Convert 7.501 742 2 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

7.501 742 2(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 2.

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 2 × 2 = 1 + 0.003 484 4;
  • 2) 0.003 484 4 × 2 = 0 + 0.006 968 8;
  • 3) 0.006 968 8 × 2 = 0 + 0.013 937 6;
  • 4) 0.013 937 6 × 2 = 0 + 0.027 875 2;
  • 5) 0.027 875 2 × 2 = 0 + 0.055 750 4;
  • 6) 0.055 750 4 × 2 = 0 + 0.111 500 8;
  • 7) 0.111 500 8 × 2 = 0 + 0.223 001 6;
  • 8) 0.223 001 6 × 2 = 0 + 0.446 003 2;
  • 9) 0.446 003 2 × 2 = 0 + 0.892 006 4;
  • 10) 0.892 006 4 × 2 = 1 + 0.784 012 8;
  • 11) 0.784 012 8 × 2 = 1 + 0.568 025 6;
  • 12) 0.568 025 6 × 2 = 1 + 0.136 051 2;
  • 13) 0.136 051 2 × 2 = 0 + 0.272 102 4;
  • 14) 0.272 102 4 × 2 = 0 + 0.544 204 8;
  • 15) 0.544 204 8 × 2 = 1 + 0.088 409 6;
  • 16) 0.088 409 6 × 2 = 0 + 0.176 819 2;
  • 17) 0.176 819 2 × 2 = 0 + 0.353 638 4;
  • 18) 0.353 638 4 × 2 = 0 + 0.707 276 8;
  • 19) 0.707 276 8 × 2 = 1 + 0.414 553 6;
  • 20) 0.414 553 6 × 2 = 0 + 0.829 107 2;
  • 21) 0.829 107 2 × 2 = 1 + 0.658 214 4;
  • 22) 0.658 214 4 × 2 = 1 + 0.316 428 8;
  • 23) 0.316 428 8 × 2 = 0 + 0.632 857 6;
  • 24) 0.632 857 6 × 2 = 1 + 0.265 715 2;
  • 25) 0.265 715 2 × 2 = 0 + 0.531 430 4;
  • 26) 0.531 430 4 × 2 = 1 + 0.062 860 8;
  • 27) 0.062 860 8 × 2 = 0 + 0.125 721 6;
  • 28) 0.125 721 6 × 2 = 0 + 0.251 443 2;
  • 29) 0.251 443 2 × 2 = 0 + 0.502 886 4;
  • 30) 0.502 886 4 × 2 = 1 + 0.005 772 8;
  • 31) 0.005 772 8 × 2 = 0 + 0.011 545 6;
  • 32) 0.011 545 6 × 2 = 0 + 0.023 091 2;
  • 33) 0.023 091 2 × 2 = 0 + 0.046 182 4;
  • 34) 0.046 182 4 × 2 = 0 + 0.092 364 8;
  • 35) 0.092 364 8 × 2 = 0 + 0.184 729 6;
  • 36) 0.184 729 6 × 2 = 0 + 0.369 459 2;
  • 37) 0.369 459 2 × 2 = 0 + 0.738 918 4;
  • 38) 0.738 918 4 × 2 = 1 + 0.477 836 8;
  • 39) 0.477 836 8 × 2 = 0 + 0.955 673 6;
  • 40) 0.955 673 6 × 2 = 1 + 0.911 347 2;
  • 41) 0.911 347 2 × 2 = 1 + 0.822 694 4;
  • 42) 0.822 694 4 × 2 = 1 + 0.645 388 8;
  • 43) 0.645 388 8 × 2 = 1 + 0.290 777 6;
  • 44) 0.290 777 6 × 2 = 0 + 0.581 555 2;
  • 45) 0.581 555 2 × 2 = 1 + 0.163 110 4;
  • 46) 0.163 110 4 × 2 = 0 + 0.326 220 8;
  • 47) 0.326 220 8 × 2 = 0 + 0.652 441 6;
  • 48) 0.652 441 6 × 2 = 1 + 0.304 883 2;
  • 49) 0.304 883 2 × 2 = 0 + 0.609 766 4;
  • 50) 0.609 766 4 × 2 = 1 + 0.219 532 8;
  • 51) 0.219 532 8 × 2 = 0 + 0.439 065 6;
  • 52) 0.439 065 6 × 2 = 0 + 0.878 131 2;
  • 53) 0.878 131 2 × 2 = 1 + 0.756 262 4;

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 2(10) =


0.1000 0000 0111 0010 0010 1101 0100 0100 0000 0101 1110 1001 0100 1(2)


5. Positive number before normalization:

7.501 742 2(10) =


111.1000 0000 0111 0010 0010 1101 0100 0100 0000 0101 1110 1001 0100 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 2(10) =


111.1000 0000 0111 0010 0010 1101 0100 0100 0000 0101 1110 1001 0100 1(2) =


111.1000 0000 0111 0010 0010 1101 0100 0100 0000 0101 1110 1001 0100 1(2) × 20 =


1.1110 0000 0001 1100 1000 1011 0101 0001 0000 0001 0111 1010 0101 001(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 0101 0001 0000 0001 0111 1010 0101 001


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 0101 0001 0000 0001 0111 1010 0101 001 =


1110 0000 0001 1100 1000 1011 0101 0001 0000 0001 0111 1010 0101


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 0101 0001 0000 0001 0111 1010 0101


Number 7.501 742 2 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 0101 0001 0000 0001 0111 1010 0101

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 0

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 1

      50
    • 1

      49
    • 0

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 1

      40
    • 1

      39
    • 1

      38
    • 0

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 0

      33
    • 0

      32
    • 1

      31
    • 0

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 1

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 1

      20
    • 0

      19
    • 0

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 0

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 1

      7
    • 0

      6
    • 1

      5
    • 0

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 1

      0

More operations of this kind:

7.501 742 1 = ? ... 7.501 742 3 = ?


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)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

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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