-0.000 282 005 78 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 78(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 78(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 78| = 0.000 282 005 78


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 78.

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 78 × 2 = 0 + 0.000 564 011 56;
  • 2) 0.000 564 011 56 × 2 = 0 + 0.001 128 023 12;
  • 3) 0.001 128 023 12 × 2 = 0 + 0.002 256 046 24;
  • 4) 0.002 256 046 24 × 2 = 0 + 0.004 512 092 48;
  • 5) 0.004 512 092 48 × 2 = 0 + 0.009 024 184 96;
  • 6) 0.009 024 184 96 × 2 = 0 + 0.018 048 369 92;
  • 7) 0.018 048 369 92 × 2 = 0 + 0.036 096 739 84;
  • 8) 0.036 096 739 84 × 2 = 0 + 0.072 193 479 68;
  • 9) 0.072 193 479 68 × 2 = 0 + 0.144 386 959 36;
  • 10) 0.144 386 959 36 × 2 = 0 + 0.288 773 918 72;
  • 11) 0.288 773 918 72 × 2 = 0 + 0.577 547 837 44;
  • 12) 0.577 547 837 44 × 2 = 1 + 0.155 095 674 88;
  • 13) 0.155 095 674 88 × 2 = 0 + 0.310 191 349 76;
  • 14) 0.310 191 349 76 × 2 = 0 + 0.620 382 699 52;
  • 15) 0.620 382 699 52 × 2 = 1 + 0.240 765 399 04;
  • 16) 0.240 765 399 04 × 2 = 0 + 0.481 530 798 08;
  • 17) 0.481 530 798 08 × 2 = 0 + 0.963 061 596 16;
  • 18) 0.963 061 596 16 × 2 = 1 + 0.926 123 192 32;
  • 19) 0.926 123 192 32 × 2 = 1 + 0.852 246 384 64;
  • 20) 0.852 246 384 64 × 2 = 1 + 0.704 492 769 28;
  • 21) 0.704 492 769 28 × 2 = 1 + 0.408 985 538 56;
  • 22) 0.408 985 538 56 × 2 = 0 + 0.817 971 077 12;
  • 23) 0.817 971 077 12 × 2 = 1 + 0.635 942 154 24;
  • 24) 0.635 942 154 24 × 2 = 1 + 0.271 884 308 48;
  • 25) 0.271 884 308 48 × 2 = 0 + 0.543 768 616 96;
  • 26) 0.543 768 616 96 × 2 = 1 + 0.087 537 233 92;
  • 27) 0.087 537 233 92 × 2 = 0 + 0.175 074 467 84;
  • 28) 0.175 074 467 84 × 2 = 0 + 0.350 148 935 68;
  • 29) 0.350 148 935 68 × 2 = 0 + 0.700 297 871 36;
  • 30) 0.700 297 871 36 × 2 = 1 + 0.400 595 742 72;
  • 31) 0.400 595 742 72 × 2 = 0 + 0.801 191 485 44;
  • 32) 0.801 191 485 44 × 2 = 1 + 0.602 382 970 88;
  • 33) 0.602 382 970 88 × 2 = 1 + 0.204 765 941 76;
  • 34) 0.204 765 941 76 × 2 = 0 + 0.409 531 883 52;
  • 35) 0.409 531 883 52 × 2 = 0 + 0.819 063 767 04;
  • 36) 0.819 063 767 04 × 2 = 1 + 0.638 127 534 08;
  • 37) 0.638 127 534 08 × 2 = 1 + 0.276 255 068 16;
  • 38) 0.276 255 068 16 × 2 = 0 + 0.552 510 136 32;
  • 39) 0.552 510 136 32 × 2 = 1 + 0.105 020 272 64;
  • 40) 0.105 020 272 64 × 2 = 0 + 0.210 040 545 28;
  • 41) 0.210 040 545 28 × 2 = 0 + 0.420 081 090 56;
  • 42) 0.420 081 090 56 × 2 = 0 + 0.840 162 181 12;
  • 43) 0.840 162 181 12 × 2 = 1 + 0.680 324 362 24;
  • 44) 0.680 324 362 24 × 2 = 1 + 0.360 648 724 48;
  • 45) 0.360 648 724 48 × 2 = 0 + 0.721 297 448 96;
  • 46) 0.721 297 448 96 × 2 = 1 + 0.442 594 897 92;
  • 47) 0.442 594 897 92 × 2 = 0 + 0.885 189 795 84;
  • 48) 0.885 189 795 84 × 2 = 1 + 0.770 379 591 68;
  • 49) 0.770 379 591 68 × 2 = 1 + 0.540 759 183 36;
  • 50) 0.540 759 183 36 × 2 = 1 + 0.081 518 366 72;
  • 51) 0.081 518 366 72 × 2 = 0 + 0.163 036 733 44;
  • 52) 0.163 036 733 44 × 2 = 0 + 0.326 073 466 88;
  • 53) 0.326 073 466 88 × 2 = 0 + 0.652 146 933 76;
  • 54) 0.652 146 933 76 × 2 = 1 + 0.304 293 867 52;
  • 55) 0.304 293 867 52 × 2 = 0 + 0.608 587 735 04;
  • 56) 0.608 587 735 04 × 2 = 1 + 0.217 175 470 08;
  • 57) 0.217 175 470 08 × 2 = 0 + 0.434 350 940 16;
  • 58) 0.434 350 940 16 × 2 = 0 + 0.868 701 880 32;
  • 59) 0.868 701 880 32 × 2 = 1 + 0.737 403 760 64;
  • 60) 0.737 403 760 64 × 2 = 1 + 0.474 807 521 28;
  • 61) 0.474 807 521 28 × 2 = 0 + 0.949 615 042 56;
  • 62) 0.949 615 042 56 × 2 = 1 + 0.899 230 085 12;
  • 63) 0.899 230 085 12 × 2 = 1 + 0.798 460 170 24;
  • 64) 0.798 460 170 24 × 2 = 1 + 0.596 920 340 48;

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 78(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1001 1010 0011 0101 1100 0101 0011 0111(2)

6. Positive number before normalization:

0.000 282 005 78(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1001 1010 0011 0101 1100 0101 0011 0111(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 78(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1001 1010 0011 0101 1100 0101 0011 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1001 1010 0011 0101 1100 0101 0011 0111(2) × 20 =

1.0010 0111 1011 0100 0101 1001 1010 0011 0101 1100 0101 0011 0111(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 0101 1001 1010 0011 0101 1100 0101 0011 0111


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 0101 1001 1010 0011 0101 1100 0101 0011 0111 =

0010 0111 1011 0100 0101 1001 1010 0011 0101 1100 0101 0011 0111


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 0101 1001 1010 0011 0101 1100 0101 0011 0111


Decimal number -0.000 282 005 78 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1001 1010 0011 0101 1100 0101 0011 0111

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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