-0.000 282 005 905 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 905 8(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
-0.000 282 005 905 8(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:

|-0.000 282 005 905 8| = 0.000 282 005 905 8


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: 0.000 282 005 905 8.

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) 0.000 282 005 905 8 × 2 = 0 + 0.000 564 011 811 6;
  • 2) 0.000 564 011 811 6 × 2 = 0 + 0.001 128 023 623 2;
  • 3) 0.001 128 023 623 2 × 2 = 0 + 0.002 256 047 246 4;
  • 4) 0.002 256 047 246 4 × 2 = 0 + 0.004 512 094 492 8;
  • 5) 0.004 512 094 492 8 × 2 = 0 + 0.009 024 188 985 6;
  • 6) 0.009 024 188 985 6 × 2 = 0 + 0.018 048 377 971 2;
  • 7) 0.018 048 377 971 2 × 2 = 0 + 0.036 096 755 942 4;
  • 8) 0.036 096 755 942 4 × 2 = 0 + 0.072 193 511 884 8;
  • 9) 0.072 193 511 884 8 × 2 = 0 + 0.144 387 023 769 6;
  • 10) 0.144 387 023 769 6 × 2 = 0 + 0.288 774 047 539 2;
  • 11) 0.288 774 047 539 2 × 2 = 0 + 0.577 548 095 078 4;
  • 12) 0.577 548 095 078 4 × 2 = 1 + 0.155 096 190 156 8;
  • 13) 0.155 096 190 156 8 × 2 = 0 + 0.310 192 380 313 6;
  • 14) 0.310 192 380 313 6 × 2 = 0 + 0.620 384 760 627 2;
  • 15) 0.620 384 760 627 2 × 2 = 1 + 0.240 769 521 254 4;
  • 16) 0.240 769 521 254 4 × 2 = 0 + 0.481 539 042 508 8;
  • 17) 0.481 539 042 508 8 × 2 = 0 + 0.963 078 085 017 6;
  • 18) 0.963 078 085 017 6 × 2 = 1 + 0.926 156 170 035 2;
  • 19) 0.926 156 170 035 2 × 2 = 1 + 0.852 312 340 070 4;
  • 20) 0.852 312 340 070 4 × 2 = 1 + 0.704 624 680 140 8;
  • 21) 0.704 624 680 140 8 × 2 = 1 + 0.409 249 360 281 6;
  • 22) 0.409 249 360 281 6 × 2 = 0 + 0.818 498 720 563 2;
  • 23) 0.818 498 720 563 2 × 2 = 1 + 0.636 997 441 126 4;
  • 24) 0.636 997 441 126 4 × 2 = 1 + 0.273 994 882 252 8;
  • 25) 0.273 994 882 252 8 × 2 = 0 + 0.547 989 764 505 6;
  • 26) 0.547 989 764 505 6 × 2 = 1 + 0.095 979 529 011 2;
  • 27) 0.095 979 529 011 2 × 2 = 0 + 0.191 959 058 022 4;
  • 28) 0.191 959 058 022 4 × 2 = 0 + 0.383 918 116 044 8;
  • 29) 0.383 918 116 044 8 × 2 = 0 + 0.767 836 232 089 6;
  • 30) 0.767 836 232 089 6 × 2 = 1 + 0.535 672 464 179 2;
  • 31) 0.535 672 464 179 2 × 2 = 1 + 0.071 344 928 358 4;
  • 32) 0.071 344 928 358 4 × 2 = 0 + 0.142 689 856 716 8;
  • 33) 0.142 689 856 716 8 × 2 = 0 + 0.285 379 713 433 6;
  • 34) 0.285 379 713 433 6 × 2 = 0 + 0.570 759 426 867 2;
  • 35) 0.570 759 426 867 2 × 2 = 1 + 0.141 518 853 734 4;
  • 36) 0.141 518 853 734 4 × 2 = 0 + 0.283 037 707 468 8;
  • 37) 0.283 037 707 468 8 × 2 = 0 + 0.566 075 414 937 6;
  • 38) 0.566 075 414 937 6 × 2 = 1 + 0.132 150 829 875 2;
  • 39) 0.132 150 829 875 2 × 2 = 0 + 0.264 301 659 750 4;
  • 40) 0.264 301 659 750 4 × 2 = 0 + 0.528 603 319 500 8;
  • 41) 0.528 603 319 500 8 × 2 = 1 + 0.057 206 639 001 6;
  • 42) 0.057 206 639 001 6 × 2 = 0 + 0.114 413 278 003 2;
  • 43) 0.114 413 278 003 2 × 2 = 0 + 0.228 826 556 006 4;
  • 44) 0.228 826 556 006 4 × 2 = 0 + 0.457 653 112 012 8;
  • 45) 0.457 653 112 012 8 × 2 = 0 + 0.915 306 224 025 6;
  • 46) 0.915 306 224 025 6 × 2 = 1 + 0.830 612 448 051 2;
  • 47) 0.830 612 448 051 2 × 2 = 1 + 0.661 224 896 102 4;
  • 48) 0.661 224 896 102 4 × 2 = 1 + 0.322 449 792 204 8;
  • 49) 0.322 449 792 204 8 × 2 = 0 + 0.644 899 584 409 6;
  • 50) 0.644 899 584 409 6 × 2 = 1 + 0.289 799 168 819 2;
  • 51) 0.289 799 168 819 2 × 2 = 0 + 0.579 598 337 638 4;
  • 52) 0.579 598 337 638 4 × 2 = 1 + 0.159 196 675 276 8;
  • 53) 0.159 196 675 276 8 × 2 = 0 + 0.318 393 350 553 6;
  • 54) 0.318 393 350 553 6 × 2 = 0 + 0.636 786 701 107 2;
  • 55) 0.636 786 701 107 2 × 2 = 1 + 0.273 573 402 214 4;
  • 56) 0.273 573 402 214 4 × 2 = 0 + 0.547 146 804 428 8;
  • 57) 0.547 146 804 428 8 × 2 = 1 + 0.094 293 608 857 6;
  • 58) 0.094 293 608 857 6 × 2 = 0 + 0.188 587 217 715 2;
  • 59) 0.188 587 217 715 2 × 2 = 0 + 0.377 174 435 430 4;
  • 60) 0.377 174 435 430 4 × 2 = 0 + 0.754 348 870 860 8;
  • 61) 0.754 348 870 860 8 × 2 = 1 + 0.508 697 741 721 6;
  • 62) 0.508 697 741 721 6 × 2 = 1 + 0.017 395 483 443 2;
  • 63) 0.017 395 483 443 2 × 2 = 0 + 0.034 790 966 886 4;
  • 64) 0.034 790 966 886 4 × 2 = 0 + 0.069 581 933 772 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:


0.000 282 005 905 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0100 1000 0111 0101 0010 1000 1100(2)

6. Positive number before normalization:

0.000 282 005 905 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0100 1000 0111 0101 0010 1000 1100(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:


0.000 282 005 905 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0100 1000 0111 0101 0010 1000 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 0100 1000 0111 0101 0010 1000 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0010 0100 1000 0111 0101 0010 1000 1100(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 0110 0010 0100 1000 0111 0101 0010 1000 1100


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 0110 0010 0100 1000 0111 0101 0010 1000 1100 =

0010 0111 1011 0100 0110 0010 0100 1000 0111 0101 0010 1000 1100


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 0110 0010 0100 1000 0111 0101 0010 1000 1100


Decimal number -0.000 282 005 905 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 0100 1000 0111 0101 0010 1000 1100

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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