0.000 000 007 13 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 007 13(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
0.000 000 007 13(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: 0.000 000 007 13.

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) 0.000 000 007 13 × 2 = 0 + 0.000 000 014 26;
  • 2) 0.000 000 014 26 × 2 = 0 + 0.000 000 028 52;
  • 3) 0.000 000 028 52 × 2 = 0 + 0.000 000 057 04;
  • 4) 0.000 000 057 04 × 2 = 0 + 0.000 000 114 08;
  • 5) 0.000 000 114 08 × 2 = 0 + 0.000 000 228 16;
  • 6) 0.000 000 228 16 × 2 = 0 + 0.000 000 456 32;
  • 7) 0.000 000 456 32 × 2 = 0 + 0.000 000 912 64;
  • 8) 0.000 000 912 64 × 2 = 0 + 0.000 001 825 28;
  • 9) 0.000 001 825 28 × 2 = 0 + 0.000 003 650 56;
  • 10) 0.000 003 650 56 × 2 = 0 + 0.000 007 301 12;
  • 11) 0.000 007 301 12 × 2 = 0 + 0.000 014 602 24;
  • 12) 0.000 014 602 24 × 2 = 0 + 0.000 029 204 48;
  • 13) 0.000 029 204 48 × 2 = 0 + 0.000 058 408 96;
  • 14) 0.000 058 408 96 × 2 = 0 + 0.000 116 817 92;
  • 15) 0.000 116 817 92 × 2 = 0 + 0.000 233 635 84;
  • 16) 0.000 233 635 84 × 2 = 0 + 0.000 467 271 68;
  • 17) 0.000 467 271 68 × 2 = 0 + 0.000 934 543 36;
  • 18) 0.000 934 543 36 × 2 = 0 + 0.001 869 086 72;
  • 19) 0.001 869 086 72 × 2 = 0 + 0.003 738 173 44;
  • 20) 0.003 738 173 44 × 2 = 0 + 0.007 476 346 88;
  • 21) 0.007 476 346 88 × 2 = 0 + 0.014 952 693 76;
  • 22) 0.014 952 693 76 × 2 = 0 + 0.029 905 387 52;
  • 23) 0.029 905 387 52 × 2 = 0 + 0.059 810 775 04;
  • 24) 0.059 810 775 04 × 2 = 0 + 0.119 621 550 08;
  • 25) 0.119 621 550 08 × 2 = 0 + 0.239 243 100 16;
  • 26) 0.239 243 100 16 × 2 = 0 + 0.478 486 200 32;
  • 27) 0.478 486 200 32 × 2 = 0 + 0.956 972 400 64;
  • 28) 0.956 972 400 64 × 2 = 1 + 0.913 944 801 28;
  • 29) 0.913 944 801 28 × 2 = 1 + 0.827 889 602 56;
  • 30) 0.827 889 602 56 × 2 = 1 + 0.655 779 205 12;
  • 31) 0.655 779 205 12 × 2 = 1 + 0.311 558 410 24;
  • 32) 0.311 558 410 24 × 2 = 0 + 0.623 116 820 48;
  • 33) 0.623 116 820 48 × 2 = 1 + 0.246 233 640 96;
  • 34) 0.246 233 640 96 × 2 = 0 + 0.492 467 281 92;
  • 35) 0.492 467 281 92 × 2 = 0 + 0.984 934 563 84;
  • 36) 0.984 934 563 84 × 2 = 1 + 0.969 869 127 68;
  • 37) 0.969 869 127 68 × 2 = 1 + 0.939 738 255 36;
  • 38) 0.939 738 255 36 × 2 = 1 + 0.879 476 510 72;
  • 39) 0.879 476 510 72 × 2 = 1 + 0.758 953 021 44;
  • 40) 0.758 953 021 44 × 2 = 1 + 0.517 906 042 88;
  • 41) 0.517 906 042 88 × 2 = 1 + 0.035 812 085 76;
  • 42) 0.035 812 085 76 × 2 = 0 + 0.071 624 171 52;
  • 43) 0.071 624 171 52 × 2 = 0 + 0.143 248 343 04;
  • 44) 0.143 248 343 04 × 2 = 0 + 0.286 496 686 08;
  • 45) 0.286 496 686 08 × 2 = 0 + 0.572 993 372 16;
  • 46) 0.572 993 372 16 × 2 = 1 + 0.145 986 744 32;
  • 47) 0.145 986 744 32 × 2 = 0 + 0.291 973 488 64;
  • 48) 0.291 973 488 64 × 2 = 0 + 0.583 946 977 28;
  • 49) 0.583 946 977 28 × 2 = 1 + 0.167 893 954 56;
  • 50) 0.167 893 954 56 × 2 = 0 + 0.335 787 909 12;
  • 51) 0.335 787 909 12 × 2 = 0 + 0.671 575 818 24;
  • 52) 0.671 575 818 24 × 2 = 1 + 0.343 151 636 48;
  • 53) 0.343 151 636 48 × 2 = 0 + 0.686 303 272 96;
  • 54) 0.686 303 272 96 × 2 = 1 + 0.372 606 545 92;
  • 55) 0.372 606 545 92 × 2 = 0 + 0.745 213 091 84;
  • 56) 0.745 213 091 84 × 2 = 1 + 0.490 426 183 68;
  • 57) 0.490 426 183 68 × 2 = 0 + 0.980 852 367 36;
  • 58) 0.980 852 367 36 × 2 = 1 + 0.961 704 734 72;
  • 59) 0.961 704 734 72 × 2 = 1 + 0.923 409 469 44;
  • 60) 0.923 409 469 44 × 2 = 1 + 0.846 818 938 88;
  • 61) 0.846 818 938 88 × 2 = 1 + 0.693 637 877 76;
  • 62) 0.693 637 877 76 × 2 = 1 + 0.387 275 755 52;
  • 63) 0.387 275 755 52 × 2 = 0 + 0.774 551 511 04;
  • 64) 0.774 551 511 04 × 2 = 1 + 0.549 103 022 08;
  • 65) 0.549 103 022 08 × 2 = 1 + 0.098 206 044 16;
  • 66) 0.098 206 044 16 × 2 = 0 + 0.196 412 088 32;
  • 67) 0.196 412 088 32 × 2 = 0 + 0.392 824 176 64;
  • 68) 0.392 824 176 64 × 2 = 0 + 0.785 648 353 28;
  • 69) 0.785 648 353 28 × 2 = 1 + 0.571 296 706 56;
  • 70) 0.571 296 706 56 × 2 = 1 + 0.142 593 413 12;
  • 71) 0.142 593 413 12 × 2 = 0 + 0.285 186 826 24;
  • 72) 0.285 186 826 24 × 2 = 0 + 0.570 373 652 48;
  • 73) 0.570 373 652 48 × 2 = 1 + 0.140 747 304 96;
  • 74) 0.140 747 304 96 × 2 = 0 + 0.281 494 609 92;
  • 75) 0.281 494 609 92 × 2 = 0 + 0.562 989 219 84;
  • 76) 0.562 989 219 84 × 2 = 1 + 0.125 978 439 68;
  • 77) 0.125 978 439 68 × 2 = 0 + 0.251 956 879 36;
  • 78) 0.251 956 879 36 × 2 = 0 + 0.503 913 758 72;
  • 79) 0.503 913 758 72 × 2 = 1 + 0.007 827 517 44;
  • 80) 0.007 827 517 44 × 2 = 0 + 0.015 655 034 88;

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:


0.000 000 007 13(10) =

0.0000 0000 0000 0000 0000 0000 0001 1110 1001 1111 1000 0100 1001 0101 0111 1101 1000 1100 1001 0010(2)

5. Positive number before normalization:

0.000 000 007 13(10) =

0.0000 0000 0000 0000 0000 0000 0001 1110 1001 1111 1000 0100 1001 0101 0111 1101 1000 1100 1001 0010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 007 13(10) =

0.0000 0000 0000 0000 0000 0000 0001 1110 1001 1111 1000 0100 1001 0101 0111 1101 1000 1100 1001 0010(2) =

0.0000 0000 0000 0000 0000 0000 0001 1110 1001 1111 1000 0100 1001 0101 0111 1101 1000 1100 1001 0010(2) × 20 =

1.1110 1001 1111 1000 0100 1001 0101 0111 1101 1000 1100 1001 0010(2) × 2-28


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): -28

Mantissa (not normalized):
1.1110 1001 1111 1000 0100 1001 0101 0111 1101 1000 1100 1001 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(11-1) - 1 =

(-28 + 1 023)(10) =

995(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;
  • 995 ÷ 2 = 497 + 1;
  • 497 ÷ 2 = 248 + 1;
  • 248 ÷ 2 = 124 + 0;
  • 124 ÷ 2 = 62 + 0;
  • 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) =

995(10) =

011 1110 0011(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. 1110 1001 1111 1000 0100 1001 0101 0111 1101 1000 1100 1001 0010 =

1110 1001 1111 1000 0100 1001 0101 0111 1101 1000 1100 1001 0010


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 0011

Mantissa (52 bits) =
1110 1001 1111 1000 0100 1001 0101 0111 1101 1000 1100 1001 0010


Decimal number 0.000 000 007 13 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0011 - 1110 1001 1111 1000 0100 1001 0101 0111 1101 1000 1100 1001 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