-0.042 419 142 308 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

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

|-0.042 419 142 308 2| = 0.042 419 142 308 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: 0.042 419 142 308 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) 0.042 419 142 308 2 × 2 = 0 + 0.084 838 284 616 4;
  • 2) 0.084 838 284 616 4 × 2 = 0 + 0.169 676 569 232 8;
  • 3) 0.169 676 569 232 8 × 2 = 0 + 0.339 353 138 465 6;
  • 4) 0.339 353 138 465 6 × 2 = 0 + 0.678 706 276 931 2;
  • 5) 0.678 706 276 931 2 × 2 = 1 + 0.357 412 553 862 4;
  • 6) 0.357 412 553 862 4 × 2 = 0 + 0.714 825 107 724 8;
  • 7) 0.714 825 107 724 8 × 2 = 1 + 0.429 650 215 449 6;
  • 8) 0.429 650 215 449 6 × 2 = 0 + 0.859 300 430 899 2;
  • 9) 0.859 300 430 899 2 × 2 = 1 + 0.718 600 861 798 4;
  • 10) 0.718 600 861 798 4 × 2 = 1 + 0.437 201 723 596 8;
  • 11) 0.437 201 723 596 8 × 2 = 0 + 0.874 403 447 193 6;
  • 12) 0.874 403 447 193 6 × 2 = 1 + 0.748 806 894 387 2;
  • 13) 0.748 806 894 387 2 × 2 = 1 + 0.497 613 788 774 4;
  • 14) 0.497 613 788 774 4 × 2 = 0 + 0.995 227 577 548 8;
  • 15) 0.995 227 577 548 8 × 2 = 1 + 0.990 455 155 097 6;
  • 16) 0.990 455 155 097 6 × 2 = 1 + 0.980 910 310 195 2;
  • 17) 0.980 910 310 195 2 × 2 = 1 + 0.961 820 620 390 4;
  • 18) 0.961 820 620 390 4 × 2 = 1 + 0.923 641 240 780 8;
  • 19) 0.923 641 240 780 8 × 2 = 1 + 0.847 282 481 561 6;
  • 20) 0.847 282 481 561 6 × 2 = 1 + 0.694 564 963 123 2;
  • 21) 0.694 564 963 123 2 × 2 = 1 + 0.389 129 926 246 4;
  • 22) 0.389 129 926 246 4 × 2 = 0 + 0.778 259 852 492 8;
  • 23) 0.778 259 852 492 8 × 2 = 1 + 0.556 519 704 985 6;
  • 24) 0.556 519 704 985 6 × 2 = 1 + 0.113 039 409 971 2;
  • 25) 0.113 039 409 971 2 × 2 = 0 + 0.226 078 819 942 4;
  • 26) 0.226 078 819 942 4 × 2 = 0 + 0.452 157 639 884 8;
  • 27) 0.452 157 639 884 8 × 2 = 0 + 0.904 315 279 769 6;
  • 28) 0.904 315 279 769 6 × 2 = 1 + 0.808 630 559 539 2;
  • 29) 0.808 630 559 539 2 × 2 = 1 + 0.617 261 119 078 4;
  • 30) 0.617 261 119 078 4 × 2 = 1 + 0.234 522 238 156 8;
  • 31) 0.234 522 238 156 8 × 2 = 0 + 0.469 044 476 313 6;
  • 32) 0.469 044 476 313 6 × 2 = 0 + 0.938 088 952 627 2;
  • 33) 0.938 088 952 627 2 × 2 = 1 + 0.876 177 905 254 4;
  • 34) 0.876 177 905 254 4 × 2 = 1 + 0.752 355 810 508 8;
  • 35) 0.752 355 810 508 8 × 2 = 1 + 0.504 711 621 017 6;
  • 36) 0.504 711 621 017 6 × 2 = 1 + 0.009 423 242 035 2;
  • 37) 0.009 423 242 035 2 × 2 = 0 + 0.018 846 484 070 4;
  • 38) 0.018 846 484 070 4 × 2 = 0 + 0.037 692 968 140 8;
  • 39) 0.037 692 968 140 8 × 2 = 0 + 0.075 385 936 281 6;
  • 40) 0.075 385 936 281 6 × 2 = 0 + 0.150 771 872 563 2;
  • 41) 0.150 771 872 563 2 × 2 = 0 + 0.301 543 745 126 4;
  • 42) 0.301 543 745 126 4 × 2 = 0 + 0.603 087 490 252 8;
  • 43) 0.603 087 490 252 8 × 2 = 1 + 0.206 174 980 505 6;
  • 44) 0.206 174 980 505 6 × 2 = 0 + 0.412 349 961 011 2;
  • 45) 0.412 349 961 011 2 × 2 = 0 + 0.824 699 922 022 4;
  • 46) 0.824 699 922 022 4 × 2 = 1 + 0.649 399 844 044 8;
  • 47) 0.649 399 844 044 8 × 2 = 1 + 0.298 799 688 089 6;
  • 48) 0.298 799 688 089 6 × 2 = 0 + 0.597 599 376 179 2;
  • 49) 0.597 599 376 179 2 × 2 = 1 + 0.195 198 752 358 4;
  • 50) 0.195 198 752 358 4 × 2 = 0 + 0.390 397 504 716 8;
  • 51) 0.390 397 504 716 8 × 2 = 0 + 0.780 795 009 433 6;
  • 52) 0.780 795 009 433 6 × 2 = 1 + 0.561 590 018 867 2;
  • 53) 0.561 590 018 867 2 × 2 = 1 + 0.123 180 037 734 4;
  • 54) 0.123 180 037 734 4 × 2 = 0 + 0.246 360 075 468 8;
  • 55) 0.246 360 075 468 8 × 2 = 0 + 0.492 720 150 937 6;
  • 56) 0.492 720 150 937 6 × 2 = 0 + 0.985 440 301 875 2;
  • 57) 0.985 440 301 875 2 × 2 = 1 + 0.970 880 603 750 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:


0.042 419 142 308 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0000 0010 0110 1001 1000 1(2)

6. Positive number before normalization:

0.042 419 142 308 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0000 0010 0110 1001 1000 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 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0000 0010 0110 1001 1000 1(2) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0000 0010 0110 1001 1000 1(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1001 1110 0000 0100 1101 0011 0001(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 0100 1101 0011 0001


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 0100 1101 0011 0001 =

0101 1011 0111 1111 0110 0011 1001 1110 0000 0100 1101 0011 0001


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 0100 1101 0011 0001


Decimal number -0.042 419 142 308 2 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 0100 1101 0011 0001

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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