-0.042 419 142 331 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 331 9(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
-0.042 419 142 331 9(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:

|-0.042 419 142 331 9| = 0.042 419 142 331 9


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: 0.042 419 142 331 9.

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) 0.042 419 142 331 9 × 2 = 0 + 0.084 838 284 663 8;
  • 2) 0.084 838 284 663 8 × 2 = 0 + 0.169 676 569 327 6;
  • 3) 0.169 676 569 327 6 × 2 = 0 + 0.339 353 138 655 2;
  • 4) 0.339 353 138 655 2 × 2 = 0 + 0.678 706 277 310 4;
  • 5) 0.678 706 277 310 4 × 2 = 1 + 0.357 412 554 620 8;
  • 6) 0.357 412 554 620 8 × 2 = 0 + 0.714 825 109 241 6;
  • 7) 0.714 825 109 241 6 × 2 = 1 + 0.429 650 218 483 2;
  • 8) 0.429 650 218 483 2 × 2 = 0 + 0.859 300 436 966 4;
  • 9) 0.859 300 436 966 4 × 2 = 1 + 0.718 600 873 932 8;
  • 10) 0.718 600 873 932 8 × 2 = 1 + 0.437 201 747 865 6;
  • 11) 0.437 201 747 865 6 × 2 = 0 + 0.874 403 495 731 2;
  • 12) 0.874 403 495 731 2 × 2 = 1 + 0.748 806 991 462 4;
  • 13) 0.748 806 991 462 4 × 2 = 1 + 0.497 613 982 924 8;
  • 14) 0.497 613 982 924 8 × 2 = 0 + 0.995 227 965 849 6;
  • 15) 0.995 227 965 849 6 × 2 = 1 + 0.990 455 931 699 2;
  • 16) 0.990 455 931 699 2 × 2 = 1 + 0.980 911 863 398 4;
  • 17) 0.980 911 863 398 4 × 2 = 1 + 0.961 823 726 796 8;
  • 18) 0.961 823 726 796 8 × 2 = 1 + 0.923 647 453 593 6;
  • 19) 0.923 647 453 593 6 × 2 = 1 + 0.847 294 907 187 2;
  • 20) 0.847 294 907 187 2 × 2 = 1 + 0.694 589 814 374 4;
  • 21) 0.694 589 814 374 4 × 2 = 1 + 0.389 179 628 748 8;
  • 22) 0.389 179 628 748 8 × 2 = 0 + 0.778 359 257 497 6;
  • 23) 0.778 359 257 497 6 × 2 = 1 + 0.556 718 514 995 2;
  • 24) 0.556 718 514 995 2 × 2 = 1 + 0.113 437 029 990 4;
  • 25) 0.113 437 029 990 4 × 2 = 0 + 0.226 874 059 980 8;
  • 26) 0.226 874 059 980 8 × 2 = 0 + 0.453 748 119 961 6;
  • 27) 0.453 748 119 961 6 × 2 = 0 + 0.907 496 239 923 2;
  • 28) 0.907 496 239 923 2 × 2 = 1 + 0.814 992 479 846 4;
  • 29) 0.814 992 479 846 4 × 2 = 1 + 0.629 984 959 692 8;
  • 30) 0.629 984 959 692 8 × 2 = 1 + 0.259 969 919 385 6;
  • 31) 0.259 969 919 385 6 × 2 = 0 + 0.519 939 838 771 2;
  • 32) 0.519 939 838 771 2 × 2 = 1 + 0.039 879 677 542 4;
  • 33) 0.039 879 677 542 4 × 2 = 0 + 0.079 759 355 084 8;
  • 34) 0.079 759 355 084 8 × 2 = 0 + 0.159 518 710 169 6;
  • 35) 0.159 518 710 169 6 × 2 = 0 + 0.319 037 420 339 2;
  • 36) 0.319 037 420 339 2 × 2 = 0 + 0.638 074 840 678 4;
  • 37) 0.638 074 840 678 4 × 2 = 1 + 0.276 149 681 356 8;
  • 38) 0.276 149 681 356 8 × 2 = 0 + 0.552 299 362 713 6;
  • 39) 0.552 299 362 713 6 × 2 = 1 + 0.104 598 725 427 2;
  • 40) 0.104 598 725 427 2 × 2 = 0 + 0.209 197 450 854 4;
  • 41) 0.209 197 450 854 4 × 2 = 0 + 0.418 394 901 708 8;
  • 42) 0.418 394 901 708 8 × 2 = 0 + 0.836 789 803 417 6;
  • 43) 0.836 789 803 417 6 × 2 = 1 + 0.673 579 606 835 2;
  • 44) 0.673 579 606 835 2 × 2 = 1 + 0.347 159 213 670 4;
  • 45) 0.347 159 213 670 4 × 2 = 0 + 0.694 318 427 340 8;
  • 46) 0.694 318 427 340 8 × 2 = 1 + 0.388 636 854 681 6;
  • 47) 0.388 636 854 681 6 × 2 = 0 + 0.777 273 709 363 2;
  • 48) 0.777 273 709 363 2 × 2 = 1 + 0.554 547 418 726 4;
  • 49) 0.554 547 418 726 4 × 2 = 1 + 0.109 094 837 452 8;
  • 50) 0.109 094 837 452 8 × 2 = 0 + 0.218 189 674 905 6;
  • 51) 0.218 189 674 905 6 × 2 = 0 + 0.436 379 349 811 2;
  • 52) 0.436 379 349 811 2 × 2 = 0 + 0.872 758 699 622 4;
  • 53) 0.872 758 699 622 4 × 2 = 1 + 0.745 517 399 244 8;
  • 54) 0.745 517 399 244 8 × 2 = 1 + 0.491 034 798 489 6;
  • 55) 0.491 034 798 489 6 × 2 = 0 + 0.982 069 596 979 2;
  • 56) 0.982 069 596 979 2 × 2 = 1 + 0.964 139 193 958 4;
  • 57) 0.964 139 193 958 4 × 2 = 1 + 0.928 278 387 916 8;

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:


0.042 419 142 331 9(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 1010 0011 0101 1000 1101 1(2)

6. Positive number before normalization:

0.042 419 142 331 9(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 1010 0011 0101 1000 1101 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:


0.042 419 142 331 9(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 1010 0011 0101 1000 1101 1(2) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 1010 0011 0101 1000 1101 1(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1010 0001 0100 0110 1011 0001 1011(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 1010 0001 0100 0110 1011 0001 1011


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 1010 0001 0100 0110 1011 0001 1011 =

0101 1011 0111 1111 0110 0011 1010 0001 0100 0110 1011 0001 1011


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 1010 0001 0100 0110 1011 0001 1011


Decimal number -0.042 419 142 331 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0101 1011 0111 1111 0110 0011 1010 0001 0100 0110 1011 0001 1011

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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