-0.000 282 005 913 63 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 63(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 63(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 63| = 0.000 282 005 913 63


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 63.

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 63 × 2 = 0 + 0.000 564 011 827 26;
  • 2) 0.000 564 011 827 26 × 2 = 0 + 0.001 128 023 654 52;
  • 3) 0.001 128 023 654 52 × 2 = 0 + 0.002 256 047 309 04;
  • 4) 0.002 256 047 309 04 × 2 = 0 + 0.004 512 094 618 08;
  • 5) 0.004 512 094 618 08 × 2 = 0 + 0.009 024 189 236 16;
  • 6) 0.009 024 189 236 16 × 2 = 0 + 0.018 048 378 472 32;
  • 7) 0.018 048 378 472 32 × 2 = 0 + 0.036 096 756 944 64;
  • 8) 0.036 096 756 944 64 × 2 = 0 + 0.072 193 513 889 28;
  • 9) 0.072 193 513 889 28 × 2 = 0 + 0.144 387 027 778 56;
  • 10) 0.144 387 027 778 56 × 2 = 0 + 0.288 774 055 557 12;
  • 11) 0.288 774 055 557 12 × 2 = 0 + 0.577 548 111 114 24;
  • 12) 0.577 548 111 114 24 × 2 = 1 + 0.155 096 222 228 48;
  • 13) 0.155 096 222 228 48 × 2 = 0 + 0.310 192 444 456 96;
  • 14) 0.310 192 444 456 96 × 2 = 0 + 0.620 384 888 913 92;
  • 15) 0.620 384 888 913 92 × 2 = 1 + 0.240 769 777 827 84;
  • 16) 0.240 769 777 827 84 × 2 = 0 + 0.481 539 555 655 68;
  • 17) 0.481 539 555 655 68 × 2 = 0 + 0.963 079 111 311 36;
  • 18) 0.963 079 111 311 36 × 2 = 1 + 0.926 158 222 622 72;
  • 19) 0.926 158 222 622 72 × 2 = 1 + 0.852 316 445 245 44;
  • 20) 0.852 316 445 245 44 × 2 = 1 + 0.704 632 890 490 88;
  • 21) 0.704 632 890 490 88 × 2 = 1 + 0.409 265 780 981 76;
  • 22) 0.409 265 780 981 76 × 2 = 0 + 0.818 531 561 963 52;
  • 23) 0.818 531 561 963 52 × 2 = 1 + 0.637 063 123 927 04;
  • 24) 0.637 063 123 927 04 × 2 = 1 + 0.274 126 247 854 08;
  • 25) 0.274 126 247 854 08 × 2 = 0 + 0.548 252 495 708 16;
  • 26) 0.548 252 495 708 16 × 2 = 1 + 0.096 504 991 416 32;
  • 27) 0.096 504 991 416 32 × 2 = 0 + 0.193 009 982 832 64;
  • 28) 0.193 009 982 832 64 × 2 = 0 + 0.386 019 965 665 28;
  • 29) 0.386 019 965 665 28 × 2 = 0 + 0.772 039 931 330 56;
  • 30) 0.772 039 931 330 56 × 2 = 1 + 0.544 079 862 661 12;
  • 31) 0.544 079 862 661 12 × 2 = 1 + 0.088 159 725 322 24;
  • 32) 0.088 159 725 322 24 × 2 = 0 + 0.176 319 450 644 48;
  • 33) 0.176 319 450 644 48 × 2 = 0 + 0.352 638 901 288 96;
  • 34) 0.352 638 901 288 96 × 2 = 0 + 0.705 277 802 577 92;
  • 35) 0.705 277 802 577 92 × 2 = 1 + 0.410 555 605 155 84;
  • 36) 0.410 555 605 155 84 × 2 = 0 + 0.821 111 210 311 68;
  • 37) 0.821 111 210 311 68 × 2 = 1 + 0.642 222 420 623 36;
  • 38) 0.642 222 420 623 36 × 2 = 1 + 0.284 444 841 246 72;
  • 39) 0.284 444 841 246 72 × 2 = 0 + 0.568 889 682 493 44;
  • 40) 0.568 889 682 493 44 × 2 = 1 + 0.137 779 364 986 88;
  • 41) 0.137 779 364 986 88 × 2 = 0 + 0.275 558 729 973 76;
  • 42) 0.275 558 729 973 76 × 2 = 0 + 0.551 117 459 947 52;
  • 43) 0.551 117 459 947 52 × 2 = 1 + 0.102 234 919 895 04;
  • 44) 0.102 234 919 895 04 × 2 = 0 + 0.204 469 839 790 08;
  • 45) 0.204 469 839 790 08 × 2 = 0 + 0.408 939 679 580 16;
  • 46) 0.408 939 679 580 16 × 2 = 0 + 0.817 879 359 160 32;
  • 47) 0.817 879 359 160 32 × 2 = 1 + 0.635 758 718 320 64;
  • 48) 0.635 758 718 320 64 × 2 = 1 + 0.271 517 436 641 28;
  • 49) 0.271 517 436 641 28 × 2 = 0 + 0.543 034 873 282 56;
  • 50) 0.543 034 873 282 56 × 2 = 1 + 0.086 069 746 565 12;
  • 51) 0.086 069 746 565 12 × 2 = 0 + 0.172 139 493 130 24;
  • 52) 0.172 139 493 130 24 × 2 = 0 + 0.344 278 986 260 48;
  • 53) 0.344 278 986 260 48 × 2 = 0 + 0.688 557 972 520 96;
  • 54) 0.688 557 972 520 96 × 2 = 1 + 0.377 115 945 041 92;
  • 55) 0.377 115 945 041 92 × 2 = 0 + 0.754 231 890 083 84;
  • 56) 0.754 231 890 083 84 × 2 = 1 + 0.508 463 780 167 68;
  • 57) 0.508 463 780 167 68 × 2 = 1 + 0.016 927 560 335 36;
  • 58) 0.016 927 560 335 36 × 2 = 0 + 0.033 855 120 670 72;
  • 59) 0.033 855 120 670 72 × 2 = 0 + 0.067 710 241 341 44;
  • 60) 0.067 710 241 341 44 × 2 = 0 + 0.135 420 482 682 88;
  • 61) 0.135 420 482 682 88 × 2 = 0 + 0.270 840 965 365 76;
  • 62) 0.270 840 965 365 76 × 2 = 0 + 0.541 681 930 731 52;
  • 63) 0.541 681 930 731 52 × 2 = 1 + 0.083 363 861 463 04;
  • 64) 0.083 363 861 463 04 × 2 = 0 + 0.166 727 722 926 08;

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 63(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0010 0011 0100 0101 1000 0010(2)

6. Positive number before normalization:

0.000 282 005 913 63(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0010 0011 0100 0101 1000 0010(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 63(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0010 0011 0100 0101 1000 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0010 0011 0100 0101 1000 0010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 0010 0011 0100 0101 1000 0010(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 1101 0010 0011 0100 0101 1000 0010


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 1101 0010 0011 0100 0101 1000 0010 =

0010 0111 1011 0100 0110 0010 1101 0010 0011 0100 0101 1000 0010


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 1101 0010 0011 0100 0101 1000 0010


Decimal number -0.000 282 005 913 63 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 0010 0011 0100 0101 1000 0010

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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