0.000 001 300 900 42 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 001 300 900 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
0.000 001 300 900 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. 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;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 001 300 900 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) 0.000 001 300 900 42 × 2 = 0 + 0.000 002 601 800 84;
  • 2) 0.000 002 601 800 84 × 2 = 0 + 0.000 005 203 601 68;
  • 3) 0.000 005 203 601 68 × 2 = 0 + 0.000 010 407 203 36;
  • 4) 0.000 010 407 203 36 × 2 = 0 + 0.000 020 814 406 72;
  • 5) 0.000 020 814 406 72 × 2 = 0 + 0.000 041 628 813 44;
  • 6) 0.000 041 628 813 44 × 2 = 0 + 0.000 083 257 626 88;
  • 7) 0.000 083 257 626 88 × 2 = 0 + 0.000 166 515 253 76;
  • 8) 0.000 166 515 253 76 × 2 = 0 + 0.000 333 030 507 52;
  • 9) 0.000 333 030 507 52 × 2 = 0 + 0.000 666 061 015 04;
  • 10) 0.000 666 061 015 04 × 2 = 0 + 0.001 332 122 030 08;
  • 11) 0.001 332 122 030 08 × 2 = 0 + 0.002 664 244 060 16;
  • 12) 0.002 664 244 060 16 × 2 = 0 + 0.005 328 488 120 32;
  • 13) 0.005 328 488 120 32 × 2 = 0 + 0.010 656 976 240 64;
  • 14) 0.010 656 976 240 64 × 2 = 0 + 0.021 313 952 481 28;
  • 15) 0.021 313 952 481 28 × 2 = 0 + 0.042 627 904 962 56;
  • 16) 0.042 627 904 962 56 × 2 = 0 + 0.085 255 809 925 12;
  • 17) 0.085 255 809 925 12 × 2 = 0 + 0.170 511 619 850 24;
  • 18) 0.170 511 619 850 24 × 2 = 0 + 0.341 023 239 700 48;
  • 19) 0.341 023 239 700 48 × 2 = 0 + 0.682 046 479 400 96;
  • 20) 0.682 046 479 400 96 × 2 = 1 + 0.364 092 958 801 92;
  • 21) 0.364 092 958 801 92 × 2 = 0 + 0.728 185 917 603 84;
  • 22) 0.728 185 917 603 84 × 2 = 1 + 0.456 371 835 207 68;
  • 23) 0.456 371 835 207 68 × 2 = 0 + 0.912 743 670 415 36;
  • 24) 0.912 743 670 415 36 × 2 = 1 + 0.825 487 340 830 72;
  • 25) 0.825 487 340 830 72 × 2 = 1 + 0.650 974 681 661 44;
  • 26) 0.650 974 681 661 44 × 2 = 1 + 0.301 949 363 322 88;
  • 27) 0.301 949 363 322 88 × 2 = 0 + 0.603 898 726 645 76;
  • 28) 0.603 898 726 645 76 × 2 = 1 + 0.207 797 453 291 52;
  • 29) 0.207 797 453 291 52 × 2 = 0 + 0.415 594 906 583 04;
  • 30) 0.415 594 906 583 04 × 2 = 0 + 0.831 189 813 166 08;
  • 31) 0.831 189 813 166 08 × 2 = 1 + 0.662 379 626 332 16;
  • 32) 0.662 379 626 332 16 × 2 = 1 + 0.324 759 252 664 32;
  • 33) 0.324 759 252 664 32 × 2 = 0 + 0.649 518 505 328 64;
  • 34) 0.649 518 505 328 64 × 2 = 1 + 0.299 037 010 657 28;
  • 35) 0.299 037 010 657 28 × 2 = 0 + 0.598 074 021 314 56;
  • 36) 0.598 074 021 314 56 × 2 = 1 + 0.196 148 042 629 12;
  • 37) 0.196 148 042 629 12 × 2 = 0 + 0.392 296 085 258 24;
  • 38) 0.392 296 085 258 24 × 2 = 0 + 0.784 592 170 516 48;
  • 39) 0.784 592 170 516 48 × 2 = 1 + 0.569 184 341 032 96;
  • 40) 0.569 184 341 032 96 × 2 = 1 + 0.138 368 682 065 92;
  • 41) 0.138 368 682 065 92 × 2 = 0 + 0.276 737 364 131 84;
  • 42) 0.276 737 364 131 84 × 2 = 0 + 0.553 474 728 263 68;
  • 43) 0.553 474 728 263 68 × 2 = 1 + 0.106 949 456 527 36;
  • 44) 0.106 949 456 527 36 × 2 = 0 + 0.213 898 913 054 72;
  • 45) 0.213 898 913 054 72 × 2 = 0 + 0.427 797 826 109 44;
  • 46) 0.427 797 826 109 44 × 2 = 0 + 0.855 595 652 218 88;
  • 47) 0.855 595 652 218 88 × 2 = 1 + 0.711 191 304 437 76;
  • 48) 0.711 191 304 437 76 × 2 = 1 + 0.422 382 608 875 52;
  • 49) 0.422 382 608 875 52 × 2 = 0 + 0.844 765 217 751 04;
  • 50) 0.844 765 217 751 04 × 2 = 1 + 0.689 530 435 502 08;
  • 51) 0.689 530 435 502 08 × 2 = 1 + 0.379 060 871 004 16;
  • 52) 0.379 060 871 004 16 × 2 = 0 + 0.758 121 742 008 32;
  • 53) 0.758 121 742 008 32 × 2 = 1 + 0.516 243 484 016 64;
  • 54) 0.516 243 484 016 64 × 2 = 1 + 0.032 486 968 033 28;
  • 55) 0.032 486 968 033 28 × 2 = 0 + 0.064 973 936 066 56;
  • 56) 0.064 973 936 066 56 × 2 = 0 + 0.129 947 872 133 12;
  • 57) 0.129 947 872 133 12 × 2 = 0 + 0.259 895 744 266 24;
  • 58) 0.259 895 744 266 24 × 2 = 0 + 0.519 791 488 532 48;
  • 59) 0.519 791 488 532 48 × 2 = 1 + 0.039 582 977 064 96;
  • 60) 0.039 582 977 064 96 × 2 = 0 + 0.079 165 954 129 92;
  • 61) 0.079 165 954 129 92 × 2 = 0 + 0.158 331 908 259 84;
  • 62) 0.158 331 908 259 84 × 2 = 0 + 0.316 663 816 519 68;
  • 63) 0.316 663 816 519 68 × 2 = 0 + 0.633 327 633 039 36;
  • 64) 0.633 327 633 039 36 × 2 = 1 + 0.266 655 266 078 72;
  • 65) 0.266 655 266 078 72 × 2 = 0 + 0.533 310 532 157 44;
  • 66) 0.533 310 532 157 44 × 2 = 1 + 0.066 621 064 314 88;
  • 67) 0.066 621 064 314 88 × 2 = 0 + 0.133 242 128 629 76;
  • 68) 0.133 242 128 629 76 × 2 = 0 + 0.266 484 257 259 52;
  • 69) 0.266 484 257 259 52 × 2 = 0 + 0.532 968 514 519 04;
  • 70) 0.532 968 514 519 04 × 2 = 1 + 0.065 937 029 038 08;
  • 71) 0.065 937 029 038 08 × 2 = 0 + 0.131 874 058 076 16;
  • 72) 0.131 874 058 076 16 × 2 = 0 + 0.263 748 116 152 32;

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:


0.000 001 300 900 42(10) =

0.0000 0000 0000 0000 0001 0101 1101 0011 0101 0011 0010 0011 0110 1100 0010 0001 0100 0100(2)

5. Positive number before normalization:

0.000 001 300 900 42(10) =

0.0000 0000 0000 0000 0001 0101 1101 0011 0101 0011 0010 0011 0110 1100 0010 0001 0100 0100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 300 900 42(10) =

0.0000 0000 0000 0000 0001 0101 1101 0011 0101 0011 0010 0011 0110 1100 0010 0001 0100 0100(2) =

0.0000 0000 0000 0000 0001 0101 1101 0011 0101 0011 0010 0011 0110 1100 0010 0001 0100 0100(2) × 20 =

1.0101 1101 0011 0101 0011 0010 0011 0110 1100 0010 0001 0100 0100(2) × 2-20


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): -20

Mantissa (not normalized):
1.0101 1101 0011 0101 0011 0010 0011 0110 1100 0010 0001 0100 0100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(11-1) - 1 =

(-20 + 1 023)(10) =

1 003(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;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1003(10) =

011 1110 1011(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0101 1101 0011 0101 0011 0010 0011 0110 1100 0010 0001 0100 0100 =

0101 1101 0011 0101 0011 0010 0011 0110 1100 0010 0001 0100 0100


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) =
011 1110 1011

Mantissa (52 bits) =
0101 1101 0011 0101 0011 0010 0011 0110 1100 0010 0001 0100 0100


Decimal number 0.000 001 300 900 42 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1011 - 0101 1101 0011 0101 0011 0010 0011 0110 1100 0010 0001 0100 0100

» Convert decimal 0.000 001 300 901 22 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