-0.235 000 000 021 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.235 000 000 021 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
-0.235 000 000 021 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:

|-0.235 000 000 021 1| = 0.235 000 000 021 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: 0.235 000 000 021 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) 0.235 000 000 021 1 × 2 = 0 + 0.470 000 000 042 2;
  • 2) 0.470 000 000 042 2 × 2 = 0 + 0.940 000 000 084 4;
  • 3) 0.940 000 000 084 4 × 2 = 1 + 0.880 000 000 168 8;
  • 4) 0.880 000 000 168 8 × 2 = 1 + 0.760 000 000 337 6;
  • 5) 0.760 000 000 337 6 × 2 = 1 + 0.520 000 000 675 2;
  • 6) 0.520 000 000 675 2 × 2 = 1 + 0.040 000 001 350 4;
  • 7) 0.040 000 001 350 4 × 2 = 0 + 0.080 000 002 700 8;
  • 8) 0.080 000 002 700 8 × 2 = 0 + 0.160 000 005 401 6;
  • 9) 0.160 000 005 401 6 × 2 = 0 + 0.320 000 010 803 2;
  • 10) 0.320 000 010 803 2 × 2 = 0 + 0.640 000 021 606 4;
  • 11) 0.640 000 021 606 4 × 2 = 1 + 0.280 000 043 212 8;
  • 12) 0.280 000 043 212 8 × 2 = 0 + 0.560 000 086 425 6;
  • 13) 0.560 000 086 425 6 × 2 = 1 + 0.120 000 172 851 2;
  • 14) 0.120 000 172 851 2 × 2 = 0 + 0.240 000 345 702 4;
  • 15) 0.240 000 345 702 4 × 2 = 0 + 0.480 000 691 404 8;
  • 16) 0.480 000 691 404 8 × 2 = 0 + 0.960 001 382 809 6;
  • 17) 0.960 001 382 809 6 × 2 = 1 + 0.920 002 765 619 2;
  • 18) 0.920 002 765 619 2 × 2 = 1 + 0.840 005 531 238 4;
  • 19) 0.840 005 531 238 4 × 2 = 1 + 0.680 011 062 476 8;
  • 20) 0.680 011 062 476 8 × 2 = 1 + 0.360 022 124 953 6;
  • 21) 0.360 022 124 953 6 × 2 = 0 + 0.720 044 249 907 2;
  • 22) 0.720 044 249 907 2 × 2 = 1 + 0.440 088 499 814 4;
  • 23) 0.440 088 499 814 4 × 2 = 0 + 0.880 176 999 628 8;
  • 24) 0.880 176 999 628 8 × 2 = 1 + 0.760 353 999 257 6;
  • 25) 0.760 353 999 257 6 × 2 = 1 + 0.520 707 998 515 2;
  • 26) 0.520 707 998 515 2 × 2 = 1 + 0.041 415 997 030 4;
  • 27) 0.041 415 997 030 4 × 2 = 0 + 0.082 831 994 060 8;
  • 28) 0.082 831 994 060 8 × 2 = 0 + 0.165 663 988 121 6;
  • 29) 0.165 663 988 121 6 × 2 = 0 + 0.331 327 976 243 2;
  • 30) 0.331 327 976 243 2 × 2 = 0 + 0.662 655 952 486 4;
  • 31) 0.662 655 952 486 4 × 2 = 1 + 0.325 311 904 972 8;
  • 32) 0.325 311 904 972 8 × 2 = 0 + 0.650 623 809 945 6;
  • 33) 0.650 623 809 945 6 × 2 = 1 + 0.301 247 619 891 2;
  • 34) 0.301 247 619 891 2 × 2 = 0 + 0.602 495 239 782 4;
  • 35) 0.602 495 239 782 4 × 2 = 1 + 0.204 990 479 564 8;
  • 36) 0.204 990 479 564 8 × 2 = 0 + 0.409 980 959 129 6;
  • 37) 0.409 980 959 129 6 × 2 = 0 + 0.819 961 918 259 2;
  • 38) 0.819 961 918 259 2 × 2 = 1 + 0.639 923 836 518 4;
  • 39) 0.639 923 836 518 4 × 2 = 1 + 0.279 847 673 036 8;
  • 40) 0.279 847 673 036 8 × 2 = 0 + 0.559 695 346 073 6;
  • 41) 0.559 695 346 073 6 × 2 = 1 + 0.119 390 692 147 2;
  • 42) 0.119 390 692 147 2 × 2 = 0 + 0.238 781 384 294 4;
  • 43) 0.238 781 384 294 4 × 2 = 0 + 0.477 562 768 588 8;
  • 44) 0.477 562 768 588 8 × 2 = 0 + 0.955 125 537 177 6;
  • 45) 0.955 125 537 177 6 × 2 = 1 + 0.910 251 074 355 2;
  • 46) 0.910 251 074 355 2 × 2 = 1 + 0.820 502 148 710 4;
  • 47) 0.820 502 148 710 4 × 2 = 1 + 0.641 004 297 420 8;
  • 48) 0.641 004 297 420 8 × 2 = 1 + 0.282 008 594 841 6;
  • 49) 0.282 008 594 841 6 × 2 = 0 + 0.564 017 189 683 2;
  • 50) 0.564 017 189 683 2 × 2 = 1 + 0.128 034 379 366 4;
  • 51) 0.128 034 379 366 4 × 2 = 0 + 0.256 068 758 732 8;
  • 52) 0.256 068 758 732 8 × 2 = 0 + 0.512 137 517 465 6;
  • 53) 0.512 137 517 465 6 × 2 = 1 + 0.024 275 034 931 2;
  • 54) 0.024 275 034 931 2 × 2 = 0 + 0.048 550 069 862 4;
  • 55) 0.048 550 069 862 4 × 2 = 0 + 0.097 100 139 724 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:


0.235 000 000 021 1(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1010 0110 1000 1111 0100 100(2)

6. Positive number before normalization:

0.235 000 000 021 1(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1010 0110 1000 1111 0100 100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.235 000 000 021 1(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1010 0110 1000 1111 0100 100(2) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1010 0110 1000 1111 0100 100(2) × 20 =

1.1110 0001 0100 0111 1010 1110 0001 0101 0011 0100 0111 1010 0100(2) × 2-3


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): -3

Mantissa (not normalized):
1.1110 0001 0100 0111 1010 1110 0001 0101 0011 0100 0111 1010 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 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) =

1020(10) =

011 1111 1100(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. 1110 0001 0100 0111 1010 1110 0001 0101 0011 0100 0111 1010 0100 =

1110 0001 0100 0111 1010 1110 0001 0101 0011 0100 0111 1010 0100


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 1100

Mantissa (52 bits) =
1110 0001 0100 0111 1010 1110 0001 0101 0011 0100 0111 1010 0100


Decimal number -0.235 000 000 021 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 1110 0001 0100 0111 1010 1110 0001 0101 0011 0100 0111 1010 0100

» Convert decimal -0.235 000 000 027 6 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