100 1000 0001 1011 0010 1101 =

0010 0111 1011 1100 0000 0001 1000 0100 1000 0001 1011 0010 1101


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 1100 0000 0001 1000 0100 1000 0001 1011 0010 1101


Decimal number -0.000 282 034 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1100 0000 0001 1000 0100 1000 0001 1011 0010 1101

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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