-0.000 084 980 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 084 980 6(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
-0.000 084 980 6(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:

|-0.000 084 980 6| = 0.000 084 980 6


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: 0.000 084 980 6.

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) 0.000 084 980 6 × 2 = 0 + 0.000 169 961 2;
  • 2) 0.000 169 961 2 × 2 = 0 + 0.000 339 922 4;
  • 3) 0.000 339 922 4 × 2 = 0 + 0.000 679 844 8;
  • 4) 0.000 679 844 8 × 2 = 0 + 0.001 359 689 6;
  • 5) 0.001 359 689 6 × 2 = 0 + 0.002 719 379 2;
  • 6) 0.002 719 379 2 × 2 = 0 + 0.005 438 758 4;
  • 7) 0.005 438 758 4 × 2 = 0 + 0.010 877 516 8;
  • 8) 0.010 877 516 8 × 2 = 0 + 0.021 755 033 6;
  • 9) 0.021 755 033 6 × 2 = 0 + 0.043 510 067 2;
  • 10) 0.043 510 067 2 × 2 = 0 + 0.087 020 134 4;
  • 11) 0.087 020 134 4 × 2 = 0 + 0.174 040 268 8;
  • 12) 0.174 040 268 8 × 2 = 0 + 0.348 080 537 6;
  • 13) 0.348 080 537 6 × 2 = 0 + 0.696 161 075 2;
  • 14) 0.696 161 075 2 × 2 = 1 + 0.392 322 150 4;
  • 15) 0.392 322 150 4 × 2 = 0 + 0.784 644 300 8;
  • 16) 0.784 644 300 8 × 2 = 1 + 0.569 288 601 6;
  • 17) 0.569 288 601 6 × 2 = 1 + 0.138 577 203 2;
  • 18) 0.138 577 203 2 × 2 = 0 + 0.277 154 406 4;
  • 19) 0.277 154 406 4 × 2 = 0 + 0.554 308 812 8;
  • 20) 0.554 308 812 8 × 2 = 1 + 0.108 617 625 6;
  • 21) 0.108 617 625 6 × 2 = 0 + 0.217 235 251 2;
  • 22) 0.217 235 251 2 × 2 = 0 + 0.434 470 502 4;
  • 23) 0.434 470 502 4 × 2 = 0 + 0.868 941 004 8;
  • 24) 0.868 941 004 8 × 2 = 1 + 0.737 882 009 6;
  • 25) 0.737 882 009 6 × 2 = 1 + 0.475 764 019 2;
  • 26) 0.475 764 019 2 × 2 = 0 + 0.951 528 038 4;
  • 27) 0.951 528 038 4 × 2 = 1 + 0.903 056 076 8;
  • 28) 0.903 056 076 8 × 2 = 1 + 0.806 112 153 6;
  • 29) 0.806 112 153 6 × 2 = 1 + 0.612 224 307 2;
  • 30) 0.612 224 307 2 × 2 = 1 + 0.224 448 614 4;
  • 31) 0.224 448 614 4 × 2 = 0 + 0.448 897 228 8;
  • 32) 0.448 897 228 8 × 2 = 0 + 0.897 794 457 6;
  • 33) 0.897 794 457 6 × 2 = 1 + 0.795 588 915 2;
  • 34) 0.795 588 915 2 × 2 = 1 + 0.591 177 830 4;
  • 35) 0.591 177 830 4 × 2 = 1 + 0.182 355 660 8;
  • 36) 0.182 355 660 8 × 2 = 0 + 0.364 711 321 6;
  • 37) 0.364 711 321 6 × 2 = 0 + 0.729 422 643 2;
  • 38) 0.729 422 643 2 × 2 = 1 + 0.458 845 286 4;
  • 39) 0.458 845 286 4 × 2 = 0 + 0.917 690 572 8;
  • 40) 0.917 690 572 8 × 2 = 1 + 0.835 381 145 6;
  • 41) 0.835 381 145 6 × 2 = 1 + 0.670 762 291 2;
  • 42) 0.670 762 291 2 × 2 = 1 + 0.341 524 582 4;
  • 43) 0.341 524 582 4 × 2 = 0 + 0.683 049 164 8;
  • 44) 0.683 049 164 8 × 2 = 1 + 0.366 098 329 6;
  • 45) 0.366 098 329 6 × 2 = 0 + 0.732 196 659 2;
  • 46) 0.732 196 659 2 × 2 = 1 + 0.464 393 318 4;
  • 47) 0.464 393 318 4 × 2 = 0 + 0.928 786 636 8;
  • 48) 0.928 786 636 8 × 2 = 1 + 0.857 573 273 6;
  • 49) 0.857 573 273 6 × 2 = 1 + 0.715 146 547 2;
  • 50) 0.715 146 547 2 × 2 = 1 + 0.430 293 094 4;
  • 51) 0.430 293 094 4 × 2 = 0 + 0.860 586 188 8;
  • 52) 0.860 586 188 8 × 2 = 1 + 0.721 172 377 6;
  • 53) 0.721 172 377 6 × 2 = 1 + 0.442 344 755 2;
  • 54) 0.442 344 755 2 × 2 = 0 + 0.884 689 510 4;
  • 55) 0.884 689 510 4 × 2 = 1 + 0.769 379 020 8;
  • 56) 0.769 379 020 8 × 2 = 1 + 0.538 758 041 6;
  • 57) 0.538 758 041 6 × 2 = 1 + 0.077 516 083 2;
  • 58) 0.077 516 083 2 × 2 = 0 + 0.155 032 166 4;
  • 59) 0.155 032 166 4 × 2 = 0 + 0.310 064 332 8;
  • 60) 0.310 064 332 8 × 2 = 0 + 0.620 128 665 6;
  • 61) 0.620 128 665 6 × 2 = 1 + 0.240 257 331 2;
  • 62) 0.240 257 331 2 × 2 = 0 + 0.480 514 662 4;
  • 63) 0.480 514 662 4 × 2 = 0 + 0.961 029 324 8;
  • 64) 0.961 029 324 8 × 2 = 1 + 0.922 058 649 6;
  • 65) 0.922 058 649 6 × 2 = 1 + 0.844 117 299 2;
  • 66) 0.844 117 299 2 × 2 = 1 + 0.688 234 598 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:


0.000 084 980 6(10) =

0.0000 0000 0000 0101 1001 0001 1011 1100 1110 0101 1101 0101 1101 1011 1000 1001 11(2)

6. Positive number before normalization:

0.000 084 980 6(10) =

0.0000 0000 0000 0101 1001 0001 1011 1100 1110 0101 1101 0101 1101 1011 1000 1001 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 084 980 6(10) =

0.0000 0000 0000 0101 1001 0001 1011 1100 1110 0101 1101 0101 1101 1011 1000 1001 11(2) =

0.0000 0000 0000 0101 1001 0001 1011 1100 1110 0101 1101 0101 1101 1011 1000 1001 11(2) × 20 =

1.0110 0100 0110 1111 0011 1001 0111 0101 0111 0110 1110 0010 0111(2) × 2-14


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): -14

Mantissa (not normalized):
1.0110 0100 0110 1111 0011 1001 0111 0101 0111 0110 1110 0010 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-14 + 2(11-1) - 1 =

(-14 + 1 023)(10) =

1 009(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;
  • 1 009 ÷ 2 = 504 + 1;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1009(10) =

011 1111 0001(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. 0110 0100 0110 1111 0011 1001 0111 0101 0111 0110 1110 0010 0111 =

0110 0100 0110 1111 0011 1001 0111 0101 0111 0110 1110 0010 0111


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 0001

Mantissa (52 bits) =
0110 0100 0110 1111 0011 1001 0111 0101 0111 0110 1110 0010 0111


Decimal number -0.000 084 980 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0001 - 0110 0100 0110 1111 0011 1001 0111 0101 0111 0110 1110 0010 0111

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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