-0.042 419 142 302 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 302 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
-0.042 419 142 302 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:

|-0.042 419 142 302 2| = 0.042 419 142 302 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: 0.042 419 142 302 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) 0.042 419 142 302 2 × 2 = 0 + 0.084 838 284 604 4;
  • 2) 0.084 838 284 604 4 × 2 = 0 + 0.169 676 569 208 8;
  • 3) 0.169 676 569 208 8 × 2 = 0 + 0.339 353 138 417 6;
  • 4) 0.339 353 138 417 6 × 2 = 0 + 0.678 706 276 835 2;
  • 5) 0.678 706 276 835 2 × 2 = 1 + 0.357 412 553 670 4;
  • 6) 0.357 412 553 670 4 × 2 = 0 + 0.714 825 107 340 8;
  • 7) 0.714 825 107 340 8 × 2 = 1 + 0.429 650 214 681 6;
  • 8) 0.429 650 214 681 6 × 2 = 0 + 0.859 300 429 363 2;
  • 9) 0.859 300 429 363 2 × 2 = 1 + 0.718 600 858 726 4;
  • 10) 0.718 600 858 726 4 × 2 = 1 + 0.437 201 717 452 8;
  • 11) 0.437 201 717 452 8 × 2 = 0 + 0.874 403 434 905 6;
  • 12) 0.874 403 434 905 6 × 2 = 1 + 0.748 806 869 811 2;
  • 13) 0.748 806 869 811 2 × 2 = 1 + 0.497 613 739 622 4;
  • 14) 0.497 613 739 622 4 × 2 = 0 + 0.995 227 479 244 8;
  • 15) 0.995 227 479 244 8 × 2 = 1 + 0.990 454 958 489 6;
  • 16) 0.990 454 958 489 6 × 2 = 1 + 0.980 909 916 979 2;
  • 17) 0.980 909 916 979 2 × 2 = 1 + 0.961 819 833 958 4;
  • 18) 0.961 819 833 958 4 × 2 = 1 + 0.923 639 667 916 8;
  • 19) 0.923 639 667 916 8 × 2 = 1 + 0.847 279 335 833 6;
  • 20) 0.847 279 335 833 6 × 2 = 1 + 0.694 558 671 667 2;
  • 21) 0.694 558 671 667 2 × 2 = 1 + 0.389 117 343 334 4;
  • 22) 0.389 117 343 334 4 × 2 = 0 + 0.778 234 686 668 8;
  • 23) 0.778 234 686 668 8 × 2 = 1 + 0.556 469 373 337 6;
  • 24) 0.556 469 373 337 6 × 2 = 1 + 0.112 938 746 675 2;
  • 25) 0.112 938 746 675 2 × 2 = 0 + 0.225 877 493 350 4;
  • 26) 0.225 877 493 350 4 × 2 = 0 + 0.451 754 986 700 8;
  • 27) 0.451 754 986 700 8 × 2 = 0 + 0.903 509 973 401 6;
  • 28) 0.903 509 973 401 6 × 2 = 1 + 0.807 019 946 803 2;
  • 29) 0.807 019 946 803 2 × 2 = 1 + 0.614 039 893 606 4;
  • 30) 0.614 039 893 606 4 × 2 = 1 + 0.228 079 787 212 8;
  • 31) 0.228 079 787 212 8 × 2 = 0 + 0.456 159 574 425 6;
  • 32) 0.456 159 574 425 6 × 2 = 0 + 0.912 319 148 851 2;
  • 33) 0.912 319 148 851 2 × 2 = 1 + 0.824 638 297 702 4;
  • 34) 0.824 638 297 702 4 × 2 = 1 + 0.649 276 595 404 8;
  • 35) 0.649 276 595 404 8 × 2 = 1 + 0.298 553 190 809 6;
  • 36) 0.298 553 190 809 6 × 2 = 0 + 0.597 106 381 619 2;
  • 37) 0.597 106 381 619 2 × 2 = 1 + 0.194 212 763 238 4;
  • 38) 0.194 212 763 238 4 × 2 = 0 + 0.388 425 526 476 8;
  • 39) 0.388 425 526 476 8 × 2 = 0 + 0.776 851 052 953 6;
  • 40) 0.776 851 052 953 6 × 2 = 1 + 0.553 702 105 907 2;
  • 41) 0.553 702 105 907 2 × 2 = 1 + 0.107 404 211 814 4;
  • 42) 0.107 404 211 814 4 × 2 = 0 + 0.214 808 423 628 8;
  • 43) 0.214 808 423 628 8 × 2 = 0 + 0.429 616 847 257 6;
  • 44) 0.429 616 847 257 6 × 2 = 0 + 0.859 233 694 515 2;
  • 45) 0.859 233 694 515 2 × 2 = 1 + 0.718 467 389 030 4;
  • 46) 0.718 467 389 030 4 × 2 = 1 + 0.436 934 778 060 8;
  • 47) 0.436 934 778 060 8 × 2 = 0 + 0.873 869 556 121 6;
  • 48) 0.873 869 556 121 6 × 2 = 1 + 0.747 739 112 243 2;
  • 49) 0.747 739 112 243 2 × 2 = 1 + 0.495 478 224 486 4;
  • 50) 0.495 478 224 486 4 × 2 = 0 + 0.990 956 448 972 8;
  • 51) 0.990 956 448 972 8 × 2 = 1 + 0.981 912 897 945 6;
  • 52) 0.981 912 897 945 6 × 2 = 1 + 0.963 825 795 891 2;
  • 53) 0.963 825 795 891 2 × 2 = 1 + 0.927 651 591 782 4;
  • 54) 0.927 651 591 782 4 × 2 = 1 + 0.855 303 183 564 8;
  • 55) 0.855 303 183 564 8 × 2 = 1 + 0.710 606 367 129 6;
  • 56) 0.710 606 367 129 6 × 2 = 1 + 0.421 212 734 259 2;
  • 57) 0.421 212 734 259 2 × 2 = 0 + 0.842 425 468 518 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:


0.042 419 142 302 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 1001 1000 1101 1011 1111 0(2)

6. Positive number before normalization:

0.042 419 142 302 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 1001 1000 1101 1011 1111 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:


0.042 419 142 302 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 1001 1000 1101 1011 1111 0(2) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 1001 1000 1101 1011 1111 0(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1001 1101 0011 0001 1011 0111 1110(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 0011 0001 1011 0111 1110


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 0011 0001 1011 0111 1110 =

0101 1011 0111 1111 0110 0011 1001 1101 0011 0001 1011 0111 1110


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 0011 0001 1011 0111 1110


Decimal number -0.042 419 142 302 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 0011 0001 1011 0111 1110

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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