-0.042 419 142 311 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 311 3(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
-0.042 419 142 311 3(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:

|-0.042 419 142 311 3| = 0.042 419 142 311 3


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: 0.042 419 142 311 3.

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) 0.042 419 142 311 3 × 2 = 0 + 0.084 838 284 622 6;
  • 2) 0.084 838 284 622 6 × 2 = 0 + 0.169 676 569 245 2;
  • 3) 0.169 676 569 245 2 × 2 = 0 + 0.339 353 138 490 4;
  • 4) 0.339 353 138 490 4 × 2 = 0 + 0.678 706 276 980 8;
  • 5) 0.678 706 276 980 8 × 2 = 1 + 0.357 412 553 961 6;
  • 6) 0.357 412 553 961 6 × 2 = 0 + 0.714 825 107 923 2;
  • 7) 0.714 825 107 923 2 × 2 = 1 + 0.429 650 215 846 4;
  • 8) 0.429 650 215 846 4 × 2 = 0 + 0.859 300 431 692 8;
  • 9) 0.859 300 431 692 8 × 2 = 1 + 0.718 600 863 385 6;
  • 10) 0.718 600 863 385 6 × 2 = 1 + 0.437 201 726 771 2;
  • 11) 0.437 201 726 771 2 × 2 = 0 + 0.874 403 453 542 4;
  • 12) 0.874 403 453 542 4 × 2 = 1 + 0.748 806 907 084 8;
  • 13) 0.748 806 907 084 8 × 2 = 1 + 0.497 613 814 169 6;
  • 14) 0.497 613 814 169 6 × 2 = 0 + 0.995 227 628 339 2;
  • 15) 0.995 227 628 339 2 × 2 = 1 + 0.990 455 256 678 4;
  • 16) 0.990 455 256 678 4 × 2 = 1 + 0.980 910 513 356 8;
  • 17) 0.980 910 513 356 8 × 2 = 1 + 0.961 821 026 713 6;
  • 18) 0.961 821 026 713 6 × 2 = 1 + 0.923 642 053 427 2;
  • 19) 0.923 642 053 427 2 × 2 = 1 + 0.847 284 106 854 4;
  • 20) 0.847 284 106 854 4 × 2 = 1 + 0.694 568 213 708 8;
  • 21) 0.694 568 213 708 8 × 2 = 1 + 0.389 136 427 417 6;
  • 22) 0.389 136 427 417 6 × 2 = 0 + 0.778 272 854 835 2;
  • 23) 0.778 272 854 835 2 × 2 = 1 + 0.556 545 709 670 4;
  • 24) 0.556 545 709 670 4 × 2 = 1 + 0.113 091 419 340 8;
  • 25) 0.113 091 419 340 8 × 2 = 0 + 0.226 182 838 681 6;
  • 26) 0.226 182 838 681 6 × 2 = 0 + 0.452 365 677 363 2;
  • 27) 0.452 365 677 363 2 × 2 = 0 + 0.904 731 354 726 4;
  • 28) 0.904 731 354 726 4 × 2 = 1 + 0.809 462 709 452 8;
  • 29) 0.809 462 709 452 8 × 2 = 1 + 0.618 925 418 905 6;
  • 30) 0.618 925 418 905 6 × 2 = 1 + 0.237 850 837 811 2;
  • 31) 0.237 850 837 811 2 × 2 = 0 + 0.475 701 675 622 4;
  • 32) 0.475 701 675 622 4 × 2 = 0 + 0.951 403 351 244 8;
  • 33) 0.951 403 351 244 8 × 2 = 1 + 0.902 806 702 489 6;
  • 34) 0.902 806 702 489 6 × 2 = 1 + 0.805 613 404 979 2;
  • 35) 0.805 613 404 979 2 × 2 = 1 + 0.611 226 809 958 4;
  • 36) 0.611 226 809 958 4 × 2 = 1 + 0.222 453 619 916 8;
  • 37) 0.222 453 619 916 8 × 2 = 0 + 0.444 907 239 833 6;
  • 38) 0.444 907 239 833 6 × 2 = 0 + 0.889 814 479 667 2;
  • 39) 0.889 814 479 667 2 × 2 = 1 + 0.779 628 959 334 4;
  • 40) 0.779 628 959 334 4 × 2 = 1 + 0.559 257 918 668 8;
  • 41) 0.559 257 918 668 8 × 2 = 1 + 0.118 515 837 337 6;
  • 42) 0.118 515 837 337 6 × 2 = 0 + 0.237 031 674 675 2;
  • 43) 0.237 031 674 675 2 × 2 = 0 + 0.474 063 349 350 4;
  • 44) 0.474 063 349 350 4 × 2 = 0 + 0.948 126 698 700 8;
  • 45) 0.948 126 698 700 8 × 2 = 1 + 0.896 253 397 401 6;
  • 46) 0.896 253 397 401 6 × 2 = 1 + 0.792 506 794 803 2;
  • 47) 0.792 506 794 803 2 × 2 = 1 + 0.585 013 589 606 4;
  • 48) 0.585 013 589 606 4 × 2 = 1 + 0.170 027 179 212 8;
  • 49) 0.170 027 179 212 8 × 2 = 0 + 0.340 054 358 425 6;
  • 50) 0.340 054 358 425 6 × 2 = 0 + 0.680 108 716 851 2;
  • 51) 0.680 108 716 851 2 × 2 = 1 + 0.360 217 433 702 4;
  • 52) 0.360 217 433 702 4 × 2 = 0 + 0.720 434 867 404 8;
  • 53) 0.720 434 867 404 8 × 2 = 1 + 0.440 869 734 809 6;
  • 54) 0.440 869 734 809 6 × 2 = 0 + 0.881 739 469 619 2;
  • 55) 0.881 739 469 619 2 × 2 = 1 + 0.763 478 939 238 4;
  • 56) 0.763 478 939 238 4 × 2 = 1 + 0.526 957 878 476 8;
  • 57) 0.526 957 878 476 8 × 2 = 1 + 0.053 915 756 953 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:


0.042 419 142 311 3(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0011 1000 1111 0010 1011 1(2)

6. Positive number before normalization:

0.042 419 142 311 3(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0011 1000 1111 0010 1011 1(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:


0.042 419 142 311 3(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0011 1000 1111 0010 1011 1(2) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0011 1000 1111 0010 1011 1(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1001 1110 0111 0001 1110 0101 0111(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 1110 0111 0001 1110 0101 0111


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 1110 0111 0001 1110 0101 0111 =

0101 1011 0111 1111 0110 0011 1001 1110 0111 0001 1110 0101 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 1010

Mantissa (52 bits) =
0101 1011 0111 1111 0110 0011 1001 1110 0111 0001 1110 0101 0111


Decimal number -0.042 419 142 311 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0101 1011 0111 1111 0110 0011 1001 1110 0111 0001 1110 0101 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