-0.042 419 142 303 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 303 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
-0.042 419 142 303 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:

|-0.042 419 142 303 2| = 0.042 419 142 303 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: 0.042 419 142 303 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) 0.042 419 142 303 2 × 2 = 0 + 0.084 838 284 606 4;
  • 2) 0.084 838 284 606 4 × 2 = 0 + 0.169 676 569 212 8;
  • 3) 0.169 676 569 212 8 × 2 = 0 + 0.339 353 138 425 6;
  • 4) 0.339 353 138 425 6 × 2 = 0 + 0.678 706 276 851 2;
  • 5) 0.678 706 276 851 2 × 2 = 1 + 0.357 412 553 702 4;
  • 6) 0.357 412 553 702 4 × 2 = 0 + 0.714 825 107 404 8;
  • 7) 0.714 825 107 404 8 × 2 = 1 + 0.429 650 214 809 6;
  • 8) 0.429 650 214 809 6 × 2 = 0 + 0.859 300 429 619 2;
  • 9) 0.859 300 429 619 2 × 2 = 1 + 0.718 600 859 238 4;
  • 10) 0.718 600 859 238 4 × 2 = 1 + 0.437 201 718 476 8;
  • 11) 0.437 201 718 476 8 × 2 = 0 + 0.874 403 436 953 6;
  • 12) 0.874 403 436 953 6 × 2 = 1 + 0.748 806 873 907 2;
  • 13) 0.748 806 873 907 2 × 2 = 1 + 0.497 613 747 814 4;
  • 14) 0.497 613 747 814 4 × 2 = 0 + 0.995 227 495 628 8;
  • 15) 0.995 227 495 628 8 × 2 = 1 + 0.990 454 991 257 6;
  • 16) 0.990 454 991 257 6 × 2 = 1 + 0.980 909 982 515 2;
  • 17) 0.980 909 982 515 2 × 2 = 1 + 0.961 819 965 030 4;
  • 18) 0.961 819 965 030 4 × 2 = 1 + 0.923 639 930 060 8;
  • 19) 0.923 639 930 060 8 × 2 = 1 + 0.847 279 860 121 6;
  • 20) 0.847 279 860 121 6 × 2 = 1 + 0.694 559 720 243 2;
  • 21) 0.694 559 720 243 2 × 2 = 1 + 0.389 119 440 486 4;
  • 22) 0.389 119 440 486 4 × 2 = 0 + 0.778 238 880 972 8;
  • 23) 0.778 238 880 972 8 × 2 = 1 + 0.556 477 761 945 6;
  • 24) 0.556 477 761 945 6 × 2 = 1 + 0.112 955 523 891 2;
  • 25) 0.112 955 523 891 2 × 2 = 0 + 0.225 911 047 782 4;
  • 26) 0.225 911 047 782 4 × 2 = 0 + 0.451 822 095 564 8;
  • 27) 0.451 822 095 564 8 × 2 = 0 + 0.903 644 191 129 6;
  • 28) 0.903 644 191 129 6 × 2 = 1 + 0.807 288 382 259 2;
  • 29) 0.807 288 382 259 2 × 2 = 1 + 0.614 576 764 518 4;
  • 30) 0.614 576 764 518 4 × 2 = 1 + 0.229 153 529 036 8;
  • 31) 0.229 153 529 036 8 × 2 = 0 + 0.458 307 058 073 6;
  • 32) 0.458 307 058 073 6 × 2 = 0 + 0.916 614 116 147 2;
  • 33) 0.916 614 116 147 2 × 2 = 1 + 0.833 228 232 294 4;
  • 34) 0.833 228 232 294 4 × 2 = 1 + 0.666 456 464 588 8;
  • 35) 0.666 456 464 588 8 × 2 = 1 + 0.332 912 929 177 6;
  • 36) 0.332 912 929 177 6 × 2 = 0 + 0.665 825 858 355 2;
  • 37) 0.665 825 858 355 2 × 2 = 1 + 0.331 651 716 710 4;
  • 38) 0.331 651 716 710 4 × 2 = 0 + 0.663 303 433 420 8;
  • 39) 0.663 303 433 420 8 × 2 = 1 + 0.326 606 866 841 6;
  • 40) 0.326 606 866 841 6 × 2 = 0 + 0.653 213 733 683 2;
  • 41) 0.653 213 733 683 2 × 2 = 1 + 0.306 427 467 366 4;
  • 42) 0.306 427 467 366 4 × 2 = 0 + 0.612 854 934 732 8;
  • 43) 0.612 854 934 732 8 × 2 = 1 + 0.225 709 869 465 6;
  • 44) 0.225 709 869 465 6 × 2 = 0 + 0.451 419 738 931 2;
  • 45) 0.451 419 738 931 2 × 2 = 0 + 0.902 839 477 862 4;
  • 46) 0.902 839 477 862 4 × 2 = 1 + 0.805 678 955 724 8;
  • 47) 0.805 678 955 724 8 × 2 = 1 + 0.611 357 911 449 6;
  • 48) 0.611 357 911 449 6 × 2 = 1 + 0.222 715 822 899 2;
  • 49) 0.222 715 822 899 2 × 2 = 0 + 0.445 431 645 798 4;
  • 50) 0.445 431 645 798 4 × 2 = 0 + 0.890 863 291 596 8;
  • 51) 0.890 863 291 596 8 × 2 = 1 + 0.781 726 583 193 6;
  • 52) 0.781 726 583 193 6 × 2 = 1 + 0.563 453 166 387 2;
  • 53) 0.563 453 166 387 2 × 2 = 1 + 0.126 906 332 774 4;
  • 54) 0.126 906 332 774 4 × 2 = 0 + 0.253 812 665 548 8;
  • 55) 0.253 812 665 548 8 × 2 = 0 + 0.507 625 331 097 6;
  • 56) 0.507 625 331 097 6 × 2 = 1 + 0.015 250 662 195 2;
  • 57) 0.015 250 662 195 2 × 2 = 0 + 0.030 501 324 390 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:


0.042 419 142 303 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 1010 1010 0111 0011 1001 0(2)

6. Positive number before normalization:

0.042 419 142 303 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 1010 1010 0111 0011 1001 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.042 419 142 303 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 1010 1010 0111 0011 1001 0(2) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 1010 1010 0111 0011 1001 0(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1001 1101 0101 0100 1110 0111 0010(2) × 2-5


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): -5

Mantissa (not normalized):
1.0101 1011 0111 1111 0110 0011 1001 1101 0101 0100 1110 0111 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 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) =

1018(10) =

011 1111 1010(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. 0101 1011 0111 1111 0110 0011 1001 1101 0101 0100 1110 0111 0010 =

0101 1011 0111 1111 0110 0011 1001 1101 0101 0100 1110 0111 0010


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 1010

Mantissa (52 bits) =
0101 1011 0111 1111 0110 0011 1001 1101 0101 0100 1110 0111 0010


Decimal number -0.042 419 142 303 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0101 1011 0111 1111 0110 0011 1001 1101 0101 0100 1110 0111 0010

» Convert decimal -0.042 419 142 298 2 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