-0.019 834 293 405 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.019 834 293 405 5(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
-0.019 834 293 405 5(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:

|-0.019 834 293 405 5| = 0.019 834 293 405 5


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: 0.019 834 293 405 5.

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) 0.019 834 293 405 5 × 2 = 0 + 0.039 668 586 811;
  • 2) 0.039 668 586 811 × 2 = 0 + 0.079 337 173 622;
  • 3) 0.079 337 173 622 × 2 = 0 + 0.158 674 347 244;
  • 4) 0.158 674 347 244 × 2 = 0 + 0.317 348 694 488;
  • 5) 0.317 348 694 488 × 2 = 0 + 0.634 697 388 976;
  • 6) 0.634 697 388 976 × 2 = 1 + 0.269 394 777 952;
  • 7) 0.269 394 777 952 × 2 = 0 + 0.538 789 555 904;
  • 8) 0.538 789 555 904 × 2 = 1 + 0.077 579 111 808;
  • 9) 0.077 579 111 808 × 2 = 0 + 0.155 158 223 616;
  • 10) 0.155 158 223 616 × 2 = 0 + 0.310 316 447 232;
  • 11) 0.310 316 447 232 × 2 = 0 + 0.620 632 894 464;
  • 12) 0.620 632 894 464 × 2 = 1 + 0.241 265 788 928;
  • 13) 0.241 265 788 928 × 2 = 0 + 0.482 531 577 856;
  • 14) 0.482 531 577 856 × 2 = 0 + 0.965 063 155 712;
  • 15) 0.965 063 155 712 × 2 = 1 + 0.930 126 311 424;
  • 16) 0.930 126 311 424 × 2 = 1 + 0.860 252 622 848;
  • 17) 0.860 252 622 848 × 2 = 1 + 0.720 505 245 696;
  • 18) 0.720 505 245 696 × 2 = 1 + 0.441 010 491 392;
  • 19) 0.441 010 491 392 × 2 = 0 + 0.882 020 982 784;
  • 20) 0.882 020 982 784 × 2 = 1 + 0.764 041 965 568;
  • 21) 0.764 041 965 568 × 2 = 1 + 0.528 083 931 136;
  • 22) 0.528 083 931 136 × 2 = 1 + 0.056 167 862 272;
  • 23) 0.056 167 862 272 × 2 = 0 + 0.112 335 724 544;
  • 24) 0.112 335 724 544 × 2 = 0 + 0.224 671 449 088;
  • 25) 0.224 671 449 088 × 2 = 0 + 0.449 342 898 176;
  • 26) 0.449 342 898 176 × 2 = 0 + 0.898 685 796 352;
  • 27) 0.898 685 796 352 × 2 = 1 + 0.797 371 592 704;
  • 28) 0.797 371 592 704 × 2 = 1 + 0.594 743 185 408;
  • 29) 0.594 743 185 408 × 2 = 1 + 0.189 486 370 816;
  • 30) 0.189 486 370 816 × 2 = 0 + 0.378 972 741 632;
  • 31) 0.378 972 741 632 × 2 = 0 + 0.757 945 483 264;
  • 32) 0.757 945 483 264 × 2 = 1 + 0.515 890 966 528;
  • 33) 0.515 890 966 528 × 2 = 1 + 0.031 781 933 056;
  • 34) 0.031 781 933 056 × 2 = 0 + 0.063 563 866 112;
  • 35) 0.063 563 866 112 × 2 = 0 + 0.127 127 732 224;
  • 36) 0.127 127 732 224 × 2 = 0 + 0.254 255 464 448;
  • 37) 0.254 255 464 448 × 2 = 0 + 0.508 510 928 896;
  • 38) 0.508 510 928 896 × 2 = 1 + 0.017 021 857 792;
  • 39) 0.017 021 857 792 × 2 = 0 + 0.034 043 715 584;
  • 40) 0.034 043 715 584 × 2 = 0 + 0.068 087 431 168;
  • 41) 0.068 087 431 168 × 2 = 0 + 0.136 174 862 336;
  • 42) 0.136 174 862 336 × 2 = 0 + 0.272 349 724 672;
  • 43) 0.272 349 724 672 × 2 = 0 + 0.544 699 449 344;
  • 44) 0.544 699 449 344 × 2 = 1 + 0.089 398 898 688;
  • 45) 0.089 398 898 688 × 2 = 0 + 0.178 797 797 376;
  • 46) 0.178 797 797 376 × 2 = 0 + 0.357 595 594 752;
  • 47) 0.357 595 594 752 × 2 = 0 + 0.715 191 189 504;
  • 48) 0.715 191 189 504 × 2 = 1 + 0.430 382 379 008;
  • 49) 0.430 382 379 008 × 2 = 0 + 0.860 764 758 016;
  • 50) 0.860 764 758 016 × 2 = 1 + 0.721 529 516 032;
  • 51) 0.721 529 516 032 × 2 = 1 + 0.443 059 032 064;
  • 52) 0.443 059 032 064 × 2 = 0 + 0.886 118 064 128;
  • 53) 0.886 118 064 128 × 2 = 1 + 0.772 236 128 256;
  • 54) 0.772 236 128 256 × 2 = 1 + 0.544 472 256 512;
  • 55) 0.544 472 256 512 × 2 = 1 + 0.088 944 513 024;
  • 56) 0.088 944 513 024 × 2 = 0 + 0.177 889 026 048;
  • 57) 0.177 889 026 048 × 2 = 0 + 0.355 778 052 096;
  • 58) 0.355 778 052 096 × 2 = 0 + 0.711 556 104 192;

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:


0.019 834 293 405 5(10) =

0.0000 0101 0001 0011 1101 1100 0011 1001 1000 0100 0001 0001 0110 1110 00(2)

6. Positive number before normalization:

0.019 834 293 405 5(10) =

0.0000 0101 0001 0011 1101 1100 0011 1001 1000 0100 0001 0001 0110 1110 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.019 834 293 405 5(10) =

0.0000 0101 0001 0011 1101 1100 0011 1001 1000 0100 0001 0001 0110 1110 00(2) =

0.0000 0101 0001 0011 1101 1100 0011 1001 1000 0100 0001 0001 0110 1110 00(2) × 20 =

1.0100 0100 1111 0111 0000 1110 0110 0001 0000 0100 0101 1011 1000(2) × 2-6


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): -6

Mantissa (not normalized):
1.0100 0100 1111 0111 0000 1110 0110 0001 0000 0100 0101 1011 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 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) =

1017(10) =

011 1111 1001(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 0100 1111 0111 0000 1110 0110 0001 0000 0100 0101 1011 1000 =

0100 0100 1111 0111 0000 1110 0110 0001 0000 0100 0101 1011 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 1001

Mantissa (52 bits) =
0100 0100 1111 0111 0000 1110 0110 0001 0000 0100 0101 1011 1000


Decimal number -0.019 834 293 405 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0100 0100 1111 0111 0000 1110 0110 0001 0000 0100 0101 1011 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