-0.000 282 005 913 16 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 16(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
-0.000 282 005 913 16(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:

|-0.000 282 005 913 16| = 0.000 282 005 913 16


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: 0.000 282 005 913 16.

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) 0.000 282 005 913 16 × 2 = 0 + 0.000 564 011 826 32;
  • 2) 0.000 564 011 826 32 × 2 = 0 + 0.001 128 023 652 64;
  • 3) 0.001 128 023 652 64 × 2 = 0 + 0.002 256 047 305 28;
  • 4) 0.002 256 047 305 28 × 2 = 0 + 0.004 512 094 610 56;
  • 5) 0.004 512 094 610 56 × 2 = 0 + 0.009 024 189 221 12;
  • 6) 0.009 024 189 221 12 × 2 = 0 + 0.018 048 378 442 24;
  • 7) 0.018 048 378 442 24 × 2 = 0 + 0.036 096 756 884 48;
  • 8) 0.036 096 756 884 48 × 2 = 0 + 0.072 193 513 768 96;
  • 9) 0.072 193 513 768 96 × 2 = 0 + 0.144 387 027 537 92;
  • 10) 0.144 387 027 537 92 × 2 = 0 + 0.288 774 055 075 84;
  • 11) 0.288 774 055 075 84 × 2 = 0 + 0.577 548 110 151 68;
  • 12) 0.577 548 110 151 68 × 2 = 1 + 0.155 096 220 303 36;
  • 13) 0.155 096 220 303 36 × 2 = 0 + 0.310 192 440 606 72;
  • 14) 0.310 192 440 606 72 × 2 = 0 + 0.620 384 881 213 44;
  • 15) 0.620 384 881 213 44 × 2 = 1 + 0.240 769 762 426 88;
  • 16) 0.240 769 762 426 88 × 2 = 0 + 0.481 539 524 853 76;
  • 17) 0.481 539 524 853 76 × 2 = 0 + 0.963 079 049 707 52;
  • 18) 0.963 079 049 707 52 × 2 = 1 + 0.926 158 099 415 04;
  • 19) 0.926 158 099 415 04 × 2 = 1 + 0.852 316 198 830 08;
  • 20) 0.852 316 198 830 08 × 2 = 1 + 0.704 632 397 660 16;
  • 21) 0.704 632 397 660 16 × 2 = 1 + 0.409 264 795 320 32;
  • 22) 0.409 264 795 320 32 × 2 = 0 + 0.818 529 590 640 64;
  • 23) 0.818 529 590 640 64 × 2 = 1 + 0.637 059 181 281 28;
  • 24) 0.637 059 181 281 28 × 2 = 1 + 0.274 118 362 562 56;
  • 25) 0.274 118 362 562 56 × 2 = 0 + 0.548 236 725 125 12;
  • 26) 0.548 236 725 125 12 × 2 = 1 + 0.096 473 450 250 24;
  • 27) 0.096 473 450 250 24 × 2 = 0 + 0.192 946 900 500 48;
  • 28) 0.192 946 900 500 48 × 2 = 0 + 0.385 893 801 000 96;
  • 29) 0.385 893 801 000 96 × 2 = 0 + 0.771 787 602 001 92;
  • 30) 0.771 787 602 001 92 × 2 = 1 + 0.543 575 204 003 84;
  • 31) 0.543 575 204 003 84 × 2 = 1 + 0.087 150 408 007 68;
  • 32) 0.087 150 408 007 68 × 2 = 0 + 0.174 300 816 015 36;
  • 33) 0.174 300 816 015 36 × 2 = 0 + 0.348 601 632 030 72;
  • 34) 0.348 601 632 030 72 × 2 = 0 + 0.697 203 264 061 44;
  • 35) 0.697 203 264 061 44 × 2 = 1 + 0.394 406 528 122 88;
  • 36) 0.394 406 528 122 88 × 2 = 0 + 0.788 813 056 245 76;
  • 37) 0.788 813 056 245 76 × 2 = 1 + 0.577 626 112 491 52;
  • 38) 0.577 626 112 491 52 × 2 = 1 + 0.155 252 224 983 04;
  • 39) 0.155 252 224 983 04 × 2 = 0 + 0.310 504 449 966 08;
  • 40) 0.310 504 449 966 08 × 2 = 0 + 0.621 008 899 932 16;
  • 41) 0.621 008 899 932 16 × 2 = 1 + 0.242 017 799 864 32;
  • 42) 0.242 017 799 864 32 × 2 = 0 + 0.484 035 599 728 64;
  • 43) 0.484 035 599 728 64 × 2 = 0 + 0.968 071 199 457 28;
  • 44) 0.968 071 199 457 28 × 2 = 1 + 0.936 142 398 914 56;
  • 45) 0.936 142 398 914 56 × 2 = 1 + 0.872 284 797 829 12;
  • 46) 0.872 284 797 829 12 × 2 = 1 + 0.744 569 595 658 24;
  • 47) 0.744 569 595 658 24 × 2 = 1 + 0.489 139 191 316 48;
  • 48) 0.489 139 191 316 48 × 2 = 0 + 0.978 278 382 632 96;
  • 49) 0.978 278 382 632 96 × 2 = 1 + 0.956 556 765 265 92;
  • 50) 0.956 556 765 265 92 × 2 = 1 + 0.913 113 530 531 84;
  • 51) 0.913 113 530 531 84 × 2 = 1 + 0.826 227 061 063 68;
  • 52) 0.826 227 061 063 68 × 2 = 1 + 0.652 454 122 127 36;
  • 53) 0.652 454 122 127 36 × 2 = 1 + 0.304 908 244 254 72;
  • 54) 0.304 908 244 254 72 × 2 = 0 + 0.609 816 488 509 44;
  • 55) 0.609 816 488 509 44 × 2 = 1 + 0.219 632 977 018 88;
  • 56) 0.219 632 977 018 88 × 2 = 0 + 0.439 265 954 037 76;
  • 57) 0.439 265 954 037 76 × 2 = 0 + 0.878 531 908 075 52;
  • 58) 0.878 531 908 075 52 × 2 = 1 + 0.757 063 816 151 04;
  • 59) 0.757 063 816 151 04 × 2 = 1 + 0.514 127 632 302 08;
  • 60) 0.514 127 632 302 08 × 2 = 1 + 0.028 255 264 604 16;
  • 61) 0.028 255 264 604 16 × 2 = 0 + 0.056 510 529 208 32;
  • 62) 0.056 510 529 208 32 × 2 = 0 + 0.113 021 058 416 64;
  • 63) 0.113 021 058 416 64 × 2 = 0 + 0.226 042 116 833 28;
  • 64) 0.226 042 116 833 28 × 2 = 0 + 0.452 084 233 666 56;

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:


0.000 282 005 913 16(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1001 1110 1111 1010 0111 0000(2)

6. Positive number before normalization:

0.000 282 005 913 16(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1001 1110 1111 1010 0111 0000(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:


0.000 282 005 913 16(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1001 1110 1111 1010 0111 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1001 1110 1111 1010 0111 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 1001 1110 1111 1010 0111 0000(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 1100 1001 1110 1111 1010 0111 0000


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 1100 1001 1110 1111 1010 0111 0000 =

0010 0111 1011 0100 0110 0010 1100 1001 1110 1111 1010 0111 0000


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 1100 1001 1110 1111 1010 0111 0000


Decimal number -0.000 282 005 913 16 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 1001 1110 1111 1010 0111 0000

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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