0.000 000 031 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 031(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
0.000 000 031(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: 0.000 000 031.

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) 0.000 000 031 × 2 = 0 + 0.000 000 062;
  • 2) 0.000 000 062 × 2 = 0 + 0.000 000 124;
  • 3) 0.000 000 124 × 2 = 0 + 0.000 000 248;
  • 4) 0.000 000 248 × 2 = 0 + 0.000 000 496;
  • 5) 0.000 000 496 × 2 = 0 + 0.000 000 992;
  • 6) 0.000 000 992 × 2 = 0 + 0.000 001 984;
  • 7) 0.000 001 984 × 2 = 0 + 0.000 003 968;
  • 8) 0.000 003 968 × 2 = 0 + 0.000 007 936;
  • 9) 0.000 007 936 × 2 = 0 + 0.000 015 872;
  • 10) 0.000 015 872 × 2 = 0 + 0.000 031 744;
  • 11) 0.000 031 744 × 2 = 0 + 0.000 063 488;
  • 12) 0.000 063 488 × 2 = 0 + 0.000 126 976;
  • 13) 0.000 126 976 × 2 = 0 + 0.000 253 952;
  • 14) 0.000 253 952 × 2 = 0 + 0.000 507 904;
  • 15) 0.000 507 904 × 2 = 0 + 0.001 015 808;
  • 16) 0.001 015 808 × 2 = 0 + 0.002 031 616;
  • 17) 0.002 031 616 × 2 = 0 + 0.004 063 232;
  • 18) 0.004 063 232 × 2 = 0 + 0.008 126 464;
  • 19) 0.008 126 464 × 2 = 0 + 0.016 252 928;
  • 20) 0.016 252 928 × 2 = 0 + 0.032 505 856;
  • 21) 0.032 505 856 × 2 = 0 + 0.065 011 712;
  • 22) 0.065 011 712 × 2 = 0 + 0.130 023 424;
  • 23) 0.130 023 424 × 2 = 0 + 0.260 046 848;
  • 24) 0.260 046 848 × 2 = 0 + 0.520 093 696;
  • 25) 0.520 093 696 × 2 = 1 + 0.040 187 392;
  • 26) 0.040 187 392 × 2 = 0 + 0.080 374 784;
  • 27) 0.080 374 784 × 2 = 0 + 0.160 749 568;
  • 28) 0.160 749 568 × 2 = 0 + 0.321 499 136;
  • 29) 0.321 499 136 × 2 = 0 + 0.642 998 272;
  • 30) 0.642 998 272 × 2 = 1 + 0.285 996 544;
  • 31) 0.285 996 544 × 2 = 0 + 0.571 993 088;
  • 32) 0.571 993 088 × 2 = 1 + 0.143 986 176;
  • 33) 0.143 986 176 × 2 = 0 + 0.287 972 352;
  • 34) 0.287 972 352 × 2 = 0 + 0.575 944 704;
  • 35) 0.575 944 704 × 2 = 1 + 0.151 889 408;
  • 36) 0.151 889 408 × 2 = 0 + 0.303 778 816;
  • 37) 0.303 778 816 × 2 = 0 + 0.607 557 632;
  • 38) 0.607 557 632 × 2 = 1 + 0.215 115 264;
  • 39) 0.215 115 264 × 2 = 0 + 0.430 230 528;
  • 40) 0.430 230 528 × 2 = 0 + 0.860 461 056;
  • 41) 0.860 461 056 × 2 = 1 + 0.720 922 112;
  • 42) 0.720 922 112 × 2 = 1 + 0.441 844 224;
  • 43) 0.441 844 224 × 2 = 0 + 0.883 688 448;
  • 44) 0.883 688 448 × 2 = 1 + 0.767 376 896;
  • 45) 0.767 376 896 × 2 = 1 + 0.534 753 792;
  • 46) 0.534 753 792 × 2 = 1 + 0.069 507 584;
  • 47) 0.069 507 584 × 2 = 0 + 0.139 015 168;
  • 48) 0.139 015 168 × 2 = 0 + 0.278 030 336;

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:


0.000 000 031(10) =

0.0000 0000 0000 0000 0000 0000 1000 0101 0010 0100 1101 1100(2)

5. Positive number before normalization:

0.000 000 031(10) =

0.0000 0000 0000 0000 0000 0000 1000 0101 0010 0100 1101 1100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 031(10) =

0.0000 0000 0000 0000 0000 0000 1000 0101 0010 0100 1101 1100(2) =

0.0000 0000 0000 0000 0000 0000 1000 0101 0010 0100 1101 1100(2) × 20 =

1.0000 1010 0100 1001 1011 100(2) × 2-25


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): -25

Mantissa (not normalized):
1.0000 1010 0100 1001 1011 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-25 + 2(8-1) - 1 =

(-25 + 127)(10) =

102(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;
  • 102 ÷ 2 = 51 + 0;
  • 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) =

102(10) =

0110 0110(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. 000 0101 0010 0100 1101 1100 =

000 0101 0010 0100 1101 1100


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 0110

Mantissa (23 bits) =
000 0101 0010 0100 1101 1100


Decimal number 0.000 000 031 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0110 - 000 0101 0010 0100 1101 1100

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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