0.000 000 110 027 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 027 5(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 110 027 5(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 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: 0.000 000 110 027 5.

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) 0.000 000 110 027 5 × 2 = 0 + 0.000 000 220 055;
  • 2) 0.000 000 220 055 × 2 = 0 + 0.000 000 440 11;
  • 3) 0.000 000 440 11 × 2 = 0 + 0.000 000 880 22;
  • 4) 0.000 000 880 22 × 2 = 0 + 0.000 001 760 44;
  • 5) 0.000 001 760 44 × 2 = 0 + 0.000 003 520 88;
  • 6) 0.000 003 520 88 × 2 = 0 + 0.000 007 041 76;
  • 7) 0.000 007 041 76 × 2 = 0 + 0.000 014 083 52;
  • 8) 0.000 014 083 52 × 2 = 0 + 0.000 028 167 04;
  • 9) 0.000 028 167 04 × 2 = 0 + 0.000 056 334 08;
  • 10) 0.000 056 334 08 × 2 = 0 + 0.000 112 668 16;
  • 11) 0.000 112 668 16 × 2 = 0 + 0.000 225 336 32;
  • 12) 0.000 225 336 32 × 2 = 0 + 0.000 450 672 64;
  • 13) 0.000 450 672 64 × 2 = 0 + 0.000 901 345 28;
  • 14) 0.000 901 345 28 × 2 = 0 + 0.001 802 690 56;
  • 15) 0.001 802 690 56 × 2 = 0 + 0.003 605 381 12;
  • 16) 0.003 605 381 12 × 2 = 0 + 0.007 210 762 24;
  • 17) 0.007 210 762 24 × 2 = 0 + 0.014 421 524 48;
  • 18) 0.014 421 524 48 × 2 = 0 + 0.028 843 048 96;
  • 19) 0.028 843 048 96 × 2 = 0 + 0.057 686 097 92;
  • 20) 0.057 686 097 92 × 2 = 0 + 0.115 372 195 84;
  • 21) 0.115 372 195 84 × 2 = 0 + 0.230 744 391 68;
  • 22) 0.230 744 391 68 × 2 = 0 + 0.461 488 783 36;
  • 23) 0.461 488 783 36 × 2 = 0 + 0.922 977 566 72;
  • 24) 0.922 977 566 72 × 2 = 1 + 0.845 955 133 44;
  • 25) 0.845 955 133 44 × 2 = 1 + 0.691 910 266 88;
  • 26) 0.691 910 266 88 × 2 = 1 + 0.383 820 533 76;
  • 27) 0.383 820 533 76 × 2 = 0 + 0.767 641 067 52;
  • 28) 0.767 641 067 52 × 2 = 1 + 0.535 282 135 04;
  • 29) 0.535 282 135 04 × 2 = 1 + 0.070 564 270 08;
  • 30) 0.070 564 270 08 × 2 = 0 + 0.141 128 540 16;
  • 31) 0.141 128 540 16 × 2 = 0 + 0.282 257 080 32;
  • 32) 0.282 257 080 32 × 2 = 0 + 0.564 514 160 64;
  • 33) 0.564 514 160 64 × 2 = 1 + 0.129 028 321 28;
  • 34) 0.129 028 321 28 × 2 = 0 + 0.258 056 642 56;
  • 35) 0.258 056 642 56 × 2 = 0 + 0.516 113 285 12;
  • 36) 0.516 113 285 12 × 2 = 1 + 0.032 226 570 24;
  • 37) 0.032 226 570 24 × 2 = 0 + 0.064 453 140 48;
  • 38) 0.064 453 140 48 × 2 = 0 + 0.128 906 280 96;
  • 39) 0.128 906 280 96 × 2 = 0 + 0.257 812 561 92;
  • 40) 0.257 812 561 92 × 2 = 0 + 0.515 625 123 84;
  • 41) 0.515 625 123 84 × 2 = 1 + 0.031 250 247 68;
  • 42) 0.031 250 247 68 × 2 = 0 + 0.062 500 495 36;
  • 43) 0.062 500 495 36 × 2 = 0 + 0.125 000 990 72;
  • 44) 0.125 000 990 72 × 2 = 0 + 0.250 001 981 44;
  • 45) 0.250 001 981 44 × 2 = 0 + 0.500 003 962 88;
  • 46) 0.500 003 962 88 × 2 = 1 + 0.000 007 925 76;
  • 47) 0.000 007 925 76 × 2 = 0 + 0.000 015 851 52;

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:


0.000 000 110 027 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1001 0000 1000 010(2)

5. Positive number before normalization:

0.000 000 110 027 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1001 0000 1000 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 110 027 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1001 0000 1000 010(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1001 0000 1000 010(2) × 20 =

1.1101 1000 1001 0000 1000 010(2) × 2-24


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): -24

Mantissa (not normalized):
1.1101 1000 1001 0000 1000 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

103(10) =

0110 0111(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. 110 1100 0100 1000 0100 0010 =

110 1100 0100 1000 0100 0010


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 0111

Mantissa (23 bits) =
110 1100 0100 1000 0100 0010


Decimal number 0.000 000 110 027 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0100 1000 0100 0010

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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