Convert 7.501 742 3 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

7.501 742 3(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (base 2) the integer part: 7.
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;
  • 7 ÷ 2 = 3 + 1;
  • 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.

7(10) =


111(2)


3. Convert to the binary (base 2) the fractional part: 0.501 742 3.

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) 0.501 742 3 × 2 = 1 + 0.003 484 6;
  • 2) 0.003 484 6 × 2 = 0 + 0.006 969 2;
  • 3) 0.006 969 2 × 2 = 0 + 0.013 938 4;
  • 4) 0.013 938 4 × 2 = 0 + 0.027 876 8;
  • 5) 0.027 876 8 × 2 = 0 + 0.055 753 6;
  • 6) 0.055 753 6 × 2 = 0 + 0.111 507 2;
  • 7) 0.111 507 2 × 2 = 0 + 0.223 014 4;
  • 8) 0.223 014 4 × 2 = 0 + 0.446 028 8;
  • 9) 0.446 028 8 × 2 = 0 + 0.892 057 6;
  • 10) 0.892 057 6 × 2 = 1 + 0.784 115 2;
  • 11) 0.784 115 2 × 2 = 1 + 0.568 230 4;
  • 12) 0.568 230 4 × 2 = 1 + 0.136 460 8;
  • 13) 0.136 460 8 × 2 = 0 + 0.272 921 6;
  • 14) 0.272 921 6 × 2 = 0 + 0.545 843 2;
  • 15) 0.545 843 2 × 2 = 1 + 0.091 686 4;
  • 16) 0.091 686 4 × 2 = 0 + 0.183 372 8;
  • 17) 0.183 372 8 × 2 = 0 + 0.366 745 6;
  • 18) 0.366 745 6 × 2 = 0 + 0.733 491 2;
  • 19) 0.733 491 2 × 2 = 1 + 0.466 982 4;
  • 20) 0.466 982 4 × 2 = 0 + 0.933 964 8;
  • 21) 0.933 964 8 × 2 = 1 + 0.867 929 6;
  • 22) 0.867 929 6 × 2 = 1 + 0.735 859 2;
  • 23) 0.735 859 2 × 2 = 1 + 0.471 718 4;
  • 24) 0.471 718 4 × 2 = 0 + 0.943 436 8;
  • 25) 0.943 436 8 × 2 = 1 + 0.886 873 6;
  • 26) 0.886 873 6 × 2 = 1 + 0.773 747 2;
  • 27) 0.773 747 2 × 2 = 1 + 0.547 494 4;
  • 28) 0.547 494 4 × 2 = 1 + 0.094 988 8;
  • 29) 0.094 988 8 × 2 = 0 + 0.189 977 6;
  • 30) 0.189 977 6 × 2 = 0 + 0.379 955 2;
  • 31) 0.379 955 2 × 2 = 0 + 0.759 910 4;
  • 32) 0.759 910 4 × 2 = 1 + 0.519 820 8;
  • 33) 0.519 820 8 × 2 = 1 + 0.039 641 6;
  • 34) 0.039 641 6 × 2 = 0 + 0.079 283 2;
  • 35) 0.079 283 2 × 2 = 0 + 0.158 566 4;
  • 36) 0.158 566 4 × 2 = 0 + 0.317 132 8;
  • 37) 0.317 132 8 × 2 = 0 + 0.634 265 6;
  • 38) 0.634 265 6 × 2 = 1 + 0.268 531 2;
  • 39) 0.268 531 2 × 2 = 0 + 0.537 062 4;
  • 40) 0.537 062 4 × 2 = 1 + 0.074 124 8;
  • 41) 0.074 124 8 × 2 = 0 + 0.148 249 6;
  • 42) 0.148 249 6 × 2 = 0 + 0.296 499 2;
  • 43) 0.296 499 2 × 2 = 0 + 0.592 998 4;
  • 44) 0.592 998 4 × 2 = 1 + 0.185 996 8;
  • 45) 0.185 996 8 × 2 = 0 + 0.371 993 6;
  • 46) 0.371 993 6 × 2 = 0 + 0.743 987 2;
  • 47) 0.743 987 2 × 2 = 1 + 0.487 974 4;
  • 48) 0.487 974 4 × 2 = 0 + 0.975 948 8;
  • 49) 0.975 948 8 × 2 = 1 + 0.951 897 6;
  • 50) 0.951 897 6 × 2 = 1 + 0.903 795 2;
  • 51) 0.903 795 2 × 2 = 1 + 0.807 590 4;
  • 52) 0.807 590 4 × 2 = 1 + 0.615 180 8;
  • 53) 0.615 180 8 × 2 = 1 + 0.230 361 6;

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...)


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:

0.501 742 3(10) =


0.1000 0000 0111 0010 0010 1110 1111 0001 1000 0101 0001 0010 1111 1(2)


5. Positive number before normalization:

7.501 742 3(10) =


111.1000 0000 0111 0010 0010 1110 1111 0001 1000 0101 0001 0010 1111 1(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left so that only one non zero digit remains to the left of it:

7.501 742 3(10) =


111.1000 0000 0111 0010 0010 1110 1111 0001 1000 0101 0001 0010 1111 1(2) =


111.1000 0000 0111 0010 0010 1110 1111 0001 1000 0101 0001 0010 1111 1(2) × 20 =


1.1110 0000 0001 1100 1000 1011 1011 1100 0110 0001 0100 0100 1011 111(2) × 22


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): 2


Mantissa (not normalized):
1.1110 0000 0001 1100 1000 1011 1011 1100 0110 0001 0100 0100 1011 111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


2 + 2(11-1) - 1 =


(2 + 1 023)(10) =


1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 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) =


1025(10) =


100 0000 0001(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. 1110 0000 0001 1100 1000 1011 1011 1100 0110 0001 0100 0100 1011 111 =


1110 0000 0001 1100 1000 1011 1011 1100 0110 0001 0100 0100 1011


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 0001


Mantissa (52 bits) =
1110 0000 0001 1100 1000 1011 1011 1100 0110 0001 0100 0100 1011


Number 7.501 742 3 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0001 - 1110 0000 0001 1100 1000 1011 1011 1100 0110 0001 0100 0100 1011

(64 bits IEEE 754)

More operations of this kind:

7.501 742 2 = ? ... 7.501 742 4 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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