-0.080 000 013 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.080 000 013 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
-0.080 000 013 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:

|-0.080 000 013 9| = 0.080 000 013 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: 0.080 000 013 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) 0.080 000 013 9 × 2 = 0 + 0.160 000 027 8;
  • 2) 0.160 000 027 8 × 2 = 0 + 0.320 000 055 6;
  • 3) 0.320 000 055 6 × 2 = 0 + 0.640 000 111 2;
  • 4) 0.640 000 111 2 × 2 = 1 + 0.280 000 222 4;
  • 5) 0.280 000 222 4 × 2 = 0 + 0.560 000 444 8;
  • 6) 0.560 000 444 8 × 2 = 1 + 0.120 000 889 6;
  • 7) 0.120 000 889 6 × 2 = 0 + 0.240 001 779 2;
  • 8) 0.240 001 779 2 × 2 = 0 + 0.480 003 558 4;
  • 9) 0.480 003 558 4 × 2 = 0 + 0.960 007 116 8;
  • 10) 0.960 007 116 8 × 2 = 1 + 0.920 014 233 6;
  • 11) 0.920 014 233 6 × 2 = 1 + 0.840 028 467 2;
  • 12) 0.840 028 467 2 × 2 = 1 + 0.680 056 934 4;
  • 13) 0.680 056 934 4 × 2 = 1 + 0.360 113 868 8;
  • 14) 0.360 113 868 8 × 2 = 0 + 0.720 227 737 6;
  • 15) 0.720 227 737 6 × 2 = 1 + 0.440 455 475 2;
  • 16) 0.440 455 475 2 × 2 = 0 + 0.880 910 950 4;
  • 17) 0.880 910 950 4 × 2 = 1 + 0.761 821 900 8;
  • 18) 0.761 821 900 8 × 2 = 1 + 0.523 643 801 6;
  • 19) 0.523 643 801 6 × 2 = 1 + 0.047 287 603 2;
  • 20) 0.047 287 603 2 × 2 = 0 + 0.094 575 206 4;
  • 21) 0.094 575 206 4 × 2 = 0 + 0.189 150 412 8;
  • 22) 0.189 150 412 8 × 2 = 0 + 0.378 300 825 6;
  • 23) 0.378 300 825 6 × 2 = 0 + 0.756 601 651 2;
  • 24) 0.756 601 651 2 × 2 = 1 + 0.513 203 302 4;
  • 25) 0.513 203 302 4 × 2 = 1 + 0.026 406 604 8;
  • 26) 0.026 406 604 8 × 2 = 0 + 0.052 813 209 6;
  • 27) 0.052 813 209 6 × 2 = 0 + 0.105 626 419 2;
  • 28) 0.105 626 419 2 × 2 = 0 + 0.211 252 838 4;
  • 29) 0.211 252 838 4 × 2 = 0 + 0.422 505 676 8;
  • 30) 0.422 505 676 8 × 2 = 0 + 0.845 011 353 6;
  • 31) 0.845 011 353 6 × 2 = 1 + 0.690 022 707 2;
  • 32) 0.690 022 707 2 × 2 = 1 + 0.380 045 414 4;
  • 33) 0.380 045 414 4 × 2 = 0 + 0.760 090 828 8;
  • 34) 0.760 090 828 8 × 2 = 1 + 0.520 181 657 6;
  • 35) 0.520 181 657 6 × 2 = 1 + 0.040 363 315 2;
  • 36) 0.040 363 315 2 × 2 = 0 + 0.080 726 630 4;
  • 37) 0.080 726 630 4 × 2 = 0 + 0.161 453 260 8;
  • 38) 0.161 453 260 8 × 2 = 0 + 0.322 906 521 6;
  • 39) 0.322 906 521 6 × 2 = 0 + 0.645 813 043 2;
  • 40) 0.645 813 043 2 × 2 = 1 + 0.291 626 086 4;
  • 41) 0.291 626 086 4 × 2 = 0 + 0.583 252 172 8;
  • 42) 0.583 252 172 8 × 2 = 1 + 0.166 504 345 6;
  • 43) 0.166 504 345 6 × 2 = 0 + 0.333 008 691 2;
  • 44) 0.333 008 691 2 × 2 = 0 + 0.666 017 382 4;
  • 45) 0.666 017 382 4 × 2 = 1 + 0.332 034 764 8;
  • 46) 0.332 034 764 8 × 2 = 0 + 0.664 069 529 6;
  • 47) 0.664 069 529 6 × 2 = 1 + 0.328 139 059 2;
  • 48) 0.328 139 059 2 × 2 = 0 + 0.656 278 118 4;
  • 49) 0.656 278 118 4 × 2 = 1 + 0.312 556 236 8;
  • 50) 0.312 556 236 8 × 2 = 0 + 0.625 112 473 6;
  • 51) 0.625 112 473 6 × 2 = 1 + 0.250 224 947 2;
  • 52) 0.250 224 947 2 × 2 = 0 + 0.500 449 894 4;
  • 53) 0.500 449 894 4 × 2 = 1 + 0.000 899 788 8;
  • 54) 0.000 899 788 8 × 2 = 0 + 0.001 799 577 6;
  • 55) 0.001 799 577 6 × 2 = 0 + 0.003 599 155 2;
  • 56) 0.003 599 155 2 × 2 = 0 + 0.007 198 310 4;

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:


0.080 000 013 9(10) =

0.0001 0100 0111 1010 1110 0001 1000 0011 0110 0001 0100 1010 1010 1000(2)

6. Positive number before normalization:

0.080 000 013 9(10) =

0.0001 0100 0111 1010 1110 0001 1000 0011 0110 0001 0100 1010 1010 1000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.080 000 013 9(10) =

0.0001 0100 0111 1010 1110 0001 1000 0011 0110 0001 0100 1010 1010 1000(2) =

0.0001 0100 0111 1010 1110 0001 1000 0011 0110 0001 0100 1010 1010 1000(2) × 20 =

1.0100 0111 1010 1110 0001 1000 0011 0110 0001 0100 1010 1010 1000(2) × 2-4


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):
1.0100 0111 1010 1110 0001 1000 0011 0110 0001 0100 1010 1010 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1019(10) =

011 1111 1011(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. 0100 0111 1010 1110 0001 1000 0011 0110 0001 0100 1010 1010 1000 =

0100 0111 1010 1110 0001 1000 0011 0110 0001 0100 1010 1010 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 1011

Mantissa (52 bits) =
0100 0111 1010 1110 0001 1000 0011 0110 0001 0100 1010 1010 1000


Decimal number -0.080 000 013 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 0100 0111 1010 1110 0001 1000 0011 0110 0001 0100 1010 1010 1000

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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