-0.180 000 000 038 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.180 000 000 038 2(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
-0.180 000 000 038 2(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:

|-0.180 000 000 038 2| = 0.180 000 000 038 2


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: 0.180 000 000 038 2.

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) 0.180 000 000 038 2 × 2 = 0 + 0.360 000 000 076 4;
  • 2) 0.360 000 000 076 4 × 2 = 0 + 0.720 000 000 152 8;
  • 3) 0.720 000 000 152 8 × 2 = 1 + 0.440 000 000 305 6;
  • 4) 0.440 000 000 305 6 × 2 = 0 + 0.880 000 000 611 2;
  • 5) 0.880 000 000 611 2 × 2 = 1 + 0.760 000 001 222 4;
  • 6) 0.760 000 001 222 4 × 2 = 1 + 0.520 000 002 444 8;
  • 7) 0.520 000 002 444 8 × 2 = 1 + 0.040 000 004 889 6;
  • 8) 0.040 000 004 889 6 × 2 = 0 + 0.080 000 009 779 2;
  • 9) 0.080 000 009 779 2 × 2 = 0 + 0.160 000 019 558 4;
  • 10) 0.160 000 019 558 4 × 2 = 0 + 0.320 000 039 116 8;
  • 11) 0.320 000 039 116 8 × 2 = 0 + 0.640 000 078 233 6;
  • 12) 0.640 000 078 233 6 × 2 = 1 + 0.280 000 156 467 2;
  • 13) 0.280 000 156 467 2 × 2 = 0 + 0.560 000 312 934 4;
  • 14) 0.560 000 312 934 4 × 2 = 1 + 0.120 000 625 868 8;
  • 15) 0.120 000 625 868 8 × 2 = 0 + 0.240 001 251 737 6;
  • 16) 0.240 001 251 737 6 × 2 = 0 + 0.480 002 503 475 2;
  • 17) 0.480 002 503 475 2 × 2 = 0 + 0.960 005 006 950 4;
  • 18) 0.960 005 006 950 4 × 2 = 1 + 0.920 010 013 900 8;
  • 19) 0.920 010 013 900 8 × 2 = 1 + 0.840 020 027 801 6;
  • 20) 0.840 020 027 801 6 × 2 = 1 + 0.680 040 055 603 2;
  • 21) 0.680 040 055 603 2 × 2 = 1 + 0.360 080 111 206 4;
  • 22) 0.360 080 111 206 4 × 2 = 0 + 0.720 160 222 412 8;
  • 23) 0.720 160 222 412 8 × 2 = 1 + 0.440 320 444 825 6;
  • 24) 0.440 320 444 825 6 × 2 = 0 + 0.880 640 889 651 2;
  • 25) 0.880 640 889 651 2 × 2 = 1 + 0.761 281 779 302 4;
  • 26) 0.761 281 779 302 4 × 2 = 1 + 0.522 563 558 604 8;
  • 27) 0.522 563 558 604 8 × 2 = 1 + 0.045 127 117 209 6;
  • 28) 0.045 127 117 209 6 × 2 = 0 + 0.090 254 234 419 2;
  • 29) 0.090 254 234 419 2 × 2 = 0 + 0.180 508 468 838 4;
  • 30) 0.180 508 468 838 4 × 2 = 0 + 0.361 016 937 676 8;
  • 31) 0.361 016 937 676 8 × 2 = 0 + 0.722 033 875 353 6;
  • 32) 0.722 033 875 353 6 × 2 = 1 + 0.444 067 750 707 2;
  • 33) 0.444 067 750 707 2 × 2 = 0 + 0.888 135 501 414 4;
  • 34) 0.888 135 501 414 4 × 2 = 1 + 0.776 271 002 828 8;
  • 35) 0.776 271 002 828 8 × 2 = 1 + 0.552 542 005 657 6;
  • 36) 0.552 542 005 657 6 × 2 = 1 + 0.105 084 011 315 2;
  • 37) 0.105 084 011 315 2 × 2 = 0 + 0.210 168 022 630 4;
  • 38) 0.210 168 022 630 4 × 2 = 0 + 0.420 336 045 260 8;
  • 39) 0.420 336 045 260 8 × 2 = 0 + 0.840 672 090 521 6;
  • 40) 0.840 672 090 521 6 × 2 = 1 + 0.681 344 181 043 2;
  • 41) 0.681 344 181 043 2 × 2 = 1 + 0.362 688 362 086 4;
  • 42) 0.362 688 362 086 4 × 2 = 0 + 0.725 376 724 172 8;
  • 43) 0.725 376 724 172 8 × 2 = 1 + 0.450 753 448 345 6;
  • 44) 0.450 753 448 345 6 × 2 = 0 + 0.901 506 896 691 2;
  • 45) 0.901 506 896 691 2 × 2 = 1 + 0.803 013 793 382 4;
  • 46) 0.803 013 793 382 4 × 2 = 1 + 0.606 027 586 764 8;
  • 47) 0.606 027 586 764 8 × 2 = 1 + 0.212 055 173 529 6;
  • 48) 0.212 055 173 529 6 × 2 = 0 + 0.424 110 347 059 2;
  • 49) 0.424 110 347 059 2 × 2 = 0 + 0.848 220 694 118 4;
  • 50) 0.848 220 694 118 4 × 2 = 1 + 0.696 441 388 236 8;
  • 51) 0.696 441 388 236 8 × 2 = 1 + 0.392 882 776 473 6;
  • 52) 0.392 882 776 473 6 × 2 = 0 + 0.785 765 552 947 2;
  • 53) 0.785 765 552 947 2 × 2 = 1 + 0.571 531 105 894 4;
  • 54) 0.571 531 105 894 4 × 2 = 1 + 0.143 062 211 788 8;
  • 55) 0.143 062 211 788 8 × 2 = 0 + 0.286 124 423 577 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:


0.180 000 000 038 2(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0001 1010 1110 0110 110(2)

6. Positive number before normalization:

0.180 000 000 038 2(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0001 1010 1110 0110 110(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:


0.180 000 000 038 2(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0001 1010 1110 0110 110(2) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0001 1010 1110 0110 110(2) × 20 =

1.0111 0000 1010 0011 1101 0111 0000 1011 1000 1101 0111 0011 0110(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):
1.0111 0000 1010 0011 1101 0111 0000 1011 1000 1101 0111 0011 0110


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. 0111 0000 1010 0011 1101 0111 0000 1011 1000 1101 0111 0011 0110 =

0111 0000 1010 0011 1101 0111 0000 1011 1000 1101 0111 0011 0110


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) =
0111 0000 1010 0011 1101 0111 0000 1011 1000 1101 0111 0011 0110


Decimal number -0.180 000 000 038 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0111 0000 1010 0011 1101 0111 0000 1011 1000 1101 0111 0011 0110

» Convert decimal -0.180 000 000 037 2 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