-0.180 000 000 037 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.180 000 000 037 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
-0.180 000 000 037 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:

|-0.180 000 000 037 2| = 0.180 000 000 037 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: 0.180 000 000 037 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) 0.180 000 000 037 2 × 2 = 0 + 0.360 000 000 074 4;
  • 2) 0.360 000 000 074 4 × 2 = 0 + 0.720 000 000 148 8;
  • 3) 0.720 000 000 148 8 × 2 = 1 + 0.440 000 000 297 6;
  • 4) 0.440 000 000 297 6 × 2 = 0 + 0.880 000 000 595 2;
  • 5) 0.880 000 000 595 2 × 2 = 1 + 0.760 000 001 190 4;
  • 6) 0.760 000 001 190 4 × 2 = 1 + 0.520 000 002 380 8;
  • 7) 0.520 000 002 380 8 × 2 = 1 + 0.040 000 004 761 6;
  • 8) 0.040 000 004 761 6 × 2 = 0 + 0.080 000 009 523 2;
  • 9) 0.080 000 009 523 2 × 2 = 0 + 0.160 000 019 046 4;
  • 10) 0.160 000 019 046 4 × 2 = 0 + 0.320 000 038 092 8;
  • 11) 0.320 000 038 092 8 × 2 = 0 + 0.640 000 076 185 6;
  • 12) 0.640 000 076 185 6 × 2 = 1 + 0.280 000 152 371 2;
  • 13) 0.280 000 152 371 2 × 2 = 0 + 0.560 000 304 742 4;
  • 14) 0.560 000 304 742 4 × 2 = 1 + 0.120 000 609 484 8;
  • 15) 0.120 000 609 484 8 × 2 = 0 + 0.240 001 218 969 6;
  • 16) 0.240 001 218 969 6 × 2 = 0 + 0.480 002 437 939 2;
  • 17) 0.480 002 437 939 2 × 2 = 0 + 0.960 004 875 878 4;
  • 18) 0.960 004 875 878 4 × 2 = 1 + 0.920 009 751 756 8;
  • 19) 0.920 009 751 756 8 × 2 = 1 + 0.840 019 503 513 6;
  • 20) 0.840 019 503 513 6 × 2 = 1 + 0.680 039 007 027 2;
  • 21) 0.680 039 007 027 2 × 2 = 1 + 0.360 078 014 054 4;
  • 22) 0.360 078 014 054 4 × 2 = 0 + 0.720 156 028 108 8;
  • 23) 0.720 156 028 108 8 × 2 = 1 + 0.440 312 056 217 6;
  • 24) 0.440 312 056 217 6 × 2 = 0 + 0.880 624 112 435 2;
  • 25) 0.880 624 112 435 2 × 2 = 1 + 0.761 248 224 870 4;
  • 26) 0.761 248 224 870 4 × 2 = 1 + 0.522 496 449 740 8;
  • 27) 0.522 496 449 740 8 × 2 = 1 + 0.044 992 899 481 6;
  • 28) 0.044 992 899 481 6 × 2 = 0 + 0.089 985 798 963 2;
  • 29) 0.089 985 798 963 2 × 2 = 0 + 0.179 971 597 926 4;
  • 30) 0.179 971 597 926 4 × 2 = 0 + 0.359 943 195 852 8;
  • 31) 0.359 943 195 852 8 × 2 = 0 + 0.719 886 391 705 6;
  • 32) 0.719 886 391 705 6 × 2 = 1 + 0.439 772 783 411 2;
  • 33) 0.439 772 783 411 2 × 2 = 0 + 0.879 545 566 822 4;
  • 34) 0.879 545 566 822 4 × 2 = 1 + 0.759 091 133 644 8;
  • 35) 0.759 091 133 644 8 × 2 = 1 + 0.518 182 267 289 6;
  • 36) 0.518 182 267 289 6 × 2 = 1 + 0.036 364 534 579 2;
  • 37) 0.036 364 534 579 2 × 2 = 0 + 0.072 729 069 158 4;
  • 38) 0.072 729 069 158 4 × 2 = 0 + 0.145 458 138 316 8;
  • 39) 0.145 458 138 316 8 × 2 = 0 + 0.290 916 276 633 6;
  • 40) 0.290 916 276 633 6 × 2 = 0 + 0.581 832 553 267 2;
  • 41) 0.581 832 553 267 2 × 2 = 1 + 0.163 665 106 534 4;
  • 42) 0.163 665 106 534 4 × 2 = 0 + 0.327 330 213 068 8;
  • 43) 0.327 330 213 068 8 × 2 = 0 + 0.654 660 426 137 6;
  • 44) 0.654 660 426 137 6 × 2 = 1 + 0.309 320 852 275 2;
  • 45) 0.309 320 852 275 2 × 2 = 0 + 0.618 641 704 550 4;
  • 46) 0.618 641 704 550 4 × 2 = 1 + 0.237 283 409 100 8;
  • 47) 0.237 283 409 100 8 × 2 = 0 + 0.474 566 818 201 6;
  • 48) 0.474 566 818 201 6 × 2 = 0 + 0.949 133 636 403 2;
  • 49) 0.949 133 636 403 2 × 2 = 1 + 0.898 267 272 806 4;
  • 50) 0.898 267 272 806 4 × 2 = 1 + 0.796 534 545 612 8;
  • 51) 0.796 534 545 612 8 × 2 = 1 + 0.593 069 091 225 6;
  • 52) 0.593 069 091 225 6 × 2 = 1 + 0.186 138 182 451 2;
  • 53) 0.186 138 182 451 2 × 2 = 0 + 0.372 276 364 902 4;
  • 54) 0.372 276 364 902 4 × 2 = 0 + 0.744 552 729 804 8;
  • 55) 0.744 552 729 804 8 × 2 = 1 + 0.489 105 459 609 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:


0.180 000 000 037 2(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0000 1001 0100 1111 001(2)

6. Positive number before normalization:

0.180 000 000 037 2(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0000 1001 0100 1111 001(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:


0.180 000 000 037 2(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0000 1001 0100 1111 001(2) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0111 0000 1001 0100 1111 001(2) × 20 =

1.0111 0000 1010 0011 1101 0111 0000 1011 1000 0100 1010 0111 1001(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 0100 1010 0111 1001


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 0100 1010 0111 1001 =

0111 0000 1010 0011 1101 0111 0000 1011 1000 0100 1010 0111 1001


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 0100 1010 0111 1001


Decimal number -0.180 000 000 037 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 0100 1010 0111 1001

» Convert decimal -0.180 000 000 038 4 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