-0.000 282 014 62 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 014 62(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
-0.000 282 014 62(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:

|-0.000 282 014 62| = 0.000 282 014 62


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: 0.000 282 014 62.

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) 0.000 282 014 62 × 2 = 0 + 0.000 564 029 24;
  • 2) 0.000 564 029 24 × 2 = 0 + 0.001 128 058 48;
  • 3) 0.001 128 058 48 × 2 = 0 + 0.002 256 116 96;
  • 4) 0.002 256 116 96 × 2 = 0 + 0.004 512 233 92;
  • 5) 0.004 512 233 92 × 2 = 0 + 0.009 024 467 84;
  • 6) 0.009 024 467 84 × 2 = 0 + 0.018 048 935 68;
  • 7) 0.018 048 935 68 × 2 = 0 + 0.036 097 871 36;
  • 8) 0.036 097 871 36 × 2 = 0 + 0.072 195 742 72;
  • 9) 0.072 195 742 72 × 2 = 0 + 0.144 391 485 44;
  • 10) 0.144 391 485 44 × 2 = 0 + 0.288 782 970 88;
  • 11) 0.288 782 970 88 × 2 = 0 + 0.577 565 941 76;
  • 12) 0.577 565 941 76 × 2 = 1 + 0.155 131 883 52;
  • 13) 0.155 131 883 52 × 2 = 0 + 0.310 263 767 04;
  • 14) 0.310 263 767 04 × 2 = 0 + 0.620 527 534 08;
  • 15) 0.620 527 534 08 × 2 = 1 + 0.241 055 068 16;
  • 16) 0.241 055 068 16 × 2 = 0 + 0.482 110 136 32;
  • 17) 0.482 110 136 32 × 2 = 0 + 0.964 220 272 64;
  • 18) 0.964 220 272 64 × 2 = 1 + 0.928 440 545 28;
  • 19) 0.928 440 545 28 × 2 = 1 + 0.856 881 090 56;
  • 20) 0.856 881 090 56 × 2 = 1 + 0.713 762 181 12;
  • 21) 0.713 762 181 12 × 2 = 1 + 0.427 524 362 24;
  • 22) 0.427 524 362 24 × 2 = 0 + 0.855 048 724 48;
  • 23) 0.855 048 724 48 × 2 = 1 + 0.710 097 448 96;
  • 24) 0.710 097 448 96 × 2 = 1 + 0.420 194 897 92;
  • 25) 0.420 194 897 92 × 2 = 0 + 0.840 389 795 84;
  • 26) 0.840 389 795 84 × 2 = 1 + 0.680 779 591 68;
  • 27) 0.680 779 591 68 × 2 = 1 + 0.361 559 183 36;
  • 28) 0.361 559 183 36 × 2 = 0 + 0.723 118 366 72;
  • 29) 0.723 118 366 72 × 2 = 1 + 0.446 236 733 44;
  • 30) 0.446 236 733 44 × 2 = 0 + 0.892 473 466 88;
  • 31) 0.892 473 466 88 × 2 = 1 + 0.784 946 933 76;
  • 32) 0.784 946 933 76 × 2 = 1 + 0.569 893 867 52;
  • 33) 0.569 893 867 52 × 2 = 1 + 0.139 787 735 04;
  • 34) 0.139 787 735 04 × 2 = 0 + 0.279 575 470 08;
  • 35) 0.279 575 470 08 × 2 = 0 + 0.559 150 940 16;
  • 36) 0.559 150 940 16 × 2 = 1 + 0.118 301 880 32;
  • 37) 0.118 301 880 32 × 2 = 0 + 0.236 603 760 64;
  • 38) 0.236 603 760 64 × 2 = 0 + 0.473 207 521 28;
  • 39) 0.473 207 521 28 × 2 = 0 + 0.946 415 042 56;
  • 40) 0.946 415 042 56 × 2 = 1 + 0.892 830 085 12;
  • 41) 0.892 830 085 12 × 2 = 1 + 0.785 660 170 24;
  • 42) 0.785 660 170 24 × 2 = 1 + 0.571 320 340 48;
  • 43) 0.571 320 340 48 × 2 = 1 + 0.142 640 680 96;
  • 44) 0.142 640 680 96 × 2 = 0 + 0.285 281 361 92;
  • 45) 0.285 281 361 92 × 2 = 0 + 0.570 562 723 84;
  • 46) 0.570 562 723 84 × 2 = 1 + 0.141 125 447 68;
  • 47) 0.141 125 447 68 × 2 = 0 + 0.282 250 895 36;
  • 48) 0.282 250 895 36 × 2 = 0 + 0.564 501 790 72;
  • 49) 0.564 501 790 72 × 2 = 1 + 0.129 003 581 44;
  • 50) 0.129 003 581 44 × 2 = 0 + 0.258 007 162 88;
  • 51) 0.258 007 162 88 × 2 = 0 + 0.516 014 325 76;
  • 52) 0.516 014 325 76 × 2 = 1 + 0.032 028 651 52;
  • 53) 0.032 028 651 52 × 2 = 0 + 0.064 057 303 04;
  • 54) 0.064 057 303 04 × 2 = 0 + 0.128 114 606 08;
  • 55) 0.128 114 606 08 × 2 = 0 + 0.256 229 212 16;
  • 56) 0.256 229 212 16 × 2 = 0 + 0.512 458 424 32;
  • 57) 0.512 458 424 32 × 2 = 1 + 0.024 916 848 64;
  • 58) 0.024 916 848 64 × 2 = 0 + 0.049 833 697 28;
  • 59) 0.049 833 697 28 × 2 = 0 + 0.099 667 394 56;
  • 60) 0.099 667 394 56 × 2 = 0 + 0.199 334 789 12;
  • 61) 0.199 334 789 12 × 2 = 0 + 0.398 669 578 24;
  • 62) 0.398 669 578 24 × 2 = 0 + 0.797 339 156 48;
  • 63) 0.797 339 156 48 × 2 = 1 + 0.594 678 312 96;
  • 64) 0.594 678 312 96 × 2 = 1 + 0.189 356 625 92;

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:


0.000 282 014 62(10) =

0.0000 0000 0001 0010 0111 1011 0110 1011 1001 0001 1110 0100 1001 0000 1000 0011(2)

6. Positive number before normalization:

0.000 282 014 62(10) =

0.0000 0000 0001 0010 0111 1011 0110 1011 1001 0001 1110 0100 1001 0000 1000 0011(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:


0.000 282 014 62(10) =

0.0000 0000 0001 0010 0111 1011 0110 1011 1001 0001 1110 0100 1001 0000 1000 0011(2) =

0.0000 0000 0001 0010 0111 1011 0110 1011 1001 0001 1110 0100 1001 0000 1000 0011(2) × 20 =

1.0010 0111 1011 0110 1011 1001 0001 1110 0100 1001 0000 1000 0011(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 0110 1011 1001 0001 1110 0100 1001 0000 1000 0011


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 0110 1011 1001 0001 1110 0100 1001 0000 1000 0011 =

0010 0111 1011 0110 1011 1001 0001 1110 0100 1001 0000 1000 0011


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 0110 1011 1001 0001 1110 0100 1001 0000 1000 0011


Decimal number -0.000 282 014 62 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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