Convert 0.000 000 001 4 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

0.000 000 001 4(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.000 000 001 4.

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) 0.000 000 001 4 × 2 = 0 + 0.000 000 002 8;
  • 2) 0.000 000 002 8 × 2 = 0 + 0.000 000 005 6;
  • 3) 0.000 000 005 6 × 2 = 0 + 0.000 000 011 2;
  • 4) 0.000 000 011 2 × 2 = 0 + 0.000 000 022 4;
  • 5) 0.000 000 022 4 × 2 = 0 + 0.000 000 044 8;
  • 6) 0.000 000 044 8 × 2 = 0 + 0.000 000 089 6;
  • 7) 0.000 000 089 6 × 2 = 0 + 0.000 000 179 2;
  • 8) 0.000 000 179 2 × 2 = 0 + 0.000 000 358 4;
  • 9) 0.000 000 358 4 × 2 = 0 + 0.000 000 716 8;
  • 10) 0.000 000 716 8 × 2 = 0 + 0.000 001 433 6;
  • 11) 0.000 001 433 6 × 2 = 0 + 0.000 002 867 2;
  • 12) 0.000 002 867 2 × 2 = 0 + 0.000 005 734 4;
  • 13) 0.000 005 734 4 × 2 = 0 + 0.000 011 468 8;
  • 14) 0.000 011 468 8 × 2 = 0 + 0.000 022 937 6;
  • 15) 0.000 022 937 6 × 2 = 0 + 0.000 045 875 2;
  • 16) 0.000 045 875 2 × 2 = 0 + 0.000 091 750 4;
  • 17) 0.000 091 750 4 × 2 = 0 + 0.000 183 500 8;
  • 18) 0.000 183 500 8 × 2 = 0 + 0.000 367 001 6;
  • 19) 0.000 367 001 6 × 2 = 0 + 0.000 734 003 2;
  • 20) 0.000 734 003 2 × 2 = 0 + 0.001 468 006 4;
  • 21) 0.001 468 006 4 × 2 = 0 + 0.002 936 012 8;
  • 22) 0.002 936 012 8 × 2 = 0 + 0.005 872 025 6;
  • 23) 0.005 872 025 6 × 2 = 0 + 0.011 744 051 2;
  • 24) 0.011 744 051 2 × 2 = 0 + 0.023 488 102 4;
  • 25) 0.023 488 102 4 × 2 = 0 + 0.046 976 204 8;
  • 26) 0.046 976 204 8 × 2 = 0 + 0.093 952 409 6;
  • 27) 0.093 952 409 6 × 2 = 0 + 0.187 904 819 2;
  • 28) 0.187 904 819 2 × 2 = 0 + 0.375 809 638 4;
  • 29) 0.375 809 638 4 × 2 = 0 + 0.751 619 276 8;
  • 30) 0.751 619 276 8 × 2 = 1 + 0.503 238 553 6;
  • 31) 0.503 238 553 6 × 2 = 1 + 0.006 477 107 2;
  • 32) 0.006 477 107 2 × 2 = 0 + 0.012 954 214 4;
  • 33) 0.012 954 214 4 × 2 = 0 + 0.025 908 428 8;
  • 34) 0.025 908 428 8 × 2 = 0 + 0.051 816 857 6;
  • 35) 0.051 816 857 6 × 2 = 0 + 0.103 633 715 2;
  • 36) 0.103 633 715 2 × 2 = 0 + 0.207 267 430 4;
  • 37) 0.207 267 430 4 × 2 = 0 + 0.414 534 860 8;
  • 38) 0.414 534 860 8 × 2 = 0 + 0.829 069 721 6;
  • 39) 0.829 069 721 6 × 2 = 1 + 0.658 139 443 2;
  • 40) 0.658 139 443 2 × 2 = 1 + 0.316 278 886 4;
  • 41) 0.316 278 886 4 × 2 = 0 + 0.632 557 772 8;
  • 42) 0.632 557 772 8 × 2 = 1 + 0.265 115 545 6;
  • 43) 0.265 115 545 6 × 2 = 0 + 0.530 231 091 2;
  • 44) 0.530 231 091 2 × 2 = 1 + 0.060 462 182 4;
  • 45) 0.060 462 182 4 × 2 = 0 + 0.120 924 364 8;
  • 46) 0.120 924 364 8 × 2 = 0 + 0.241 848 729 6;
  • 47) 0.241 848 729 6 × 2 = 0 + 0.483 697 459 2;
  • 48) 0.483 697 459 2 × 2 = 0 + 0.967 394 918 4;
  • 49) 0.967 394 918 4 × 2 = 1 + 0.934 789 836 8;
  • 50) 0.934 789 836 8 × 2 = 1 + 0.869 579 673 6;
  • 51) 0.869 579 673 6 × 2 = 1 + 0.739 159 347 2;
  • 52) 0.739 159 347 2 × 2 = 1 + 0.478 318 694 4;
  • 53) 0.478 318 694 4 × 2 = 0 + 0.956 637 388 8;

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:

0.000 000 001 4(10) =


0.0000 0000 0000 0000 0000 0000 0000 0110 0000 0011 0101 0000 1111 0(2)


5. Positive number before normalization:

0.000 000 001 4(10) =


0.0000 0000 0000 0000 0000 0000 0000 0110 0000 0011 0101 0000 1111 0(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 30 positions to the right so that only one non zero digit remains to the left of it:

0.000 000 001 4(10) =


0.0000 0000 0000 0000 0000 0000 0000 0110 0000 0011 0101 0000 1111 0(2) =


0.0000 0000 0000 0000 0000 0000 0000 0110 0000 0011 0101 0000 1111 0(2) × 20 =


1.1000 0000 1101 0100 0011 110(2) × 2-30


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign: 0 (a positive number)


Exponent (unadjusted): -30


Mantissa (not normalized):
1.1000 0000 1101 0100 0011 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


-30 + 2(8-1) - 1 =


(-30 + 127)(10) =


97(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:

  • division = quotient + remainder;
  • 97 ÷ 2 = 48 + 1;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =


97(10) =


0110 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 23 bits, only if necessary (not the case here).

Mantissa (normalized) =


1. 100 0000 0110 1010 0001 1110 =


100 0000 0110 1010 0001 1110


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

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


Exponent (8 bits) =
0110 0001


Mantissa (23 bits) =
100 0000 0110 1010 0001 1110


Number 0.000 000 001 4 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 0110 0001 - 100 0000 0110 1010 0001 1110

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 1

      23
  • Mantissa (23 bits):

    • 1

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 0

      10
    • 1

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 1

      2
    • 1

      1
    • 0

      0

More operations of this kind:

0.000 000 001 3 = ? ... 0.000 000 001 5 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =


    1 - 1000 0011 - 100 1010 1100 0110 1010 0111