-0.235 000 000 018 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.235 000 000 018 3(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
-0.235 000 000 018 3(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:

|-0.235 000 000 018 3| = 0.235 000 000 018 3


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: 0.235 000 000 018 3.

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) 0.235 000 000 018 3 × 2 = 0 + 0.470 000 000 036 6;
  • 2) 0.470 000 000 036 6 × 2 = 0 + 0.940 000 000 073 2;
  • 3) 0.940 000 000 073 2 × 2 = 1 + 0.880 000 000 146 4;
  • 4) 0.880 000 000 146 4 × 2 = 1 + 0.760 000 000 292 8;
  • 5) 0.760 000 000 292 8 × 2 = 1 + 0.520 000 000 585 6;
  • 6) 0.520 000 000 585 6 × 2 = 1 + 0.040 000 001 171 2;
  • 7) 0.040 000 001 171 2 × 2 = 0 + 0.080 000 002 342 4;
  • 8) 0.080 000 002 342 4 × 2 = 0 + 0.160 000 004 684 8;
  • 9) 0.160 000 004 684 8 × 2 = 0 + 0.320 000 009 369 6;
  • 10) 0.320 000 009 369 6 × 2 = 0 + 0.640 000 018 739 2;
  • 11) 0.640 000 018 739 2 × 2 = 1 + 0.280 000 037 478 4;
  • 12) 0.280 000 037 478 4 × 2 = 0 + 0.560 000 074 956 8;
  • 13) 0.560 000 074 956 8 × 2 = 1 + 0.120 000 149 913 6;
  • 14) 0.120 000 149 913 6 × 2 = 0 + 0.240 000 299 827 2;
  • 15) 0.240 000 299 827 2 × 2 = 0 + 0.480 000 599 654 4;
  • 16) 0.480 000 599 654 4 × 2 = 0 + 0.960 001 199 308 8;
  • 17) 0.960 001 199 308 8 × 2 = 1 + 0.920 002 398 617 6;
  • 18) 0.920 002 398 617 6 × 2 = 1 + 0.840 004 797 235 2;
  • 19) 0.840 004 797 235 2 × 2 = 1 + 0.680 009 594 470 4;
  • 20) 0.680 009 594 470 4 × 2 = 1 + 0.360 019 188 940 8;
  • 21) 0.360 019 188 940 8 × 2 = 0 + 0.720 038 377 881 6;
  • 22) 0.720 038 377 881 6 × 2 = 1 + 0.440 076 755 763 2;
  • 23) 0.440 076 755 763 2 × 2 = 0 + 0.880 153 511 526 4;
  • 24) 0.880 153 511 526 4 × 2 = 1 + 0.760 307 023 052 8;
  • 25) 0.760 307 023 052 8 × 2 = 1 + 0.520 614 046 105 6;
  • 26) 0.520 614 046 105 6 × 2 = 1 + 0.041 228 092 211 2;
  • 27) 0.041 228 092 211 2 × 2 = 0 + 0.082 456 184 422 4;
  • 28) 0.082 456 184 422 4 × 2 = 0 + 0.164 912 368 844 8;
  • 29) 0.164 912 368 844 8 × 2 = 0 + 0.329 824 737 689 6;
  • 30) 0.329 824 737 689 6 × 2 = 0 + 0.659 649 475 379 2;
  • 31) 0.659 649 475 379 2 × 2 = 1 + 0.319 298 950 758 4;
  • 32) 0.319 298 950 758 4 × 2 = 0 + 0.638 597 901 516 8;
  • 33) 0.638 597 901 516 8 × 2 = 1 + 0.277 195 803 033 6;
  • 34) 0.277 195 803 033 6 × 2 = 0 + 0.554 391 606 067 2;
  • 35) 0.554 391 606 067 2 × 2 = 1 + 0.108 783 212 134 4;
  • 36) 0.108 783 212 134 4 × 2 = 0 + 0.217 566 424 268 8;
  • 37) 0.217 566 424 268 8 × 2 = 0 + 0.435 132 848 537 6;
  • 38) 0.435 132 848 537 6 × 2 = 0 + 0.870 265 697 075 2;
  • 39) 0.870 265 697 075 2 × 2 = 1 + 0.740 531 394 150 4;
  • 40) 0.740 531 394 150 4 × 2 = 1 + 0.481 062 788 300 8;
  • 41) 0.481 062 788 300 8 × 2 = 0 + 0.962 125 576 601 6;
  • 42) 0.962 125 576 601 6 × 2 = 1 + 0.924 251 153 203 2;
  • 43) 0.924 251 153 203 2 × 2 = 1 + 0.848 502 306 406 4;
  • 44) 0.848 502 306 406 4 × 2 = 1 + 0.697 004 612 812 8;
  • 45) 0.697 004 612 812 8 × 2 = 1 + 0.394 009 225 625 6;
  • 46) 0.394 009 225 625 6 × 2 = 0 + 0.788 018 451 251 2;
  • 47) 0.788 018 451 251 2 × 2 = 1 + 0.576 036 902 502 4;
  • 48) 0.576 036 902 502 4 × 2 = 1 + 0.152 073 805 004 8;
  • 49) 0.152 073 805 004 8 × 2 = 0 + 0.304 147 610 009 6;
  • 50) 0.304 147 610 009 6 × 2 = 0 + 0.608 295 220 019 2;
  • 51) 0.608 295 220 019 2 × 2 = 1 + 0.216 590 440 038 4;
  • 52) 0.216 590 440 038 4 × 2 = 0 + 0.433 180 880 076 8;
  • 53) 0.433 180 880 076 8 × 2 = 0 + 0.866 361 760 153 6;
  • 54) 0.866 361 760 153 6 × 2 = 1 + 0.732 723 520 307 2;
  • 55) 0.732 723 520 307 2 × 2 = 1 + 0.465 447 040 614 4;

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:


0.235 000 000 018 3(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1010 0011 0111 1011 0010 011(2)

6. Positive number before normalization:

0.235 000 000 018 3(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1010 0011 0111 1011 0010 011(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:


0.235 000 000 018 3(10) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1010 0011 0111 1011 0010 011(2) =

0.0011 1100 0010 1000 1111 0101 1100 0010 1010 0011 0111 1011 0010 011(2) × 20 =

1.1110 0001 0100 0111 1010 1110 0001 0101 0001 1011 1101 1001 0011(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 0001 1011 1101 1001 0011


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 0001 1011 1101 1001 0011 =

1110 0001 0100 0111 1010 1110 0001 0101 0001 1011 1101 1001 0011


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 0001 1011 1101 1001 0011


Decimal number -0.235 000 000 018 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 1110 0001 0100 0111 1010 1110 0001 0101 0001 1011 1101 1001 0011

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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