-0.180 000 000 029 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.180 000 000 029 4(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
-0.180 000 000 029 4(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:

|-0.180 000 000 029 4| = 0.180 000 000 029 4


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: 0.180 000 000 029 4.

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) 0.180 000 000 029 4 × 2 = 0 + 0.360 000 000 058 8;
  • 2) 0.360 000 000 058 8 × 2 = 0 + 0.720 000 000 117 6;
  • 3) 0.720 000 000 117 6 × 2 = 1 + 0.440 000 000 235 2;
  • 4) 0.440 000 000 235 2 × 2 = 0 + 0.880 000 000 470 4;
  • 5) 0.880 000 000 470 4 × 2 = 1 + 0.760 000 000 940 8;
  • 6) 0.760 000 000 940 8 × 2 = 1 + 0.520 000 001 881 6;
  • 7) 0.520 000 001 881 6 × 2 = 1 + 0.040 000 003 763 2;
  • 8) 0.040 000 003 763 2 × 2 = 0 + 0.080 000 007 526 4;
  • 9) 0.080 000 007 526 4 × 2 = 0 + 0.160 000 015 052 8;
  • 10) 0.160 000 015 052 8 × 2 = 0 + 0.320 000 030 105 6;
  • 11) 0.320 000 030 105 6 × 2 = 0 + 0.640 000 060 211 2;
  • 12) 0.640 000 060 211 2 × 2 = 1 + 0.280 000 120 422 4;
  • 13) 0.280 000 120 422 4 × 2 = 0 + 0.560 000 240 844 8;
  • 14) 0.560 000 240 844 8 × 2 = 1 + 0.120 000 481 689 6;
  • 15) 0.120 000 481 689 6 × 2 = 0 + 0.240 000 963 379 2;
  • 16) 0.240 000 963 379 2 × 2 = 0 + 0.480 001 926 758 4;
  • 17) 0.480 001 926 758 4 × 2 = 0 + 0.960 003 853 516 8;
  • 18) 0.960 003 853 516 8 × 2 = 1 + 0.920 007 707 033 6;
  • 19) 0.920 007 707 033 6 × 2 = 1 + 0.840 015 414 067 2;
  • 20) 0.840 015 414 067 2 × 2 = 1 + 0.680 030 828 134 4;
  • 21) 0.680 030 828 134 4 × 2 = 1 + 0.360 061 656 268 8;
  • 22) 0.360 061 656 268 8 × 2 = 0 + 0.720 123 312 537 6;
  • 23) 0.720 123 312 537 6 × 2 = 1 + 0.440 246 625 075 2;
  • 24) 0.440 246 625 075 2 × 2 = 0 + 0.880 493 250 150 4;
  • 25) 0.880 493 250 150 4 × 2 = 1 + 0.760 986 500 300 8;
  • 26) 0.760 986 500 300 8 × 2 = 1 + 0.521 973 000 601 6;
  • 27) 0.521 973 000 601 6 × 2 = 1 + 0.043 946 001 203 2;
  • 28) 0.043 946 001 203 2 × 2 = 0 + 0.087 892 002 406 4;
  • 29) 0.087 892 002 406 4 × 2 = 0 + 0.175 784 004 812 8;
  • 30) 0.175 784 004 812 8 × 2 = 0 + 0.351 568 009 625 6;
  • 31) 0.351 568 009 625 6 × 2 = 0 + 0.703 136 019 251 2;
  • 32) 0.703 136 019 251 2 × 2 = 1 + 0.406 272 038 502 4;
  • 33) 0.406 272 038 502 4 × 2 = 0 + 0.812 544 077 004 8;
  • 34) 0.812 544 077 004 8 × 2 = 1 + 0.625 088 154 009 6;
  • 35) 0.625 088 154 009 6 × 2 = 1 + 0.250 176 308 019 2;
  • 36) 0.250 176 308 019 2 × 2 = 0 + 0.500 352 616 038 4;
  • 37) 0.500 352 616 038 4 × 2 = 1 + 0.000 705 232 076 8;
  • 38) 0.000 705 232 076 8 × 2 = 0 + 0.001 410 464 153 6;
  • 39) 0.001 410 464 153 6 × 2 = 0 + 0.002 820 928 307 2;
  • 40) 0.002 820 928 307 2 × 2 = 0 + 0.005 641 856 614 4;
  • 41) 0.005 641 856 614 4 × 2 = 0 + 0.011 283 713 228 8;
  • 42) 0.011 283 713 228 8 × 2 = 0 + 0.022 567 426 457 6;
  • 43) 0.022 567 426 457 6 × 2 = 0 + 0.045 134 852 915 2;
  • 44) 0.045 134 852 915 2 × 2 = 0 + 0.090 269 705 830 4;
  • 45) 0.090 269 705 830 4 × 2 = 0 + 0.180 539 411 660 8;
  • 46) 0.180 539 411 660 8 × 2 = 0 + 0.361 078 823 321 6;
  • 47) 0.361 078 823 321 6 × 2 = 0 + 0.722 157 646 643 2;
  • 48) 0.722 157 646 643 2 × 2 = 1 + 0.444 315 293 286 4;
  • 49) 0.444 315 293 286 4 × 2 = 0 + 0.888 630 586 572 8;
  • 50) 0.888 630 586 572 8 × 2 = 1 + 0.777 261 173 145 6;
  • 51) 0.777 261 173 145 6 × 2 = 1 + 0.554 522 346 291 2;
  • 52) 0.554 522 346 291 2 × 2 = 1 + 0.109 044 692 582 4;
  • 53) 0.109 044 692 582 4 × 2 = 0 + 0.218 089 385 164 8;
  • 54) 0.218 089 385 164 8 × 2 = 0 + 0.436 178 770 329 6;
  • 55) 0.436 178 770 329 6 × 2 = 0 + 0.872 357 540 659 2;

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:


0.180 000 000 029 4(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1000 0000 0001 0111 000(2)

6. Positive number before normalization:

0.180 000 000 029 4(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1000 0000 0001 0111 000(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:


0.180 000 000 029 4(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1000 0000 0001 0111 000(2) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1000 0000 0001 0111 000(2) × 20 =

1.0111 0000 1010 0011 1101 0111 0000 1011 0100 0000 0000 1011 1000(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 0100 0000 0000 1011 1000


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 0100 0000 0000 1011 1000 =

0111 0000 1010 0011 1101 0111 0000 1011 0100 0000 0000 1011 1000


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 0100 0000 0000 1011 1000


Decimal number -0.180 000 000 029 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0111 0000 1010 0011 1101 0111 0000 1011 0100 0000 0000 1011 1000

» Convert decimal -0.180 000 000 036 1 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