-0.100 000 000 000 110 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.100 000 000 000 110 1(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
-0.100 000 000 000 110 1(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:

|-0.100 000 000 000 110 1| = 0.100 000 000 000 110 1


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: 0.100 000 000 000 110 1.

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) 0.100 000 000 000 110 1 × 2 = 0 + 0.200 000 000 000 220 2;
  • 2) 0.200 000 000 000 220 2 × 2 = 0 + 0.400 000 000 000 440 4;
  • 3) 0.400 000 000 000 440 4 × 2 = 0 + 0.800 000 000 000 880 8;
  • 4) 0.800 000 000 000 880 8 × 2 = 1 + 0.600 000 000 001 761 6;
  • 5) 0.600 000 000 001 761 6 × 2 = 1 + 0.200 000 000 003 523 2;
  • 6) 0.200 000 000 003 523 2 × 2 = 0 + 0.400 000 000 007 046 4;
  • 7) 0.400 000 000 007 046 4 × 2 = 0 + 0.800 000 000 014 092 8;
  • 8) 0.800 000 000 014 092 8 × 2 = 1 + 0.600 000 000 028 185 6;
  • 9) 0.600 000 000 028 185 6 × 2 = 1 + 0.200 000 000 056 371 2;
  • 10) 0.200 000 000 056 371 2 × 2 = 0 + 0.400 000 000 112 742 4;
  • 11) 0.400 000 000 112 742 4 × 2 = 0 + 0.800 000 000 225 484 8;
  • 12) 0.800 000 000 225 484 8 × 2 = 1 + 0.600 000 000 450 969 6;
  • 13) 0.600 000 000 450 969 6 × 2 = 1 + 0.200 000 000 901 939 2;
  • 14) 0.200 000 000 901 939 2 × 2 = 0 + 0.400 000 001 803 878 4;
  • 15) 0.400 000 001 803 878 4 × 2 = 0 + 0.800 000 003 607 756 8;
  • 16) 0.800 000 003 607 756 8 × 2 = 1 + 0.600 000 007 215 513 6;
  • 17) 0.600 000 007 215 513 6 × 2 = 1 + 0.200 000 014 431 027 2;
  • 18) 0.200 000 014 431 027 2 × 2 = 0 + 0.400 000 028 862 054 4;
  • 19) 0.400 000 028 862 054 4 × 2 = 0 + 0.800 000 057 724 108 8;
  • 20) 0.800 000 057 724 108 8 × 2 = 1 + 0.600 000 115 448 217 6;
  • 21) 0.600 000 115 448 217 6 × 2 = 1 + 0.200 000 230 896 435 2;
  • 22) 0.200 000 230 896 435 2 × 2 = 0 + 0.400 000 461 792 870 4;
  • 23) 0.400 000 461 792 870 4 × 2 = 0 + 0.800 000 923 585 740 8;
  • 24) 0.800 000 923 585 740 8 × 2 = 1 + 0.600 001 847 171 481 6;
  • 25) 0.600 001 847 171 481 6 × 2 = 1 + 0.200 003 694 342 963 2;
  • 26) 0.200 003 694 342 963 2 × 2 = 0 + 0.400 007 388 685 926 4;
  • 27) 0.400 007 388 685 926 4 × 2 = 0 + 0.800 014 777 371 852 8;
  • 28) 0.800 014 777 371 852 8 × 2 = 1 + 0.600 029 554 743 705 6;
  • 29) 0.600 029 554 743 705 6 × 2 = 1 + 0.200 059 109 487 411 2;
  • 30) 0.200 059 109 487 411 2 × 2 = 0 + 0.400 118 218 974 822 4;
  • 31) 0.400 118 218 974 822 4 × 2 = 0 + 0.800 236 437 949 644 8;
  • 32) 0.800 236 437 949 644 8 × 2 = 1 + 0.600 472 875 899 289 6;
  • 33) 0.600 472 875 899 289 6 × 2 = 1 + 0.200 945 751 798 579 2;
  • 34) 0.200 945 751 798 579 2 × 2 = 0 + 0.401 891 503 597 158 4;
  • 35) 0.401 891 503 597 158 4 × 2 = 0 + 0.803 783 007 194 316 8;
  • 36) 0.803 783 007 194 316 8 × 2 = 1 + 0.607 566 014 388 633 6;
  • 37) 0.607 566 014 388 633 6 × 2 = 1 + 0.215 132 028 777 267 2;
  • 38) 0.215 132 028 777 267 2 × 2 = 0 + 0.430 264 057 554 534 4;
  • 39) 0.430 264 057 554 534 4 × 2 = 0 + 0.860 528 115 109 068 8;
  • 40) 0.860 528 115 109 068 8 × 2 = 1 + 0.721 056 230 218 137 6;
  • 41) 0.721 056 230 218 137 6 × 2 = 1 + 0.442 112 460 436 275 2;
  • 42) 0.442 112 460 436 275 2 × 2 = 0 + 0.884 224 920 872 550 4;
  • 43) 0.884 224 920 872 550 4 × 2 = 1 + 0.768 449 841 745 100 8;
  • 44) 0.768 449 841 745 100 8 × 2 = 1 + 0.536 899 683 490 201 6;
  • 45) 0.536 899 683 490 201 6 × 2 = 1 + 0.073 799 366 980 403 2;
  • 46) 0.073 799 366 980 403 2 × 2 = 0 + 0.147 598 733 960 806 4;
  • 47) 0.147 598 733 960 806 4 × 2 = 0 + 0.295 197 467 921 612 8;
  • 48) 0.295 197 467 921 612 8 × 2 = 0 + 0.590 394 935 843 225 6;
  • 49) 0.590 394 935 843 225 6 × 2 = 1 + 0.180 789 871 686 451 2;
  • 50) 0.180 789 871 686 451 2 × 2 = 0 + 0.361 579 743 372 902 4;
  • 51) 0.361 579 743 372 902 4 × 2 = 0 + 0.723 159 486 745 804 8;
  • 52) 0.723 159 486 745 804 8 × 2 = 1 + 0.446 318 973 491 609 6;
  • 53) 0.446 318 973 491 609 6 × 2 = 0 + 0.892 637 946 983 219 2;
  • 54) 0.892 637 946 983 219 2 × 2 = 1 + 0.785 275 893 966 438 4;
  • 55) 0.785 275 893 966 438 4 × 2 = 1 + 0.570 551 787 932 876 8;
  • 56) 0.570 551 787 932 876 8 × 2 = 1 + 0.141 103 575 865 753 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:


0.100 000 000 000 110 1(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 1000 1001 0111(2)

6. Positive number before normalization:

0.100 000 000 000 110 1(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 1000 1001 0111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.100 000 000 000 110 1(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 1000 1001 0111(2) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 1000 1001 0111(2) × 20 =

1.1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 1000 1001 0111(2) × 2-4


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):
1.1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 1000 1001 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1019(10) =

011 1111 1011(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. 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 1000 1001 0111 =

1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 1000 1001 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 1011

Mantissa (52 bits) =
1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 1000 1001 0111


Decimal number -0.100 000 000 000 110 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 1000 1001 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