-0.100 000 000 000 107 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.100 000 000 000 107 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
-0.100 000 000 000 107 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:

|-0.100 000 000 000 107 1| = 0.100 000 000 000 107 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: 0.100 000 000 000 107 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) 0.100 000 000 000 107 1 × 2 = 0 + 0.200 000 000 000 214 2;
  • 2) 0.200 000 000 000 214 2 × 2 = 0 + 0.400 000 000 000 428 4;
  • 3) 0.400 000 000 000 428 4 × 2 = 0 + 0.800 000 000 000 856 8;
  • 4) 0.800 000 000 000 856 8 × 2 = 1 + 0.600 000 000 001 713 6;
  • 5) 0.600 000 000 001 713 6 × 2 = 1 + 0.200 000 000 003 427 2;
  • 6) 0.200 000 000 003 427 2 × 2 = 0 + 0.400 000 000 006 854 4;
  • 7) 0.400 000 000 006 854 4 × 2 = 0 + 0.800 000 000 013 708 8;
  • 8) 0.800 000 000 013 708 8 × 2 = 1 + 0.600 000 000 027 417 6;
  • 9) 0.600 000 000 027 417 6 × 2 = 1 + 0.200 000 000 054 835 2;
  • 10) 0.200 000 000 054 835 2 × 2 = 0 + 0.400 000 000 109 670 4;
  • 11) 0.400 000 000 109 670 4 × 2 = 0 + 0.800 000 000 219 340 8;
  • 12) 0.800 000 000 219 340 8 × 2 = 1 + 0.600 000 000 438 681 6;
  • 13) 0.600 000 000 438 681 6 × 2 = 1 + 0.200 000 000 877 363 2;
  • 14) 0.200 000 000 877 363 2 × 2 = 0 + 0.400 000 001 754 726 4;
  • 15) 0.400 000 001 754 726 4 × 2 = 0 + 0.800 000 003 509 452 8;
  • 16) 0.800 000 003 509 452 8 × 2 = 1 + 0.600 000 007 018 905 6;
  • 17) 0.600 000 007 018 905 6 × 2 = 1 + 0.200 000 014 037 811 2;
  • 18) 0.200 000 014 037 811 2 × 2 = 0 + 0.400 000 028 075 622 4;
  • 19) 0.400 000 028 075 622 4 × 2 = 0 + 0.800 000 056 151 244 8;
  • 20) 0.800 000 056 151 244 8 × 2 = 1 + 0.600 000 112 302 489 6;
  • 21) 0.600 000 112 302 489 6 × 2 = 1 + 0.200 000 224 604 979 2;
  • 22) 0.200 000 224 604 979 2 × 2 = 0 + 0.400 000 449 209 958 4;
  • 23) 0.400 000 449 209 958 4 × 2 = 0 + 0.800 000 898 419 916 8;
  • 24) 0.800 000 898 419 916 8 × 2 = 1 + 0.600 001 796 839 833 6;
  • 25) 0.600 001 796 839 833 6 × 2 = 1 + 0.200 003 593 679 667 2;
  • 26) 0.200 003 593 679 667 2 × 2 = 0 + 0.400 007 187 359 334 4;
  • 27) 0.400 007 187 359 334 4 × 2 = 0 + 0.800 014 374 718 668 8;
  • 28) 0.800 014 374 718 668 8 × 2 = 1 + 0.600 028 749 437 337 6;
  • 29) 0.600 028 749 437 337 6 × 2 = 1 + 0.200 057 498 874 675 2;
  • 30) 0.200 057 498 874 675 2 × 2 = 0 + 0.400 114 997 749 350 4;
  • 31) 0.400 114 997 749 350 4 × 2 = 0 + 0.800 229 995 498 700 8;
  • 32) 0.800 229 995 498 700 8 × 2 = 1 + 0.600 459 990 997 401 6;
  • 33) 0.600 459 990 997 401 6 × 2 = 1 + 0.200 919 981 994 803 2;
  • 34) 0.200 919 981 994 803 2 × 2 = 0 + 0.401 839 963 989 606 4;
  • 35) 0.401 839 963 989 606 4 × 2 = 0 + 0.803 679 927 979 212 8;
  • 36) 0.803 679 927 979 212 8 × 2 = 1 + 0.607 359 855 958 425 6;
  • 37) 0.607 359 855 958 425 6 × 2 = 1 + 0.214 719 711 916 851 2;
  • 38) 0.214 719 711 916 851 2 × 2 = 0 + 0.429 439 423 833 702 4;
  • 39) 0.429 439 423 833 702 4 × 2 = 0 + 0.858 878 847 667 404 8;
  • 40) 0.858 878 847 667 404 8 × 2 = 1 + 0.717 757 695 334 809 6;
  • 41) 0.717 757 695 334 809 6 × 2 = 1 + 0.435 515 390 669 619 2;
  • 42) 0.435 515 390 669 619 2 × 2 = 0 + 0.871 030 781 339 238 4;
  • 43) 0.871 030 781 339 238 4 × 2 = 1 + 0.742 061 562 678 476 8;
  • 44) 0.742 061 562 678 476 8 × 2 = 1 + 0.484 123 125 356 953 6;
  • 45) 0.484 123 125 356 953 6 × 2 = 0 + 0.968 246 250 713 907 2;
  • 46) 0.968 246 250 713 907 2 × 2 = 1 + 0.936 492 501 427 814 4;
  • 47) 0.936 492 501 427 814 4 × 2 = 1 + 0.872 985 002 855 628 8;
  • 48) 0.872 985 002 855 628 8 × 2 = 1 + 0.745 970 005 711 257 6;
  • 49) 0.745 970 005 711 257 6 × 2 = 1 + 0.491 940 011 422 515 2;
  • 50) 0.491 940 011 422 515 2 × 2 = 0 + 0.983 880 022 845 030 4;
  • 51) 0.983 880 022 845 030 4 × 2 = 1 + 0.967 760 045 690 060 8;
  • 52) 0.967 760 045 690 060 8 × 2 = 1 + 0.935 520 091 380 121 6;
  • 53) 0.935 520 091 380 121 6 × 2 = 1 + 0.871 040 182 760 243 2;
  • 54) 0.871 040 182 760 243 2 × 2 = 1 + 0.742 080 365 520 486 4;
  • 55) 0.742 080 365 520 486 4 × 2 = 1 + 0.484 160 731 040 972 8;
  • 56) 0.484 160 731 040 972 8 × 2 = 0 + 0.968 321 462 081 945 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:


0.100 000 000 000 107 1(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 0111 1011 1110(2)

6. Positive number before normalization:

0.100 000 000 000 107 1(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 0111 1011 1110(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:


0.100 000 000 000 107 1(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 0111 1011 1110(2) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 0111 1011 1110(2) × 20 =

1.1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 0111 1011 1110(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 0111 1011 1110


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 0111 1011 1110 =

1001 1001 1001 1001 1001 1001 1001 1001 1001 1011 0111 1011 1110


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 0111 1011 1110


Decimal number -0.100 000 000 000 107 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 0111 1011 1110

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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