-0.180 000 000 026 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.180 000 000 026 9(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
-0.180 000 000 026 9(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:

|-0.180 000 000 026 9| = 0.180 000 000 026 9


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: 0.180 000 000 026 9.

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) 0.180 000 000 026 9 × 2 = 0 + 0.360 000 000 053 8;
  • 2) 0.360 000 000 053 8 × 2 = 0 + 0.720 000 000 107 6;
  • 3) 0.720 000 000 107 6 × 2 = 1 + 0.440 000 000 215 2;
  • 4) 0.440 000 000 215 2 × 2 = 0 + 0.880 000 000 430 4;
  • 5) 0.880 000 000 430 4 × 2 = 1 + 0.760 000 000 860 8;
  • 6) 0.760 000 000 860 8 × 2 = 1 + 0.520 000 001 721 6;
  • 7) 0.520 000 001 721 6 × 2 = 1 + 0.040 000 003 443 2;
  • 8) 0.040 000 003 443 2 × 2 = 0 + 0.080 000 006 886 4;
  • 9) 0.080 000 006 886 4 × 2 = 0 + 0.160 000 013 772 8;
  • 10) 0.160 000 013 772 8 × 2 = 0 + 0.320 000 027 545 6;
  • 11) 0.320 000 027 545 6 × 2 = 0 + 0.640 000 055 091 2;
  • 12) 0.640 000 055 091 2 × 2 = 1 + 0.280 000 110 182 4;
  • 13) 0.280 000 110 182 4 × 2 = 0 + 0.560 000 220 364 8;
  • 14) 0.560 000 220 364 8 × 2 = 1 + 0.120 000 440 729 6;
  • 15) 0.120 000 440 729 6 × 2 = 0 + 0.240 000 881 459 2;
  • 16) 0.240 000 881 459 2 × 2 = 0 + 0.480 001 762 918 4;
  • 17) 0.480 001 762 918 4 × 2 = 0 + 0.960 003 525 836 8;
  • 18) 0.960 003 525 836 8 × 2 = 1 + 0.920 007 051 673 6;
  • 19) 0.920 007 051 673 6 × 2 = 1 + 0.840 014 103 347 2;
  • 20) 0.840 014 103 347 2 × 2 = 1 + 0.680 028 206 694 4;
  • 21) 0.680 028 206 694 4 × 2 = 1 + 0.360 056 413 388 8;
  • 22) 0.360 056 413 388 8 × 2 = 0 + 0.720 112 826 777 6;
  • 23) 0.720 112 826 777 6 × 2 = 1 + 0.440 225 653 555 2;
  • 24) 0.440 225 653 555 2 × 2 = 0 + 0.880 451 307 110 4;
  • 25) 0.880 451 307 110 4 × 2 = 1 + 0.760 902 614 220 8;
  • 26) 0.760 902 614 220 8 × 2 = 1 + 0.521 805 228 441 6;
  • 27) 0.521 805 228 441 6 × 2 = 1 + 0.043 610 456 883 2;
  • 28) 0.043 610 456 883 2 × 2 = 0 + 0.087 220 913 766 4;
  • 29) 0.087 220 913 766 4 × 2 = 0 + 0.174 441 827 532 8;
  • 30) 0.174 441 827 532 8 × 2 = 0 + 0.348 883 655 065 6;
  • 31) 0.348 883 655 065 6 × 2 = 0 + 0.697 767 310 131 2;
  • 32) 0.697 767 310 131 2 × 2 = 1 + 0.395 534 620 262 4;
  • 33) 0.395 534 620 262 4 × 2 = 0 + 0.791 069 240 524 8;
  • 34) 0.791 069 240 524 8 × 2 = 1 + 0.582 138 481 049 6;
  • 35) 0.582 138 481 049 6 × 2 = 1 + 0.164 276 962 099 2;
  • 36) 0.164 276 962 099 2 × 2 = 0 + 0.328 553 924 198 4;
  • 37) 0.328 553 924 198 4 × 2 = 0 + 0.657 107 848 396 8;
  • 38) 0.657 107 848 396 8 × 2 = 1 + 0.314 215 696 793 6;
  • 39) 0.314 215 696 793 6 × 2 = 0 + 0.628 431 393 587 2;
  • 40) 0.628 431 393 587 2 × 2 = 1 + 0.256 862 787 174 4;
  • 41) 0.256 862 787 174 4 × 2 = 0 + 0.513 725 574 348 8;
  • 42) 0.513 725 574 348 8 × 2 = 1 + 0.027 451 148 697 6;
  • 43) 0.027 451 148 697 6 × 2 = 0 + 0.054 902 297 395 2;
  • 44) 0.054 902 297 395 2 × 2 = 0 + 0.109 804 594 790 4;
  • 45) 0.109 804 594 790 4 × 2 = 0 + 0.219 609 189 580 8;
  • 46) 0.219 609 189 580 8 × 2 = 0 + 0.439 218 379 161 6;
  • 47) 0.439 218 379 161 6 × 2 = 0 + 0.878 436 758 323 2;
  • 48) 0.878 436 758 323 2 × 2 = 1 + 0.756 873 516 646 4;
  • 49) 0.756 873 516 646 4 × 2 = 1 + 0.513 747 033 292 8;
  • 50) 0.513 747 033 292 8 × 2 = 1 + 0.027 494 066 585 6;
  • 51) 0.027 494 066 585 6 × 2 = 0 + 0.054 988 133 171 2;
  • 52) 0.054 988 133 171 2 × 2 = 0 + 0.109 976 266 342 4;
  • 53) 0.109 976 266 342 4 × 2 = 0 + 0.219 952 532 684 8;
  • 54) 0.219 952 532 684 8 × 2 = 0 + 0.439 905 065 369 6;
  • 55) 0.439 905 065 369 6 × 2 = 0 + 0.879 810 130 739 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:


0.180 000 000 026 9(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0101 0100 0001 1100 000(2)

6. Positive number before normalization:

0.180 000 000 026 9(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0101 0100 0001 1100 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:


0.180 000 000 026 9(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0101 0100 0001 1100 000(2) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 0101 0100 0001 1100 000(2) × 20 =

1.0111 0000 1010 0011 1101 0111 0000 1011 0010 1010 0000 1110 0000(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 0010 1010 0000 1110 0000


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 0010 1010 0000 1110 0000 =

0111 0000 1010 0011 1101 0111 0000 1011 0010 1010 0000 1110 0000


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 0010 1010 0000 1110 0000


Decimal number -0.180 000 000 026 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0111 0000 1010 0011 1101 0111 0000 1011 0010 1010 0000 1110 0000

» Convert decimal -0.180 000 000 032 9 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