-0.042 419 142 330 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 330 4(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
-0.042 419 142 330 4(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:

|-0.042 419 142 330 4| = 0.042 419 142 330 4


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: 0.042 419 142 330 4.

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) 0.042 419 142 330 4 × 2 = 0 + 0.084 838 284 660 8;
  • 2) 0.084 838 284 660 8 × 2 = 0 + 0.169 676 569 321 6;
  • 3) 0.169 676 569 321 6 × 2 = 0 + 0.339 353 138 643 2;
  • 4) 0.339 353 138 643 2 × 2 = 0 + 0.678 706 277 286 4;
  • 5) 0.678 706 277 286 4 × 2 = 1 + 0.357 412 554 572 8;
  • 6) 0.357 412 554 572 8 × 2 = 0 + 0.714 825 109 145 6;
  • 7) 0.714 825 109 145 6 × 2 = 1 + 0.429 650 218 291 2;
  • 8) 0.429 650 218 291 2 × 2 = 0 + 0.859 300 436 582 4;
  • 9) 0.859 300 436 582 4 × 2 = 1 + 0.718 600 873 164 8;
  • 10) 0.718 600 873 164 8 × 2 = 1 + 0.437 201 746 329 6;
  • 11) 0.437 201 746 329 6 × 2 = 0 + 0.874 403 492 659 2;
  • 12) 0.874 403 492 659 2 × 2 = 1 + 0.748 806 985 318 4;
  • 13) 0.748 806 985 318 4 × 2 = 1 + 0.497 613 970 636 8;
  • 14) 0.497 613 970 636 8 × 2 = 0 + 0.995 227 941 273 6;
  • 15) 0.995 227 941 273 6 × 2 = 1 + 0.990 455 882 547 2;
  • 16) 0.990 455 882 547 2 × 2 = 1 + 0.980 911 765 094 4;
  • 17) 0.980 911 765 094 4 × 2 = 1 + 0.961 823 530 188 8;
  • 18) 0.961 823 530 188 8 × 2 = 1 + 0.923 647 060 377 6;
  • 19) 0.923 647 060 377 6 × 2 = 1 + 0.847 294 120 755 2;
  • 20) 0.847 294 120 755 2 × 2 = 1 + 0.694 588 241 510 4;
  • 21) 0.694 588 241 510 4 × 2 = 1 + 0.389 176 483 020 8;
  • 22) 0.389 176 483 020 8 × 2 = 0 + 0.778 352 966 041 6;
  • 23) 0.778 352 966 041 6 × 2 = 1 + 0.556 705 932 083 2;
  • 24) 0.556 705 932 083 2 × 2 = 1 + 0.113 411 864 166 4;
  • 25) 0.113 411 864 166 4 × 2 = 0 + 0.226 823 728 332 8;
  • 26) 0.226 823 728 332 8 × 2 = 0 + 0.453 647 456 665 6;
  • 27) 0.453 647 456 665 6 × 2 = 0 + 0.907 294 913 331 2;
  • 28) 0.907 294 913 331 2 × 2 = 1 + 0.814 589 826 662 4;
  • 29) 0.814 589 826 662 4 × 2 = 1 + 0.629 179 653 324 8;
  • 30) 0.629 179 653 324 8 × 2 = 1 + 0.258 359 306 649 6;
  • 31) 0.258 359 306 649 6 × 2 = 0 + 0.516 718 613 299 2;
  • 32) 0.516 718 613 299 2 × 2 = 1 + 0.033 437 226 598 4;
  • 33) 0.033 437 226 598 4 × 2 = 0 + 0.066 874 453 196 8;
  • 34) 0.066 874 453 196 8 × 2 = 0 + 0.133 748 906 393 6;
  • 35) 0.133 748 906 393 6 × 2 = 0 + 0.267 497 812 787 2;
  • 36) 0.267 497 812 787 2 × 2 = 0 + 0.534 995 625 574 4;
  • 37) 0.534 995 625 574 4 × 2 = 1 + 0.069 991 251 148 8;
  • 38) 0.069 991 251 148 8 × 2 = 0 + 0.139 982 502 297 6;
  • 39) 0.139 982 502 297 6 × 2 = 0 + 0.279 965 004 595 2;
  • 40) 0.279 965 004 595 2 × 2 = 0 + 0.559 930 009 190 4;
  • 41) 0.559 930 009 190 4 × 2 = 1 + 0.119 860 018 380 8;
  • 42) 0.119 860 018 380 8 × 2 = 0 + 0.239 720 036 761 6;
  • 43) 0.239 720 036 761 6 × 2 = 0 + 0.479 440 073 523 2;
  • 44) 0.479 440 073 523 2 × 2 = 0 + 0.958 880 147 046 4;
  • 45) 0.958 880 147 046 4 × 2 = 1 + 0.917 760 294 092 8;
  • 46) 0.917 760 294 092 8 × 2 = 1 + 0.835 520 588 185 6;
  • 47) 0.835 520 588 185 6 × 2 = 1 + 0.671 041 176 371 2;
  • 48) 0.671 041 176 371 2 × 2 = 1 + 0.342 082 352 742 4;
  • 49) 0.342 082 352 742 4 × 2 = 0 + 0.684 164 705 484 8;
  • 50) 0.684 164 705 484 8 × 2 = 1 + 0.368 329 410 969 6;
  • 51) 0.368 329 410 969 6 × 2 = 0 + 0.736 658 821 939 2;
  • 52) 0.736 658 821 939 2 × 2 = 1 + 0.473 317 643 878 4;
  • 53) 0.473 317 643 878 4 × 2 = 0 + 0.946 635 287 756 8;
  • 54) 0.946 635 287 756 8 × 2 = 1 + 0.893 270 575 513 6;
  • 55) 0.893 270 575 513 6 × 2 = 1 + 0.786 541 151 027 2;
  • 56) 0.786 541 151 027 2 × 2 = 1 + 0.573 082 302 054 4;
  • 57) 0.573 082 302 054 4 × 2 = 1 + 0.146 164 604 108 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:


0.042 419 142 330 4(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 1000 1000 1111 0101 0111 1(2)

6. Positive number before normalization:

0.042 419 142 330 4(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 1000 1000 1111 0101 0111 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:


0.042 419 142 330 4(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 1000 1000 1111 0101 0111 1(2) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 1000 1000 1111 0101 0111 1(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1010 0001 0001 0001 1110 1010 1111(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 0001 0001 1110 1010 1111


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 0001 0001 1110 1010 1111 =

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


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 0001 0001 1110 1010 1111


Decimal number -0.042 419 142 330 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

» Convert decimal -0.042 419 142 336 1 to 64 bit double precision IEEE 754 binary floating point representation standard

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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