-0.199 999 999 996 42 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.199 999 999 996 42(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
-0.199 999 999 996 42(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:

|-0.199 999 999 996 42| = 0.199 999 999 996 42


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: 0.199 999 999 996 42.

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) 0.199 999 999 996 42 × 2 = 0 + 0.399 999 999 992 84;
  • 2) 0.399 999 999 992 84 × 2 = 0 + 0.799 999 999 985 68;
  • 3) 0.799 999 999 985 68 × 2 = 1 + 0.599 999 999 971 36;
  • 4) 0.599 999 999 971 36 × 2 = 1 + 0.199 999 999 942 72;
  • 5) 0.199 999 999 942 72 × 2 = 0 + 0.399 999 999 885 44;
  • 6) 0.399 999 999 885 44 × 2 = 0 + 0.799 999 999 770 88;
  • 7) 0.799 999 999 770 88 × 2 = 1 + 0.599 999 999 541 76;
  • 8) 0.599 999 999 541 76 × 2 = 1 + 0.199 999 999 083 52;
  • 9) 0.199 999 999 083 52 × 2 = 0 + 0.399 999 998 167 04;
  • 10) 0.399 999 998 167 04 × 2 = 0 + 0.799 999 996 334 08;
  • 11) 0.799 999 996 334 08 × 2 = 1 + 0.599 999 992 668 16;
  • 12) 0.599 999 992 668 16 × 2 = 1 + 0.199 999 985 336 32;
  • 13) 0.199 999 985 336 32 × 2 = 0 + 0.399 999 970 672 64;
  • 14) 0.399 999 970 672 64 × 2 = 0 + 0.799 999 941 345 28;
  • 15) 0.799 999 941 345 28 × 2 = 1 + 0.599 999 882 690 56;
  • 16) 0.599 999 882 690 56 × 2 = 1 + 0.199 999 765 381 12;
  • 17) 0.199 999 765 381 12 × 2 = 0 + 0.399 999 530 762 24;
  • 18) 0.399 999 530 762 24 × 2 = 0 + 0.799 999 061 524 48;
  • 19) 0.799 999 061 524 48 × 2 = 1 + 0.599 998 123 048 96;
  • 20) 0.599 998 123 048 96 × 2 = 1 + 0.199 996 246 097 92;
  • 21) 0.199 996 246 097 92 × 2 = 0 + 0.399 992 492 195 84;
  • 22) 0.399 992 492 195 84 × 2 = 0 + 0.799 984 984 391 68;
  • 23) 0.799 984 984 391 68 × 2 = 1 + 0.599 969 968 783 36;
  • 24) 0.599 969 968 783 36 × 2 = 1 + 0.199 939 937 566 72;
  • 25) 0.199 939 937 566 72 × 2 = 0 + 0.399 879 875 133 44;
  • 26) 0.399 879 875 133 44 × 2 = 0 + 0.799 759 750 266 88;
  • 27) 0.799 759 750 266 88 × 2 = 1 + 0.599 519 500 533 76;
  • 28) 0.599 519 500 533 76 × 2 = 1 + 0.199 039 001 067 52;
  • 29) 0.199 039 001 067 52 × 2 = 0 + 0.398 078 002 135 04;
  • 30) 0.398 078 002 135 04 × 2 = 0 + 0.796 156 004 270 08;
  • 31) 0.796 156 004 270 08 × 2 = 1 + 0.592 312 008 540 16;
  • 32) 0.592 312 008 540 16 × 2 = 1 + 0.184 624 017 080 32;
  • 33) 0.184 624 017 080 32 × 2 = 0 + 0.369 248 034 160 64;
  • 34) 0.369 248 034 160 64 × 2 = 0 + 0.738 496 068 321 28;
  • 35) 0.738 496 068 321 28 × 2 = 1 + 0.476 992 136 642 56;
  • 36) 0.476 992 136 642 56 × 2 = 0 + 0.953 984 273 285 12;
  • 37) 0.953 984 273 285 12 × 2 = 1 + 0.907 968 546 570 24;
  • 38) 0.907 968 546 570 24 × 2 = 1 + 0.815 937 093 140 48;
  • 39) 0.815 937 093 140 48 × 2 = 1 + 0.631 874 186 280 96;
  • 40) 0.631 874 186 280 96 × 2 = 1 + 0.263 748 372 561 92;
  • 41) 0.263 748 372 561 92 × 2 = 0 + 0.527 496 745 123 84;
  • 42) 0.527 496 745 123 84 × 2 = 1 + 0.054 993 490 247 68;
  • 43) 0.054 993 490 247 68 × 2 = 0 + 0.109 986 980 495 36;
  • 44) 0.109 986 980 495 36 × 2 = 0 + 0.219 973 960 990 72;
  • 45) 0.219 973 960 990 72 × 2 = 0 + 0.439 947 921 981 44;
  • 46) 0.439 947 921 981 44 × 2 = 0 + 0.879 895 843 962 88;
  • 47) 0.879 895 843 962 88 × 2 = 1 + 0.759 791 687 925 76;
  • 48) 0.759 791 687 925 76 × 2 = 1 + 0.519 583 375 851 52;
  • 49) 0.519 583 375 851 52 × 2 = 1 + 0.039 166 751 703 04;
  • 50) 0.039 166 751 703 04 × 2 = 0 + 0.078 333 503 406 08;
  • 51) 0.078 333 503 406 08 × 2 = 0 + 0.156 667 006 812 16;
  • 52) 0.156 667 006 812 16 × 2 = 0 + 0.313 334 013 624 32;
  • 53) 0.313 334 013 624 32 × 2 = 0 + 0.626 668 027 248 64;
  • 54) 0.626 668 027 248 64 × 2 = 1 + 0.253 336 054 497 28;
  • 55) 0.253 336 054 497 28 × 2 = 0 + 0.506 672 108 994 56;

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:


0.199 999 999 996 42(10) =

0.0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 0100 0011 1000 010(2)

6. Positive number before normalization:

0.199 999 999 996 42(10) =

0.0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 0100 0011 1000 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.199 999 999 996 42(10) =

0.0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 0100 0011 1000 010(2) =

0.0011 0011 0011 0011 0011 0011 0011 0011 0010 1111 0100 0011 1000 010(2) × 20 =

1.1001 1001 1001 1001 1001 1001 1001 1001 0111 1010 0001 1100 0010(2) × 2-3


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): -3

Mantissa (not normalized):
1.1001 1001 1001 1001 1001 1001 1001 1001 0111 1010 0001 1100 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1020(10) =

011 1111 1100(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. 1001 1001 1001 1001 1001 1001 1001 1001 0111 1010 0001 1100 0010 =

1001 1001 1001 1001 1001 1001 1001 1001 0111 1010 0001 1100 0010


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 1100

Mantissa (52 bits) =
1001 1001 1001 1001 1001 1001 1001 1001 0111 1010 0001 1100 0010


Decimal number -0.199 999 999 996 42 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 1001 1001 1001 1001 1001 1001 1001 1001 0111 1010 0001 1100 0010

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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