-0.000 282 005 910 42 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 910 42(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
-0.000 282 005 910 42(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:

|-0.000 282 005 910 42| = 0.000 282 005 910 42


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: 0.000 282 005 910 42.

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) 0.000 282 005 910 42 × 2 = 0 + 0.000 564 011 820 84;
  • 2) 0.000 564 011 820 84 × 2 = 0 + 0.001 128 023 641 68;
  • 3) 0.001 128 023 641 68 × 2 = 0 + 0.002 256 047 283 36;
  • 4) 0.002 256 047 283 36 × 2 = 0 + 0.004 512 094 566 72;
  • 5) 0.004 512 094 566 72 × 2 = 0 + 0.009 024 189 133 44;
  • 6) 0.009 024 189 133 44 × 2 = 0 + 0.018 048 378 266 88;
  • 7) 0.018 048 378 266 88 × 2 = 0 + 0.036 096 756 533 76;
  • 8) 0.036 096 756 533 76 × 2 = 0 + 0.072 193 513 067 52;
  • 9) 0.072 193 513 067 52 × 2 = 0 + 0.144 387 026 135 04;
  • 10) 0.144 387 026 135 04 × 2 = 0 + 0.288 774 052 270 08;
  • 11) 0.288 774 052 270 08 × 2 = 0 + 0.577 548 104 540 16;
  • 12) 0.577 548 104 540 16 × 2 = 1 + 0.155 096 209 080 32;
  • 13) 0.155 096 209 080 32 × 2 = 0 + 0.310 192 418 160 64;
  • 14) 0.310 192 418 160 64 × 2 = 0 + 0.620 384 836 321 28;
  • 15) 0.620 384 836 321 28 × 2 = 1 + 0.240 769 672 642 56;
  • 16) 0.240 769 672 642 56 × 2 = 0 + 0.481 539 345 285 12;
  • 17) 0.481 539 345 285 12 × 2 = 0 + 0.963 078 690 570 24;
  • 18) 0.963 078 690 570 24 × 2 = 1 + 0.926 157 381 140 48;
  • 19) 0.926 157 381 140 48 × 2 = 1 + 0.852 314 762 280 96;
  • 20) 0.852 314 762 280 96 × 2 = 1 + 0.704 629 524 561 92;
  • 21) 0.704 629 524 561 92 × 2 = 1 + 0.409 259 049 123 84;
  • 22) 0.409 259 049 123 84 × 2 = 0 + 0.818 518 098 247 68;
  • 23) 0.818 518 098 247 68 × 2 = 1 + 0.637 036 196 495 36;
  • 24) 0.637 036 196 495 36 × 2 = 1 + 0.274 072 392 990 72;
  • 25) 0.274 072 392 990 72 × 2 = 0 + 0.548 144 785 981 44;
  • 26) 0.548 144 785 981 44 × 2 = 1 + 0.096 289 571 962 88;
  • 27) 0.096 289 571 962 88 × 2 = 0 + 0.192 579 143 925 76;
  • 28) 0.192 579 143 925 76 × 2 = 0 + 0.385 158 287 851 52;
  • 29) 0.385 158 287 851 52 × 2 = 0 + 0.770 316 575 703 04;
  • 30) 0.770 316 575 703 04 × 2 = 1 + 0.540 633 151 406 08;
  • 31) 0.540 633 151 406 08 × 2 = 1 + 0.081 266 302 812 16;
  • 32) 0.081 266 302 812 16 × 2 = 0 + 0.162 532 605 624 32;
  • 33) 0.162 532 605 624 32 × 2 = 0 + 0.325 065 211 248 64;
  • 34) 0.325 065 211 248 64 × 2 = 0 + 0.650 130 422 497 28;
  • 35) 0.650 130 422 497 28 × 2 = 1 + 0.300 260 844 994 56;
  • 36) 0.300 260 844 994 56 × 2 = 0 + 0.600 521 689 989 12;
  • 37) 0.600 521 689 989 12 × 2 = 1 + 0.201 043 379 978 24;
  • 38) 0.201 043 379 978 24 × 2 = 0 + 0.402 086 759 956 48;
  • 39) 0.402 086 759 956 48 × 2 = 0 + 0.804 173 519 912 96;
  • 40) 0.804 173 519 912 96 × 2 = 1 + 0.608 347 039 825 92;
  • 41) 0.608 347 039 825 92 × 2 = 1 + 0.216 694 079 651 84;
  • 42) 0.216 694 079 651 84 × 2 = 0 + 0.433 388 159 303 68;
  • 43) 0.433 388 159 303 68 × 2 = 0 + 0.866 776 318 607 36;
  • 44) 0.866 776 318 607 36 × 2 = 1 + 0.733 552 637 214 72;
  • 45) 0.733 552 637 214 72 × 2 = 1 + 0.467 105 274 429 44;
  • 46) 0.467 105 274 429 44 × 2 = 0 + 0.934 210 548 858 88;
  • 47) 0.934 210 548 858 88 × 2 = 1 + 0.868 421 097 717 76;
  • 48) 0.868 421 097 717 76 × 2 = 1 + 0.736 842 195 435 52;
  • 49) 0.736 842 195 435 52 × 2 = 1 + 0.473 684 390 871 04;
  • 50) 0.473 684 390 871 04 × 2 = 0 + 0.947 368 781 742 08;
  • 51) 0.947 368 781 742 08 × 2 = 1 + 0.894 737 563 484 16;
  • 52) 0.894 737 563 484 16 × 2 = 1 + 0.789 475 126 968 32;
  • 53) 0.789 475 126 968 32 × 2 = 1 + 0.578 950 253 936 64;
  • 54) 0.578 950 253 936 64 × 2 = 1 + 0.157 900 507 873 28;
  • 55) 0.157 900 507 873 28 × 2 = 0 + 0.315 801 015 746 56;
  • 56) 0.315 801 015 746 56 × 2 = 0 + 0.631 602 031 493 12;
  • 57) 0.631 602 031 493 12 × 2 = 1 + 0.263 204 062 986 24;
  • 58) 0.263 204 062 986 24 × 2 = 0 + 0.526 408 125 972 48;
  • 59) 0.526 408 125 972 48 × 2 = 1 + 0.052 816 251 944 96;
  • 60) 0.052 816 251 944 96 × 2 = 0 + 0.105 632 503 889 92;
  • 61) 0.105 632 503 889 92 × 2 = 0 + 0.211 265 007 779 84;
  • 62) 0.211 265 007 779 84 × 2 = 0 + 0.422 530 015 559 68;
  • 63) 0.422 530 015 559 68 × 2 = 0 + 0.845 060 031 119 36;
  • 64) 0.845 060 031 119 36 × 2 = 1 + 0.690 120 062 238 72;

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:


0.000 282 005 910 42(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1001 1011 1011 1100 1010 0001(2)

6. Positive number before normalization:

0.000 282 005 910 42(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1001 1011 1011 1100 1010 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:


0.000 282 005 910 42(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1001 1011 1011 1100 1010 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1001 1001 1011 1011 1100 1010 0001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1001 1001 1011 1011 1100 1010 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 0100 0110 0010 1001 1001 1011 1011 1100 1010 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 0100 0110 0010 1001 1001 1011 1011 1100 1010 0001 =

0010 0111 1011 0100 0110 0010 1001 1001 1011 1011 1100 1010 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 0100 0110 0010 1001 1001 1011 1011 1100 1010 0001


Decimal number -0.000 282 005 910 42 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1001 1001 1011 1011 1100 1010 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