-0.365 481 236 45 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.365 481 236 45(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 45(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 45| = 0.365 481 236 45


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 45.

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 45 × 2 = 0 + 0.730 962 472 9;
  • 2) 0.730 962 472 9 × 2 = 1 + 0.461 924 945 8;
  • 3) 0.461 924 945 8 × 2 = 0 + 0.923 849 891 6;
  • 4) 0.923 849 891 6 × 2 = 1 + 0.847 699 783 2;
  • 5) 0.847 699 783 2 × 2 = 1 + 0.695 399 566 4;
  • 6) 0.695 399 566 4 × 2 = 1 + 0.390 799 132 8;
  • 7) 0.390 799 132 8 × 2 = 0 + 0.781 598 265 6;
  • 8) 0.781 598 265 6 × 2 = 1 + 0.563 196 531 2;
  • 9) 0.563 196 531 2 × 2 = 1 + 0.126 393 062 4;
  • 10) 0.126 393 062 4 × 2 = 0 + 0.252 786 124 8;
  • 11) 0.252 786 124 8 × 2 = 0 + 0.505 572 249 6;
  • 12) 0.505 572 249 6 × 2 = 1 + 0.011 144 499 2;
  • 13) 0.011 144 499 2 × 2 = 0 + 0.022 288 998 4;
  • 14) 0.022 288 998 4 × 2 = 0 + 0.044 577 996 8;
  • 15) 0.044 577 996 8 × 2 = 0 + 0.089 155 993 6;
  • 16) 0.089 155 993 6 × 2 = 0 + 0.178 311 987 2;
  • 17) 0.178 311 987 2 × 2 = 0 + 0.356 623 974 4;
  • 18) 0.356 623 974 4 × 2 = 0 + 0.713 247 948 8;
  • 19) 0.713 247 948 8 × 2 = 1 + 0.426 495 897 6;
  • 20) 0.426 495 897 6 × 2 = 0 + 0.852 991 795 2;
  • 21) 0.852 991 795 2 × 2 = 1 + 0.705 983 590 4;
  • 22) 0.705 983 590 4 × 2 = 1 + 0.411 967 180 8;
  • 23) 0.411 967 180 8 × 2 = 0 + 0.823 934 361 6;
  • 24) 0.823 934 361 6 × 2 = 1 + 0.647 868 723 2;
  • 25) 0.647 868 723 2 × 2 = 1 + 0.295 737 446 4;
  • 26) 0.295 737 446 4 × 2 = 0 + 0.591 474 892 8;
  • 27) 0.591 474 892 8 × 2 = 1 + 0.182 949 785 6;
  • 28) 0.182 949 785 6 × 2 = 0 + 0.365 899 571 2;
  • 29) 0.365 899 571 2 × 2 = 0 + 0.731 799 142 4;
  • 30) 0.731 799 142 4 × 2 = 1 + 0.463 598 284 8;
  • 31) 0.463 598 284 8 × 2 = 0 + 0.927 196 569 6;
  • 32) 0.927 196 569 6 × 2 = 1 + 0.854 393 139 2;
  • 33) 0.854 393 139 2 × 2 = 1 + 0.708 786 278 4;
  • 34) 0.708 786 278 4 × 2 = 1 + 0.417 572 556 8;
  • 35) 0.417 572 556 8 × 2 = 0 + 0.835 145 113 6;
  • 36) 0.835 145 113 6 × 2 = 1 + 0.670 290 227 2;
  • 37) 0.670 290 227 2 × 2 = 1 + 0.340 580 454 4;
  • 38) 0.340 580 454 4 × 2 = 0 + 0.681 160 908 8;
  • 39) 0.681 160 908 8 × 2 = 1 + 0.362 321 817 6;
  • 40) 0.362 321 817 6 × 2 = 0 + 0.724 643 635 2;
  • 41) 0.724 643 635 2 × 2 = 1 + 0.449 287 270 4;
  • 42) 0.449 287 270 4 × 2 = 0 + 0.898 574 540 8;
  • 43) 0.898 574 540 8 × 2 = 1 + 0.797 149 081 6;
  • 44) 0.797 149 081 6 × 2 = 1 + 0.594 298 163 2;
  • 45) 0.594 298 163 2 × 2 = 1 + 0.188 596 326 4;
  • 46) 0.188 596 326 4 × 2 = 0 + 0.377 192 652 8;
  • 47) 0.377 192 652 8 × 2 = 0 + 0.754 385 305 6;
  • 48) 0.754 385 305 6 × 2 = 1 + 0.508 770 611 2;
  • 49) 0.508 770 611 2 × 2 = 1 + 0.017 541 222 4;
  • 50) 0.017 541 222 4 × 2 = 0 + 0.035 082 444 8;
  • 51) 0.035 082 444 8 × 2 = 0 + 0.070 164 889 6;
  • 52) 0.070 164 889 6 × 2 = 0 + 0.140 329 779 2;
  • 53) 0.140 329 779 2 × 2 = 0 + 0.280 659 558 4;
  • 54) 0.280 659 558 4 × 2 = 0 + 0.561 319 116 8;

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 45(10) =

0.0101 1101 1001 0000 0010 1101 1010 0101 1101 1010 1011 1001 1000 00(2)

6. Positive number before normalization:

0.365 481 236 45(10) =

0.0101 1101 1001 0000 0010 1101 1010 0101 1101 1010 1011 1001 1000 00(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 45(10) =

0.0101 1101 1001 0000 0010 1101 1010 0101 1101 1010 1011 1001 1000 00(2) =

0.0101 1101 1001 0000 0010 1101 1010 0101 1101 1010 1011 1001 1000 00(2) × 20 =

1.0111 0110 0100 0000 1011 0110 1001 0111 0110 1010 1110 0110 0000(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 0111 0110 1010 1110 0110 0000


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 0111 0110 1010 1110 0110 0000 =

0111 0110 0100 0000 1011 0110 1001 0111 0110 1010 1110 0110 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 1101

Mantissa (52 bits) =
0111 0110 0100 0000 1011 0110 1001 0111 0110 1010 1110 0110 0000


Decimal number -0.365 481 236 45 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 0111 0110 0100 0000 1011 0110 1001 0111 0110 1010 1110 0110 0000

» Convert decimal -0.365 481 236 56 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