-0.000 282 009 84 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 009 84(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
-0.000 282 009 84(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:

|-0.000 282 009 84| = 0.000 282 009 84


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: 0.000 282 009 84.

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) 0.000 282 009 84 × 2 = 0 + 0.000 564 019 68;
  • 2) 0.000 564 019 68 × 2 = 0 + 0.001 128 039 36;
  • 3) 0.001 128 039 36 × 2 = 0 + 0.002 256 078 72;
  • 4) 0.002 256 078 72 × 2 = 0 + 0.004 512 157 44;
  • 5) 0.004 512 157 44 × 2 = 0 + 0.009 024 314 88;
  • 6) 0.009 024 314 88 × 2 = 0 + 0.018 048 629 76;
  • 7) 0.018 048 629 76 × 2 = 0 + 0.036 097 259 52;
  • 8) 0.036 097 259 52 × 2 = 0 + 0.072 194 519 04;
  • 9) 0.072 194 519 04 × 2 = 0 + 0.144 389 038 08;
  • 10) 0.144 389 038 08 × 2 = 0 + 0.288 778 076 16;
  • 11) 0.288 778 076 16 × 2 = 0 + 0.577 556 152 32;
  • 12) 0.577 556 152 32 × 2 = 1 + 0.155 112 304 64;
  • 13) 0.155 112 304 64 × 2 = 0 + 0.310 224 609 28;
  • 14) 0.310 224 609 28 × 2 = 0 + 0.620 449 218 56;
  • 15) 0.620 449 218 56 × 2 = 1 + 0.240 898 437 12;
  • 16) 0.240 898 437 12 × 2 = 0 + 0.481 796 874 24;
  • 17) 0.481 796 874 24 × 2 = 0 + 0.963 593 748 48;
  • 18) 0.963 593 748 48 × 2 = 1 + 0.927 187 496 96;
  • 19) 0.927 187 496 96 × 2 = 1 + 0.854 374 993 92;
  • 20) 0.854 374 993 92 × 2 = 1 + 0.708 749 987 84;
  • 21) 0.708 749 987 84 × 2 = 1 + 0.417 499 975 68;
  • 22) 0.417 499 975 68 × 2 = 0 + 0.834 999 951 36;
  • 23) 0.834 999 951 36 × 2 = 1 + 0.669 999 902 72;
  • 24) 0.669 999 902 72 × 2 = 1 + 0.339 999 805 44;
  • 25) 0.339 999 805 44 × 2 = 0 + 0.679 999 610 88;
  • 26) 0.679 999 610 88 × 2 = 1 + 0.359 999 221 76;
  • 27) 0.359 999 221 76 × 2 = 0 + 0.719 998 443 52;
  • 28) 0.719 998 443 52 × 2 = 1 + 0.439 996 887 04;
  • 29) 0.439 996 887 04 × 2 = 0 + 0.879 993 774 08;
  • 30) 0.879 993 774 08 × 2 = 1 + 0.759 987 548 16;
  • 31) 0.759 987 548 16 × 2 = 1 + 0.519 975 096 32;
  • 32) 0.519 975 096 32 × 2 = 1 + 0.039 950 192 64;
  • 33) 0.039 950 192 64 × 2 = 0 + 0.079 900 385 28;
  • 34) 0.079 900 385 28 × 2 = 0 + 0.159 800 770 56;
  • 35) 0.159 800 770 56 × 2 = 0 + 0.319 601 541 12;
  • 36) 0.319 601 541 12 × 2 = 0 + 0.639 203 082 24;
  • 37) 0.639 203 082 24 × 2 = 1 + 0.278 406 164 48;
  • 38) 0.278 406 164 48 × 2 = 0 + 0.556 812 328 96;
  • 39) 0.556 812 328 96 × 2 = 1 + 0.113 624 657 92;
  • 40) 0.113 624 657 92 × 2 = 0 + 0.227 249 315 84;
  • 41) 0.227 249 315 84 × 2 = 0 + 0.454 498 631 68;
  • 42) 0.454 498 631 68 × 2 = 0 + 0.908 997 263 36;
  • 43) 0.908 997 263 36 × 2 = 1 + 0.817 994 526 72;
  • 44) 0.817 994 526 72 × 2 = 1 + 0.635 989 053 44;
  • 45) 0.635 989 053 44 × 2 = 1 + 0.271 978 106 88;
  • 46) 0.271 978 106 88 × 2 = 0 + 0.543 956 213 76;
  • 47) 0.543 956 213 76 × 2 = 1 + 0.087 912 427 52;
  • 48) 0.087 912 427 52 × 2 = 0 + 0.175 824 855 04;
  • 49) 0.175 824 855 04 × 2 = 0 + 0.351 649 710 08;
  • 50) 0.351 649 710 08 × 2 = 0 + 0.703 299 420 16;
  • 51) 0.703 299 420 16 × 2 = 1 + 0.406 598 840 32;
  • 52) 0.406 598 840 32 × 2 = 0 + 0.813 197 680 64;
  • 53) 0.813 197 680 64 × 2 = 1 + 0.626 395 361 28;
  • 54) 0.626 395 361 28 × 2 = 1 + 0.252 790 722 56;
  • 55) 0.252 790 722 56 × 2 = 0 + 0.505 581 445 12;
  • 56) 0.505 581 445 12 × 2 = 1 + 0.011 162 890 24;
  • 57) 0.011 162 890 24 × 2 = 0 + 0.022 325 780 48;
  • 58) 0.022 325 780 48 × 2 = 0 + 0.044 651 560 96;
  • 59) 0.044 651 560 96 × 2 = 0 + 0.089 303 121 92;
  • 60) 0.089 303 121 92 × 2 = 0 + 0.178 606 243 84;
  • 61) 0.178 606 243 84 × 2 = 0 + 0.357 212 487 68;
  • 62) 0.357 212 487 68 × 2 = 0 + 0.714 424 975 36;
  • 63) 0.714 424 975 36 × 2 = 1 + 0.428 849 950 72;
  • 64) 0.428 849 950 72 × 2 = 0 + 0.857 699 901 44;

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:


0.000 282 009 84(10) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0000 1010 0011 1010 0010 1101 0000 0010(2)

6. Positive number before normalization:

0.000 282 009 84(10) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0000 1010 0011 1010 0010 1101 0000 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:


0.000 282 009 84(10) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0000 1010 0011 1010 0010 1101 0000 0010(2) =

0.0000 0000 0001 0010 0111 1011 0101 0111 0000 1010 0011 1010 0010 1101 0000 0010(2) × 20 =

1.0010 0111 1011 0101 0111 0000 1010 0011 1010 0010 1101 0000 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 0101 0111 0000 1010 0011 1010 0010 1101 0000 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 0101 0111 0000 1010 0011 1010 0010 1101 0000 0010 =

0010 0111 1011 0101 0111 0000 1010 0011 1010 0010 1101 0000 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 0101 0111 0000 1010 0011 1010 0010 1101 0000 0010


Decimal number -0.000 282 009 84 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0111 0000 1010 0011 1010 0010 1101 0000 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