-0.185 000 000 017 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.185 000 000 017 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
-0.185 000 000 017 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:

|-0.185 000 000 017 2| = 0.185 000 000 017 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: 0.185 000 000 017 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) 0.185 000 000 017 2 × 2 = 0 + 0.370 000 000 034 4;
  • 2) 0.370 000 000 034 4 × 2 = 0 + 0.740 000 000 068 8;
  • 3) 0.740 000 000 068 8 × 2 = 1 + 0.480 000 000 137 6;
  • 4) 0.480 000 000 137 6 × 2 = 0 + 0.960 000 000 275 2;
  • 5) 0.960 000 000 275 2 × 2 = 1 + 0.920 000 000 550 4;
  • 6) 0.920 000 000 550 4 × 2 = 1 + 0.840 000 001 100 8;
  • 7) 0.840 000 001 100 8 × 2 = 1 + 0.680 000 002 201 6;
  • 8) 0.680 000 002 201 6 × 2 = 1 + 0.360 000 004 403 2;
  • 9) 0.360 000 004 403 2 × 2 = 0 + 0.720 000 008 806 4;
  • 10) 0.720 000 008 806 4 × 2 = 1 + 0.440 000 017 612 8;
  • 11) 0.440 000 017 612 8 × 2 = 0 + 0.880 000 035 225 6;
  • 12) 0.880 000 035 225 6 × 2 = 1 + 0.760 000 070 451 2;
  • 13) 0.760 000 070 451 2 × 2 = 1 + 0.520 000 140 902 4;
  • 14) 0.520 000 140 902 4 × 2 = 1 + 0.040 000 281 804 8;
  • 15) 0.040 000 281 804 8 × 2 = 0 + 0.080 000 563 609 6;
  • 16) 0.080 000 563 609 6 × 2 = 0 + 0.160 001 127 219 2;
  • 17) 0.160 001 127 219 2 × 2 = 0 + 0.320 002 254 438 4;
  • 18) 0.320 002 254 438 4 × 2 = 0 + 0.640 004 508 876 8;
  • 19) 0.640 004 508 876 8 × 2 = 1 + 0.280 009 017 753 6;
  • 20) 0.280 009 017 753 6 × 2 = 0 + 0.560 018 035 507 2;
  • 21) 0.560 018 035 507 2 × 2 = 1 + 0.120 036 071 014 4;
  • 22) 0.120 036 071 014 4 × 2 = 0 + 0.240 072 142 028 8;
  • 23) 0.240 072 142 028 8 × 2 = 0 + 0.480 144 284 057 6;
  • 24) 0.480 144 284 057 6 × 2 = 0 + 0.960 288 568 115 2;
  • 25) 0.960 288 568 115 2 × 2 = 1 + 0.920 577 136 230 4;
  • 26) 0.920 577 136 230 4 × 2 = 1 + 0.841 154 272 460 8;
  • 27) 0.841 154 272 460 8 × 2 = 1 + 0.682 308 544 921 6;
  • 28) 0.682 308 544 921 6 × 2 = 1 + 0.364 617 089 843 2;
  • 29) 0.364 617 089 843 2 × 2 = 0 + 0.729 234 179 686 4;
  • 30) 0.729 234 179 686 4 × 2 = 1 + 0.458 468 359 372 8;
  • 31) 0.458 468 359 372 8 × 2 = 0 + 0.916 936 718 745 6;
  • 32) 0.916 936 718 745 6 × 2 = 1 + 0.833 873 437 491 2;
  • 33) 0.833 873 437 491 2 × 2 = 1 + 0.667 746 874 982 4;
  • 34) 0.667 746 874 982 4 × 2 = 1 + 0.335 493 749 964 8;
  • 35) 0.335 493 749 964 8 × 2 = 0 + 0.670 987 499 929 6;
  • 36) 0.670 987 499 929 6 × 2 = 1 + 0.341 974 999 859 2;
  • 37) 0.341 974 999 859 2 × 2 = 0 + 0.683 949 999 718 4;
  • 38) 0.683 949 999 718 4 × 2 = 1 + 0.367 899 999 436 8;
  • 39) 0.367 899 999 436 8 × 2 = 0 + 0.735 799 998 873 6;
  • 40) 0.735 799 998 873 6 × 2 = 1 + 0.471 599 997 747 2;
  • 41) 0.471 599 997 747 2 × 2 = 0 + 0.943 199 995 494 4;
  • 42) 0.943 199 995 494 4 × 2 = 1 + 0.886 399 990 988 8;
  • 43) 0.886 399 990 988 8 × 2 = 1 + 0.772 799 981 977 6;
  • 44) 0.772 799 981 977 6 × 2 = 1 + 0.545 599 963 955 2;
  • 45) 0.545 599 963 955 2 × 2 = 1 + 0.091 199 927 910 4;
  • 46) 0.091 199 927 910 4 × 2 = 0 + 0.182 399 855 820 8;
  • 47) 0.182 399 855 820 8 × 2 = 0 + 0.364 799 711 641 6;
  • 48) 0.364 799 711 641 6 × 2 = 0 + 0.729 599 423 283 2;
  • 49) 0.729 599 423 283 2 × 2 = 1 + 0.459 198 846 566 4;
  • 50) 0.459 198 846 566 4 × 2 = 0 + 0.918 397 693 132 8;
  • 51) 0.918 397 693 132 8 × 2 = 1 + 0.836 795 386 265 6;
  • 52) 0.836 795 386 265 6 × 2 = 1 + 0.673 590 772 531 2;
  • 53) 0.673 590 772 531 2 × 2 = 1 + 0.347 181 545 062 4;
  • 54) 0.347 181 545 062 4 × 2 = 0 + 0.694 363 090 124 8;
  • 55) 0.694 363 090 124 8 × 2 = 1 + 0.388 726 180 249 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:


0.185 000 000 017 2(10) =

0.0010 1111 0101 1100 0010 1000 1111 0101 1101 0101 0111 1000 1011 101(2)

6. Positive number before normalization:

0.185 000 000 017 2(10) =

0.0010 1111 0101 1100 0010 1000 1111 0101 1101 0101 0111 1000 1011 101(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:


0.185 000 000 017 2(10) =

0.0010 1111 0101 1100 0010 1000 1111 0101 1101 0101 0111 1000 1011 101(2) =

0.0010 1111 0101 1100 0010 1000 1111 0101 1101 0101 0111 1000 1011 101(2) × 20 =

1.0111 1010 1110 0001 0100 0111 1010 1110 1010 1011 1100 0101 1101(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 1010 1110 0001 0100 0111 1010 1110 1010 1011 1100 0101 1101


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 1010 1110 0001 0100 0111 1010 1110 1010 1011 1100 0101 1101 =

0111 1010 1110 0001 0100 0111 1010 1110 1010 1011 1100 0101 1101


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 1010 1110 0001 0100 0111 1010 1110 1010 1011 1100 0101 1101


Decimal number -0.185 000 000 017 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

» Convert decimal -0.185 000 000 020 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