204.120 999 999 999 980 902 53 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 204.120 999 999 999 980 902 53(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
204.120 999 999 999 980 902 53(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. First, convert to binary (in base 2) the integer part: 204.
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;
  • 204 ÷ 2 = 102 + 0;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. 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.

204(10) =

1100 1100(2)


3. Convert to binary (base 2) the fractional part: 0.120 999 999 999 980 902 53.

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) 0.120 999 999 999 980 902 53 × 2 = 0 + 0.241 999 999 999 961 805 06;
  • 2) 0.241 999 999 999 961 805 06 × 2 = 0 + 0.483 999 999 999 923 610 12;
  • 3) 0.483 999 999 999 923 610 12 × 2 = 0 + 0.967 999 999 999 847 220 24;
  • 4) 0.967 999 999 999 847 220 24 × 2 = 1 + 0.935 999 999 999 694 440 48;
  • 5) 0.935 999 999 999 694 440 48 × 2 = 1 + 0.871 999 999 999 388 880 96;
  • 6) 0.871 999 999 999 388 880 96 × 2 = 1 + 0.743 999 999 998 777 761 92;
  • 7) 0.743 999 999 998 777 761 92 × 2 = 1 + 0.487 999 999 997 555 523 84;
  • 8) 0.487 999 999 997 555 523 84 × 2 = 0 + 0.975 999 999 995 111 047 68;
  • 9) 0.975 999 999 995 111 047 68 × 2 = 1 + 0.951 999 999 990 222 095 36;
  • 10) 0.951 999 999 990 222 095 36 × 2 = 1 + 0.903 999 999 980 444 190 72;
  • 11) 0.903 999 999 980 444 190 72 × 2 = 1 + 0.807 999 999 960 888 381 44;
  • 12) 0.807 999 999 960 888 381 44 × 2 = 1 + 0.615 999 999 921 776 762 88;
  • 13) 0.615 999 999 921 776 762 88 × 2 = 1 + 0.231 999 999 843 553 525 76;
  • 14) 0.231 999 999 843 553 525 76 × 2 = 0 + 0.463 999 999 687 107 051 52;
  • 15) 0.463 999 999 687 107 051 52 × 2 = 0 + 0.927 999 999 374 214 103 04;
  • 16) 0.927 999 999 374 214 103 04 × 2 = 1 + 0.855 999 998 748 428 206 08;
  • 17) 0.855 999 998 748 428 206 08 × 2 = 1 + 0.711 999 997 496 856 412 16;
  • 18) 0.711 999 997 496 856 412 16 × 2 = 1 + 0.423 999 994 993 712 824 32;
  • 19) 0.423 999 994 993 712 824 32 × 2 = 0 + 0.847 999 989 987 425 648 64;
  • 20) 0.847 999 989 987 425 648 64 × 2 = 1 + 0.695 999 979 974 851 297 28;
  • 21) 0.695 999 979 974 851 297 28 × 2 = 1 + 0.391 999 959 949 702 594 56;
  • 22) 0.391 999 959 949 702 594 56 × 2 = 0 + 0.783 999 919 899 405 189 12;
  • 23) 0.783 999 919 899 405 189 12 × 2 = 1 + 0.567 999 839 798 810 378 24;
  • 24) 0.567 999 839 798 810 378 24 × 2 = 1 + 0.135 999 679 597 620 756 48;
  • 25) 0.135 999 679 597 620 756 48 × 2 = 0 + 0.271 999 359 195 241 512 96;
  • 26) 0.271 999 359 195 241 512 96 × 2 = 0 + 0.543 998 718 390 483 025 92;
  • 27) 0.543 998 718 390 483 025 92 × 2 = 1 + 0.087 997 436 780 966 051 84;
  • 28) 0.087 997 436 780 966 051 84 × 2 = 0 + 0.175 994 873 561 932 103 68;
  • 29) 0.175 994 873 561 932 103 68 × 2 = 0 + 0.351 989 747 123 864 207 36;
  • 30) 0.351 989 747 123 864 207 36 × 2 = 0 + 0.703 979 494 247 728 414 72;
  • 31) 0.703 979 494 247 728 414 72 × 2 = 1 + 0.407 958 988 495 456 829 44;
  • 32) 0.407 958 988 495 456 829 44 × 2 = 0 + 0.815 917 976 990 913 658 88;
  • 33) 0.815 917 976 990 913 658 88 × 2 = 1 + 0.631 835 953 981 827 317 76;
  • 34) 0.631 835 953 981 827 317 76 × 2 = 1 + 0.263 671 907 963 654 635 52;
  • 35) 0.263 671 907 963 654 635 52 × 2 = 0 + 0.527 343 815 927 309 271 04;
  • 36) 0.527 343 815 927 309 271 04 × 2 = 1 + 0.054 687 631 854 618 542 08;
  • 37) 0.054 687 631 854 618 542 08 × 2 = 0 + 0.109 375 263 709 237 084 16;
  • 38) 0.109 375 263 709 237 084 16 × 2 = 0 + 0.218 750 527 418 474 168 32;
  • 39) 0.218 750 527 418 474 168 32 × 2 = 0 + 0.437 501 054 836 948 336 64;
  • 40) 0.437 501 054 836 948 336 64 × 2 = 0 + 0.875 002 109 673 896 673 28;
  • 41) 0.875 002 109 673 896 673 28 × 2 = 1 + 0.750 004 219 347 793 346 56;
  • 42) 0.750 004 219 347 793 346 56 × 2 = 1 + 0.500 008 438 695 586 693 12;
  • 43) 0.500 008 438 695 586 693 12 × 2 = 1 + 0.000 016 877 391 173 386 24;
  • 44) 0.000 016 877 391 173 386 24 × 2 = 0 + 0.000 033 754 782 346 772 48;
  • 45) 0.000 033 754 782 346 772 48 × 2 = 0 + 0.000 067 509 564 693 544 96;
  • 46) 0.000 067 509 564 693 544 96 × 2 = 0 + 0.000 135 019 129 387 089 92;
  • 47) 0.000 135 019 129 387 089 92 × 2 = 0 + 0.000 270 038 258 774 179 84;
  • 48) 0.000 270 038 258 774 179 84 × 2 = 0 + 0.000 540 076 517 548 359 68;
  • 49) 0.000 540 076 517 548 359 68 × 2 = 0 + 0.001 080 153 035 096 719 36;
  • 50) 0.001 080 153 035 096 719 36 × 2 = 0 + 0.002 160 306 070 193 438 72;
  • 51) 0.002 160 306 070 193 438 72 × 2 = 0 + 0.004 320 612 140 386 877 44;
  • 52) 0.004 320 612 140 386 877 44 × 2 = 0 + 0.008 641 224 280 773 754 88;
  • 53) 0.008 641 224 280 773 754 88 × 2 = 0 + 0.017 282 448 561 547 509 76;

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).


4. 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:


0.120 999 999 999 980 902 53(10) =

0.0001 1110 1111 1001 1101 1011 0010 0010 1101 0000 1110 0000 0000 0(2)

5. Positive number before normalization:

204.120 999 999 999 980 902 53(10) =

1100 1100.0001 1110 1111 1001 1101 1011 0010 0010 1101 0000 1110 0000 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


204.120 999 999 999 980 902 53(10) =

1100 1100.0001 1110 1111 1001 1101 1011 0010 0010 1101 0000 1110 0000 0000 0(2) =

1100 1100.0001 1110 1111 1001 1101 1011 0010 0010 1101 0000 1110 0000 0000 0(2) × 20 =

1.1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1100 0000 0000(2) × 27


7. 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 0 (a positive number)

Exponent (unadjusted): 7

Mantissa (not normalized):
1.1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1100 0000 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

7 + 2(11-1) - 1 =

(7 + 1 023)(10) =

1 030(10)


9. 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;
  • 1 030 ÷ 2 = 515 + 0;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1030(10) =

100 0000 0110(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1100 0000 0000 =

1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1100


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0110

Mantissa (52 bits) =
1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1100


Decimal number 204.120 999 999 999 980 902 53 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0110 - 1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1100

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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