-0.000 282 004 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 004 6(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
-0.000 282 004 6(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:

|-0.000 282 004 6| = 0.000 282 004 6


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: 0.000 282 004 6.

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) 0.000 282 004 6 × 2 = 0 + 0.000 564 009 2;
  • 2) 0.000 564 009 2 × 2 = 0 + 0.001 128 018 4;
  • 3) 0.001 128 018 4 × 2 = 0 + 0.002 256 036 8;
  • 4) 0.002 256 036 8 × 2 = 0 + 0.004 512 073 6;
  • 5) 0.004 512 073 6 × 2 = 0 + 0.009 024 147 2;
  • 6) 0.009 024 147 2 × 2 = 0 + 0.018 048 294 4;
  • 7) 0.018 048 294 4 × 2 = 0 + 0.036 096 588 8;
  • 8) 0.036 096 588 8 × 2 = 0 + 0.072 193 177 6;
  • 9) 0.072 193 177 6 × 2 = 0 + 0.144 386 355 2;
  • 10) 0.144 386 355 2 × 2 = 0 + 0.288 772 710 4;
  • 11) 0.288 772 710 4 × 2 = 0 + 0.577 545 420 8;
  • 12) 0.577 545 420 8 × 2 = 1 + 0.155 090 841 6;
  • 13) 0.155 090 841 6 × 2 = 0 + 0.310 181 683 2;
  • 14) 0.310 181 683 2 × 2 = 0 + 0.620 363 366 4;
  • 15) 0.620 363 366 4 × 2 = 1 + 0.240 726 732 8;
  • 16) 0.240 726 732 8 × 2 = 0 + 0.481 453 465 6;
  • 17) 0.481 453 465 6 × 2 = 0 + 0.962 906 931 2;
  • 18) 0.962 906 931 2 × 2 = 1 + 0.925 813 862 4;
  • 19) 0.925 813 862 4 × 2 = 1 + 0.851 627 724 8;
  • 20) 0.851 627 724 8 × 2 = 1 + 0.703 255 449 6;
  • 21) 0.703 255 449 6 × 2 = 1 + 0.406 510 899 2;
  • 22) 0.406 510 899 2 × 2 = 0 + 0.813 021 798 4;
  • 23) 0.813 021 798 4 × 2 = 1 + 0.626 043 596 8;
  • 24) 0.626 043 596 8 × 2 = 1 + 0.252 087 193 6;
  • 25) 0.252 087 193 6 × 2 = 0 + 0.504 174 387 2;
  • 26) 0.504 174 387 2 × 2 = 1 + 0.008 348 774 4;
  • 27) 0.008 348 774 4 × 2 = 0 + 0.016 697 548 8;
  • 28) 0.016 697 548 8 × 2 = 0 + 0.033 395 097 6;
  • 29) 0.033 395 097 6 × 2 = 0 + 0.066 790 195 2;
  • 30) 0.066 790 195 2 × 2 = 0 + 0.133 580 390 4;
  • 31) 0.133 580 390 4 × 2 = 0 + 0.267 160 780 8;
  • 32) 0.267 160 780 8 × 2 = 0 + 0.534 321 561 6;
  • 33) 0.534 321 561 6 × 2 = 1 + 0.068 643 123 2;
  • 34) 0.068 643 123 2 × 2 = 0 + 0.137 286 246 4;
  • 35) 0.137 286 246 4 × 2 = 0 + 0.274 572 492 8;
  • 36) 0.274 572 492 8 × 2 = 0 + 0.549 144 985 6;
  • 37) 0.549 144 985 6 × 2 = 1 + 0.098 289 971 2;
  • 38) 0.098 289 971 2 × 2 = 0 + 0.196 579 942 4;
  • 39) 0.196 579 942 4 × 2 = 0 + 0.393 159 884 8;
  • 40) 0.393 159 884 8 × 2 = 0 + 0.786 319 769 6;
  • 41) 0.786 319 769 6 × 2 = 1 + 0.572 639 539 2;
  • 42) 0.572 639 539 2 × 2 = 1 + 0.145 279 078 4;
  • 43) 0.145 279 078 4 × 2 = 0 + 0.290 558 156 8;
  • 44) 0.290 558 156 8 × 2 = 0 + 0.581 116 313 6;
  • 45) 0.581 116 313 6 × 2 = 1 + 0.162 232 627 2;
  • 46) 0.162 232 627 2 × 2 = 0 + 0.324 465 254 4;
  • 47) 0.324 465 254 4 × 2 = 0 + 0.648 930 508 8;
  • 48) 0.648 930 508 8 × 2 = 1 + 0.297 861 017 6;
  • 49) 0.297 861 017 6 × 2 = 0 + 0.595 722 035 2;
  • 50) 0.595 722 035 2 × 2 = 1 + 0.191 444 070 4;
  • 51) 0.191 444 070 4 × 2 = 0 + 0.382 888 140 8;
  • 52) 0.382 888 140 8 × 2 = 0 + 0.765 776 281 6;
  • 53) 0.765 776 281 6 × 2 = 1 + 0.531 552 563 2;
  • 54) 0.531 552 563 2 × 2 = 1 + 0.063 105 126 4;
  • 55) 0.063 105 126 4 × 2 = 0 + 0.126 210 252 8;
  • 56) 0.126 210 252 8 × 2 = 0 + 0.252 420 505 6;
  • 57) 0.252 420 505 6 × 2 = 0 + 0.504 841 011 2;
  • 58) 0.504 841 011 2 × 2 = 1 + 0.009 682 022 4;
  • 59) 0.009 682 022 4 × 2 = 0 + 0.019 364 044 8;
  • 60) 0.019 364 044 8 × 2 = 0 + 0.038 728 089 6;
  • 61) 0.038 728 089 6 × 2 = 0 + 0.077 456 179 2;
  • 62) 0.077 456 179 2 × 2 = 0 + 0.154 912 358 4;
  • 63) 0.154 912 358 4 × 2 = 0 + 0.309 824 716 8;
  • 64) 0.309 824 716 8 × 2 = 0 + 0.619 649 433 6;

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:


0.000 282 004 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0000 1000 1000 1100 1001 0100 1100 0100 0000(2)

6. Positive number before normalization:

0.000 282 004 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0000 1000 1000 1100 1001 0100 1100 0100 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:


0.000 282 004 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0000 1000 1000 1100 1001 0100 1100 0100 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0000 1000 1000 1100 1001 0100 1100 0100 0000(2) × 20 =

1.0010 0111 1011 0100 0000 1000 1000 1100 1001 0100 1100 0100 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 0000 1000 1000 1100 1001 0100 1100 0100 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 0000 1000 1000 1100 1001 0100 1100 0100 0000 =

0010 0111 1011 0100 0000 1000 1000 1100 1001 0100 1100 0100 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 0000 1000 1000 1100 1001 0100 1100 0100 0000


Decimal number -0.000 282 004 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0000 1000 1000 1100 1001 0100 1100 0100 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