-0.000 282 017 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 017 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
-0.000 282 017 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:

|-0.000 282 017 2| = 0.000 282 017 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: 0.000 282 017 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) 0.000 282 017 2 × 2 = 0 + 0.000 564 034 4;
  • 2) 0.000 564 034 4 × 2 = 0 + 0.001 128 068 8;
  • 3) 0.001 128 068 8 × 2 = 0 + 0.002 256 137 6;
  • 4) 0.002 256 137 6 × 2 = 0 + 0.004 512 275 2;
  • 5) 0.004 512 275 2 × 2 = 0 + 0.009 024 550 4;
  • 6) 0.009 024 550 4 × 2 = 0 + 0.018 049 100 8;
  • 7) 0.018 049 100 8 × 2 = 0 + 0.036 098 201 6;
  • 8) 0.036 098 201 6 × 2 = 0 + 0.072 196 403 2;
  • 9) 0.072 196 403 2 × 2 = 0 + 0.144 392 806 4;
  • 10) 0.144 392 806 4 × 2 = 0 + 0.288 785 612 8;
  • 11) 0.288 785 612 8 × 2 = 0 + 0.577 571 225 6;
  • 12) 0.577 571 225 6 × 2 = 1 + 0.155 142 451 2;
  • 13) 0.155 142 451 2 × 2 = 0 + 0.310 284 902 4;
  • 14) 0.310 284 902 4 × 2 = 0 + 0.620 569 804 8;
  • 15) 0.620 569 804 8 × 2 = 1 + 0.241 139 609 6;
  • 16) 0.241 139 609 6 × 2 = 0 + 0.482 279 219 2;
  • 17) 0.482 279 219 2 × 2 = 0 + 0.964 558 438 4;
  • 18) 0.964 558 438 4 × 2 = 1 + 0.929 116 876 8;
  • 19) 0.929 116 876 8 × 2 = 1 + 0.858 233 753 6;
  • 20) 0.858 233 753 6 × 2 = 1 + 0.716 467 507 2;
  • 21) 0.716 467 507 2 × 2 = 1 + 0.432 935 014 4;
  • 22) 0.432 935 014 4 × 2 = 0 + 0.865 870 028 8;
  • 23) 0.865 870 028 8 × 2 = 1 + 0.731 740 057 6;
  • 24) 0.731 740 057 6 × 2 = 1 + 0.463 480 115 2;
  • 25) 0.463 480 115 2 × 2 = 0 + 0.926 960 230 4;
  • 26) 0.926 960 230 4 × 2 = 1 + 0.853 920 460 8;
  • 27) 0.853 920 460 8 × 2 = 1 + 0.707 840 921 6;
  • 28) 0.707 840 921 6 × 2 = 1 + 0.415 681 843 2;
  • 29) 0.415 681 843 2 × 2 = 0 + 0.831 363 686 4;
  • 30) 0.831 363 686 4 × 2 = 1 + 0.662 727 372 8;
  • 31) 0.662 727 372 8 × 2 = 1 + 0.325 454 745 6;
  • 32) 0.325 454 745 6 × 2 = 0 + 0.650 909 491 2;
  • 33) 0.650 909 491 2 × 2 = 1 + 0.301 818 982 4;
  • 34) 0.301 818 982 4 × 2 = 0 + 0.603 637 964 8;
  • 35) 0.603 637 964 8 × 2 = 1 + 0.207 275 929 6;
  • 36) 0.207 275 929 6 × 2 = 0 + 0.414 551 859 2;
  • 37) 0.414 551 859 2 × 2 = 0 + 0.829 103 718 4;
  • 38) 0.829 103 718 4 × 2 = 1 + 0.658 207 436 8;
  • 39) 0.658 207 436 8 × 2 = 1 + 0.316 414 873 6;
  • 40) 0.316 414 873 6 × 2 = 0 + 0.632 829 747 2;
  • 41) 0.632 829 747 2 × 2 = 1 + 0.265 659 494 4;
  • 42) 0.265 659 494 4 × 2 = 0 + 0.531 318 988 8;
  • 43) 0.531 318 988 8 × 2 = 1 + 0.062 637 977 6;
  • 44) 0.062 637 977 6 × 2 = 0 + 0.125 275 955 2;
  • 45) 0.125 275 955 2 × 2 = 0 + 0.250 551 910 4;
  • 46) 0.250 551 910 4 × 2 = 0 + 0.501 103 820 8;
  • 47) 0.501 103 820 8 × 2 = 1 + 0.002 207 641 6;
  • 48) 0.002 207 641 6 × 2 = 0 + 0.004 415 283 2;
  • 49) 0.004 415 283 2 × 2 = 0 + 0.008 830 566 4;
  • 50) 0.008 830 566 4 × 2 = 0 + 0.017 661 132 8;
  • 51) 0.017 661 132 8 × 2 = 0 + 0.035 322 265 6;
  • 52) 0.035 322 265 6 × 2 = 0 + 0.070 644 531 2;
  • 53) 0.070 644 531 2 × 2 = 0 + 0.141 289 062 4;
  • 54) 0.141 289 062 4 × 2 = 0 + 0.282 578 124 8;
  • 55) 0.282 578 124 8 × 2 = 0 + 0.565 156 249 6;
  • 56) 0.565 156 249 6 × 2 = 1 + 0.130 312 499 2;
  • 57) 0.130 312 499 2 × 2 = 0 + 0.260 624 998 4;
  • 58) 0.260 624 998 4 × 2 = 0 + 0.521 249 996 8;
  • 59) 0.521 249 996 8 × 2 = 1 + 0.042 499 993 6;
  • 60) 0.042 499 993 6 × 2 = 0 + 0.084 999 987 2;
  • 61) 0.084 999 987 2 × 2 = 0 + 0.169 999 974 4;
  • 62) 0.169 999 974 4 × 2 = 0 + 0.339 999 948 8;
  • 63) 0.339 999 948 8 × 2 = 0 + 0.679 999 897 6;
  • 64) 0.679 999 897 6 × 2 = 1 + 0.359 999 795 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:


0.000 282 017 2(10) =

0.0000 0000 0001 0010 0111 1011 0111 0110 1010 0110 1010 0010 0000 0001 0010 0001(2)

6. Positive number before normalization:

0.000 282 017 2(10) =

0.0000 0000 0001 0010 0111 1011 0111 0110 1010 0110 1010 0010 0000 0001 0010 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:


0.000 282 017 2(10) =

0.0000 0000 0001 0010 0111 1011 0111 0110 1010 0110 1010 0010 0000 0001 0010 0001(2) =

0.0000 0000 0001 0010 0111 1011 0111 0110 1010 0110 1010 0010 0000 0001 0010 0001(2) × 20 =

1.0010 0111 1011 0111 0110 1010 0110 1010 0010 0000 0001 0010 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 0111 1011 0111 0110 1010 0110 1010 0010 0000 0001 0010 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 0111 1011 0111 0110 1010 0110 1010 0010 0000 0001 0010 0001 =

0010 0111 1011 0111 0110 1010 0110 1010 0010 0000 0001 0010 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 0111 1011 0111 0110 1010 0110 1010 0010 0000 0001 0010 0001


Decimal number -0.000 282 017 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0111 0110 1010 0110 1010 0010 0000 0001 0010 0001

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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