-0.365 481 236 13 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.365 481 236 13(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
-0.365 481 236 13(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:

|-0.365 481 236 13| = 0.365 481 236 13


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: 0.365 481 236 13.

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) 0.365 481 236 13 × 2 = 0 + 0.730 962 472 26;
  • 2) 0.730 962 472 26 × 2 = 1 + 0.461 924 944 52;
  • 3) 0.461 924 944 52 × 2 = 0 + 0.923 849 889 04;
  • 4) 0.923 849 889 04 × 2 = 1 + 0.847 699 778 08;
  • 5) 0.847 699 778 08 × 2 = 1 + 0.695 399 556 16;
  • 6) 0.695 399 556 16 × 2 = 1 + 0.390 799 112 32;
  • 7) 0.390 799 112 32 × 2 = 0 + 0.781 598 224 64;
  • 8) 0.781 598 224 64 × 2 = 1 + 0.563 196 449 28;
  • 9) 0.563 196 449 28 × 2 = 1 + 0.126 392 898 56;
  • 10) 0.126 392 898 56 × 2 = 0 + 0.252 785 797 12;
  • 11) 0.252 785 797 12 × 2 = 0 + 0.505 571 594 24;
  • 12) 0.505 571 594 24 × 2 = 1 + 0.011 143 188 48;
  • 13) 0.011 143 188 48 × 2 = 0 + 0.022 286 376 96;
  • 14) 0.022 286 376 96 × 2 = 0 + 0.044 572 753 92;
  • 15) 0.044 572 753 92 × 2 = 0 + 0.089 145 507 84;
  • 16) 0.089 145 507 84 × 2 = 0 + 0.178 291 015 68;
  • 17) 0.178 291 015 68 × 2 = 0 + 0.356 582 031 36;
  • 18) 0.356 582 031 36 × 2 = 0 + 0.713 164 062 72;
  • 19) 0.713 164 062 72 × 2 = 1 + 0.426 328 125 44;
  • 20) 0.426 328 125 44 × 2 = 0 + 0.852 656 250 88;
  • 21) 0.852 656 250 88 × 2 = 1 + 0.705 312 501 76;
  • 22) 0.705 312 501 76 × 2 = 1 + 0.410 625 003 52;
  • 23) 0.410 625 003 52 × 2 = 0 + 0.821 250 007 04;
  • 24) 0.821 250 007 04 × 2 = 1 + 0.642 500 014 08;
  • 25) 0.642 500 014 08 × 2 = 1 + 0.285 000 028 16;
  • 26) 0.285 000 028 16 × 2 = 0 + 0.570 000 056 32;
  • 27) 0.570 000 056 32 × 2 = 1 + 0.140 000 112 64;
  • 28) 0.140 000 112 64 × 2 = 0 + 0.280 000 225 28;
  • 29) 0.280 000 225 28 × 2 = 0 + 0.560 000 450 56;
  • 30) 0.560 000 450 56 × 2 = 1 + 0.120 000 901 12;
  • 31) 0.120 000 901 12 × 2 = 0 + 0.240 001 802 24;
  • 32) 0.240 001 802 24 × 2 = 0 + 0.480 003 604 48;
  • 33) 0.480 003 604 48 × 2 = 0 + 0.960 007 208 96;
  • 34) 0.960 007 208 96 × 2 = 1 + 0.920 014 417 92;
  • 35) 0.920 014 417 92 × 2 = 1 + 0.840 028 835 84;
  • 36) 0.840 028 835 84 × 2 = 1 + 0.680 057 671 68;
  • 37) 0.680 057 671 68 × 2 = 1 + 0.360 115 343 36;
  • 38) 0.360 115 343 36 × 2 = 0 + 0.720 230 686 72;
  • 39) 0.720 230 686 72 × 2 = 1 + 0.440 461 373 44;
  • 40) 0.440 461 373 44 × 2 = 0 + 0.880 922 746 88;
  • 41) 0.880 922 746 88 × 2 = 1 + 0.761 845 493 76;
  • 42) 0.761 845 493 76 × 2 = 1 + 0.523 690 987 52;
  • 43) 0.523 690 987 52 × 2 = 1 + 0.047 381 975 04;
  • 44) 0.047 381 975 04 × 2 = 0 + 0.094 763 950 08;
  • 45) 0.094 763 950 08 × 2 = 0 + 0.189 527 900 16;
  • 46) 0.189 527 900 16 × 2 = 0 + 0.379 055 800 32;
  • 47) 0.379 055 800 32 × 2 = 0 + 0.758 111 600 64;
  • 48) 0.758 111 600 64 × 2 = 1 + 0.516 223 201 28;
  • 49) 0.516 223 201 28 × 2 = 1 + 0.032 446 402 56;
  • 50) 0.032 446 402 56 × 2 = 0 + 0.064 892 805 12;
  • 51) 0.064 892 805 12 × 2 = 0 + 0.129 785 610 24;
  • 52) 0.129 785 610 24 × 2 = 0 + 0.259 571 220 48;
  • 53) 0.259 571 220 48 × 2 = 0 + 0.519 142 440 96;
  • 54) 0.519 142 440 96 × 2 = 1 + 0.038 284 881 92;

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:


0.365 481 236 13(10) =

0.0101 1101 1001 0000 0010 1101 1010 0100 0111 1010 1110 0001 1000 01(2)

6. Positive number before normalization:

0.365 481 236 13(10) =

0.0101 1101 1001 0000 0010 1101 1010 0100 0111 1010 1110 0001 1000 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.365 481 236 13(10) =

0.0101 1101 1001 0000 0010 1101 1010 0100 0111 1010 1110 0001 1000 01(2) =

0.0101 1101 1001 0000 0010 1101 1010 0100 0111 1010 1110 0001 1000 01(2) × 20 =

1.0111 0110 0100 0000 1011 0110 1001 0001 1110 1011 1000 0110 0001(2) × 2-2


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): -2

Mantissa (not normalized):
1.0111 0110 0100 0000 1011 0110 1001 0001 1110 1011 1000 0110 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-2 + 2(11-1) - 1 =

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1021(10) =

011 1111 1101(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. 0111 0110 0100 0000 1011 0110 1001 0001 1110 1011 1000 0110 0001 =

0111 0110 0100 0000 1011 0110 1001 0001 1110 1011 1000 0110 0001


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 1101

Mantissa (52 bits) =
0111 0110 0100 0000 1011 0110 1001 0001 1110 1011 1000 0110 0001


Decimal number -0.365 481 236 13 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 0111 0110 0100 0000 1011 0110 1001 0001 1110 1011 1000 0110 0001

» Convert decimal -0.365 481 236 71 to 64 bit double precision IEEE 754 binary floating point representation standard

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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