-0.000 282 006 04 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 04(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
-0.000 282 006 04(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:

|-0.000 282 006 04| = 0.000 282 006 04


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: 0.000 282 006 04.

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) 0.000 282 006 04 × 2 = 0 + 0.000 564 012 08;
  • 2) 0.000 564 012 08 × 2 = 0 + 0.001 128 024 16;
  • 3) 0.001 128 024 16 × 2 = 0 + 0.002 256 048 32;
  • 4) 0.002 256 048 32 × 2 = 0 + 0.004 512 096 64;
  • 5) 0.004 512 096 64 × 2 = 0 + 0.009 024 193 28;
  • 6) 0.009 024 193 28 × 2 = 0 + 0.018 048 386 56;
  • 7) 0.018 048 386 56 × 2 = 0 + 0.036 096 773 12;
  • 8) 0.036 096 773 12 × 2 = 0 + 0.072 193 546 24;
  • 9) 0.072 193 546 24 × 2 = 0 + 0.144 387 092 48;
  • 10) 0.144 387 092 48 × 2 = 0 + 0.288 774 184 96;
  • 11) 0.288 774 184 96 × 2 = 0 + 0.577 548 369 92;
  • 12) 0.577 548 369 92 × 2 = 1 + 0.155 096 739 84;
  • 13) 0.155 096 739 84 × 2 = 0 + 0.310 193 479 68;
  • 14) 0.310 193 479 68 × 2 = 0 + 0.620 386 959 36;
  • 15) 0.620 386 959 36 × 2 = 1 + 0.240 773 918 72;
  • 16) 0.240 773 918 72 × 2 = 0 + 0.481 547 837 44;
  • 17) 0.481 547 837 44 × 2 = 0 + 0.963 095 674 88;
  • 18) 0.963 095 674 88 × 2 = 1 + 0.926 191 349 76;
  • 19) 0.926 191 349 76 × 2 = 1 + 0.852 382 699 52;
  • 20) 0.852 382 699 52 × 2 = 1 + 0.704 765 399 04;
  • 21) 0.704 765 399 04 × 2 = 1 + 0.409 530 798 08;
  • 22) 0.409 530 798 08 × 2 = 0 + 0.819 061 596 16;
  • 23) 0.819 061 596 16 × 2 = 1 + 0.638 123 192 32;
  • 24) 0.638 123 192 32 × 2 = 1 + 0.276 246 384 64;
  • 25) 0.276 246 384 64 × 2 = 0 + 0.552 492 769 28;
  • 26) 0.552 492 769 28 × 2 = 1 + 0.104 985 538 56;
  • 27) 0.104 985 538 56 × 2 = 0 + 0.209 971 077 12;
  • 28) 0.209 971 077 12 × 2 = 0 + 0.419 942 154 24;
  • 29) 0.419 942 154 24 × 2 = 0 + 0.839 884 308 48;
  • 30) 0.839 884 308 48 × 2 = 1 + 0.679 768 616 96;
  • 31) 0.679 768 616 96 × 2 = 1 + 0.359 537 233 92;
  • 32) 0.359 537 233 92 × 2 = 0 + 0.719 074 467 84;
  • 33) 0.719 074 467 84 × 2 = 1 + 0.438 148 935 68;
  • 34) 0.438 148 935 68 × 2 = 0 + 0.876 297 871 36;
  • 35) 0.876 297 871 36 × 2 = 1 + 0.752 595 742 72;
  • 36) 0.752 595 742 72 × 2 = 1 + 0.505 191 485 44;
  • 37) 0.505 191 485 44 × 2 = 1 + 0.010 382 970 88;
  • 38) 0.010 382 970 88 × 2 = 0 + 0.020 765 941 76;
  • 39) 0.020 765 941 76 × 2 = 0 + 0.041 531 883 52;
  • 40) 0.041 531 883 52 × 2 = 0 + 0.083 063 767 04;
  • 41) 0.083 063 767 04 × 2 = 0 + 0.166 127 534 08;
  • 42) 0.166 127 534 08 × 2 = 0 + 0.332 255 068 16;
  • 43) 0.332 255 068 16 × 2 = 0 + 0.664 510 136 32;
  • 44) 0.664 510 136 32 × 2 = 1 + 0.329 020 272 64;
  • 45) 0.329 020 272 64 × 2 = 0 + 0.658 040 545 28;
  • 46) 0.658 040 545 28 × 2 = 1 + 0.316 081 090 56;
  • 47) 0.316 081 090 56 × 2 = 0 + 0.632 162 181 12;
  • 48) 0.632 162 181 12 × 2 = 1 + 0.264 324 362 24;
  • 49) 0.264 324 362 24 × 2 = 0 + 0.528 648 724 48;
  • 50) 0.528 648 724 48 × 2 = 1 + 0.057 297 448 96;
  • 51) 0.057 297 448 96 × 2 = 0 + 0.114 594 897 92;
  • 52) 0.114 594 897 92 × 2 = 0 + 0.229 189 795 84;
  • 53) 0.229 189 795 84 × 2 = 0 + 0.458 379 591 68;
  • 54) 0.458 379 591 68 × 2 = 0 + 0.916 759 183 36;
  • 55) 0.916 759 183 36 × 2 = 1 + 0.833 518 366 72;
  • 56) 0.833 518 366 72 × 2 = 1 + 0.667 036 733 44;
  • 57) 0.667 036 733 44 × 2 = 1 + 0.334 073 466 88;
  • 58) 0.334 073 466 88 × 2 = 0 + 0.668 146 933 76;
  • 59) 0.668 146 933 76 × 2 = 1 + 0.336 293 867 52;
  • 60) 0.336 293 867 52 × 2 = 0 + 0.672 587 735 04;
  • 61) 0.672 587 735 04 × 2 = 1 + 0.345 175 470 08;
  • 62) 0.345 175 470 08 × 2 = 0 + 0.690 350 940 16;
  • 63) 0.690 350 940 16 × 2 = 1 + 0.380 701 880 32;
  • 64) 0.380 701 880 32 × 2 = 0 + 0.761 403 760 64;

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:


0.000 282 006 04(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 1000 0001 0101 0100 0011 1010 1010(2)

6. Positive number before normalization:

0.000 282 006 04(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 1000 0001 0101 0100 0011 1010 1010(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:


0.000 282 006 04(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 1000 0001 0101 0100 0011 1010 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 1000 0001 0101 0100 0011 1010 1010(2) × 20 =

1.0010 0111 1011 0100 0110 1011 1000 0001 0101 0100 0011 1010 1010(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 1011 1000 0001 0101 0100 0011 1010 1010


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 1011 1000 0001 0101 0100 0011 1010 1010 =

0010 0111 1011 0100 0110 1011 1000 0001 0101 0100 0011 1010 1010


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 1011 1000 0001 0101 0100 0011 1010 1010


Decimal number -0.000 282 006 04 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1011 1000 0001 0101 0100 0011 1010 1010

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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