-0.042 419 142 308 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 308 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
-0.042 419 142 308 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:

|-0.042 419 142 308 3| = 0.042 419 142 308 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: 0.042 419 142 308 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) 0.042 419 142 308 3 × 2 = 0 + 0.084 838 284 616 6;
  • 2) 0.084 838 284 616 6 × 2 = 0 + 0.169 676 569 233 2;
  • 3) 0.169 676 569 233 2 × 2 = 0 + 0.339 353 138 466 4;
  • 4) 0.339 353 138 466 4 × 2 = 0 + 0.678 706 276 932 8;
  • 5) 0.678 706 276 932 8 × 2 = 1 + 0.357 412 553 865 6;
  • 6) 0.357 412 553 865 6 × 2 = 0 + 0.714 825 107 731 2;
  • 7) 0.714 825 107 731 2 × 2 = 1 + 0.429 650 215 462 4;
  • 8) 0.429 650 215 462 4 × 2 = 0 + 0.859 300 430 924 8;
  • 9) 0.859 300 430 924 8 × 2 = 1 + 0.718 600 861 849 6;
  • 10) 0.718 600 861 849 6 × 2 = 1 + 0.437 201 723 699 2;
  • 11) 0.437 201 723 699 2 × 2 = 0 + 0.874 403 447 398 4;
  • 12) 0.874 403 447 398 4 × 2 = 1 + 0.748 806 894 796 8;
  • 13) 0.748 806 894 796 8 × 2 = 1 + 0.497 613 789 593 6;
  • 14) 0.497 613 789 593 6 × 2 = 0 + 0.995 227 579 187 2;
  • 15) 0.995 227 579 187 2 × 2 = 1 + 0.990 455 158 374 4;
  • 16) 0.990 455 158 374 4 × 2 = 1 + 0.980 910 316 748 8;
  • 17) 0.980 910 316 748 8 × 2 = 1 + 0.961 820 633 497 6;
  • 18) 0.961 820 633 497 6 × 2 = 1 + 0.923 641 266 995 2;
  • 19) 0.923 641 266 995 2 × 2 = 1 + 0.847 282 533 990 4;
  • 20) 0.847 282 533 990 4 × 2 = 1 + 0.694 565 067 980 8;
  • 21) 0.694 565 067 980 8 × 2 = 1 + 0.389 130 135 961 6;
  • 22) 0.389 130 135 961 6 × 2 = 0 + 0.778 260 271 923 2;
  • 23) 0.778 260 271 923 2 × 2 = 1 + 0.556 520 543 846 4;
  • 24) 0.556 520 543 846 4 × 2 = 1 + 0.113 041 087 692 8;
  • 25) 0.113 041 087 692 8 × 2 = 0 + 0.226 082 175 385 6;
  • 26) 0.226 082 175 385 6 × 2 = 0 + 0.452 164 350 771 2;
  • 27) 0.452 164 350 771 2 × 2 = 0 + 0.904 328 701 542 4;
  • 28) 0.904 328 701 542 4 × 2 = 1 + 0.808 657 403 084 8;
  • 29) 0.808 657 403 084 8 × 2 = 1 + 0.617 314 806 169 6;
  • 30) 0.617 314 806 169 6 × 2 = 1 + 0.234 629 612 339 2;
  • 31) 0.234 629 612 339 2 × 2 = 0 + 0.469 259 224 678 4;
  • 32) 0.469 259 224 678 4 × 2 = 0 + 0.938 518 449 356 8;
  • 33) 0.938 518 449 356 8 × 2 = 1 + 0.877 036 898 713 6;
  • 34) 0.877 036 898 713 6 × 2 = 1 + 0.754 073 797 427 2;
  • 35) 0.754 073 797 427 2 × 2 = 1 + 0.508 147 594 854 4;
  • 36) 0.508 147 594 854 4 × 2 = 1 + 0.016 295 189 708 8;
  • 37) 0.016 295 189 708 8 × 2 = 0 + 0.032 590 379 417 6;
  • 38) 0.032 590 379 417 6 × 2 = 0 + 0.065 180 758 835 2;
  • 39) 0.065 180 758 835 2 × 2 = 0 + 0.130 361 517 670 4;
  • 40) 0.130 361 517 670 4 × 2 = 0 + 0.260 723 035 340 8;
  • 41) 0.260 723 035 340 8 × 2 = 0 + 0.521 446 070 681 6;
  • 42) 0.521 446 070 681 6 × 2 = 1 + 0.042 892 141 363 2;
  • 43) 0.042 892 141 363 2 × 2 = 0 + 0.085 784 282 726 4;
  • 44) 0.085 784 282 726 4 × 2 = 0 + 0.171 568 565 452 8;
  • 45) 0.171 568 565 452 8 × 2 = 0 + 0.343 137 130 905 6;
  • 46) 0.343 137 130 905 6 × 2 = 0 + 0.686 274 261 811 2;
  • 47) 0.686 274 261 811 2 × 2 = 1 + 0.372 548 523 622 4;
  • 48) 0.372 548 523 622 4 × 2 = 0 + 0.745 097 047 244 8;
  • 49) 0.745 097 047 244 8 × 2 = 1 + 0.490 194 094 489 6;
  • 50) 0.490 194 094 489 6 × 2 = 0 + 0.980 388 188 979 2;
  • 51) 0.980 388 188 979 2 × 2 = 1 + 0.960 776 377 958 4;
  • 52) 0.960 776 377 958 4 × 2 = 1 + 0.921 552 755 916 8;
  • 53) 0.921 552 755 916 8 × 2 = 1 + 0.843 105 511 833 6;
  • 54) 0.843 105 511 833 6 × 2 = 1 + 0.686 211 023 667 2;
  • 55) 0.686 211 023 667 2 × 2 = 1 + 0.372 422 047 334 4;
  • 56) 0.372 422 047 334 4 × 2 = 0 + 0.744 844 094 668 8;
  • 57) 0.744 844 094 668 8 × 2 = 1 + 0.489 688 189 337 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:


0.042 419 142 308 3(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0000 0100 0010 1011 1110 1(2)

6. Positive number before normalization:

0.042 419 142 308 3(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0000 0100 0010 1011 1110 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:


0.042 419 142 308 3(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0000 0100 0010 1011 1110 1(2) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0000 0100 0010 1011 1110 1(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1001 1110 0000 1000 0101 0111 1101(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 0000 1000 0101 0111 1101


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 0000 1000 0101 0111 1101 =

0101 1011 0111 1111 0110 0011 1001 1110 0000 1000 0101 0111 1101


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 0000 1000 0101 0111 1101


Decimal number -0.042 419 142 308 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 0000 1000 0101 0111 1101

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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