0.000 000 014 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 014 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
0.000 000 014 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: 0.000 000 014 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) 0.000 000 014 5 × 2 = 0 + 0.000 000 029;
  • 2) 0.000 000 029 × 2 = 0 + 0.000 000 058;
  • 3) 0.000 000 058 × 2 = 0 + 0.000 000 116;
  • 4) 0.000 000 116 × 2 = 0 + 0.000 000 232;
  • 5) 0.000 000 232 × 2 = 0 + 0.000 000 464;
  • 6) 0.000 000 464 × 2 = 0 + 0.000 000 928;
  • 7) 0.000 000 928 × 2 = 0 + 0.000 001 856;
  • 8) 0.000 001 856 × 2 = 0 + 0.000 003 712;
  • 9) 0.000 003 712 × 2 = 0 + 0.000 007 424;
  • 10) 0.000 007 424 × 2 = 0 + 0.000 014 848;
  • 11) 0.000 014 848 × 2 = 0 + 0.000 029 696;
  • 12) 0.000 029 696 × 2 = 0 + 0.000 059 392;
  • 13) 0.000 059 392 × 2 = 0 + 0.000 118 784;
  • 14) 0.000 118 784 × 2 = 0 + 0.000 237 568;
  • 15) 0.000 237 568 × 2 = 0 + 0.000 475 136;
  • 16) 0.000 475 136 × 2 = 0 + 0.000 950 272;
  • 17) 0.000 950 272 × 2 = 0 + 0.001 900 544;
  • 18) 0.001 900 544 × 2 = 0 + 0.003 801 088;
  • 19) 0.003 801 088 × 2 = 0 + 0.007 602 176;
  • 20) 0.007 602 176 × 2 = 0 + 0.015 204 352;
  • 21) 0.015 204 352 × 2 = 0 + 0.030 408 704;
  • 22) 0.030 408 704 × 2 = 0 + 0.060 817 408;
  • 23) 0.060 817 408 × 2 = 0 + 0.121 634 816;
  • 24) 0.121 634 816 × 2 = 0 + 0.243 269 632;
  • 25) 0.243 269 632 × 2 = 0 + 0.486 539 264;
  • 26) 0.486 539 264 × 2 = 0 + 0.973 078 528;
  • 27) 0.973 078 528 × 2 = 1 + 0.946 157 056;
  • 28) 0.946 157 056 × 2 = 1 + 0.892 314 112;
  • 29) 0.892 314 112 × 2 = 1 + 0.784 628 224;
  • 30) 0.784 628 224 × 2 = 1 + 0.569 256 448;
  • 31) 0.569 256 448 × 2 = 1 + 0.138 512 896;
  • 32) 0.138 512 896 × 2 = 0 + 0.277 025 792;
  • 33) 0.277 025 792 × 2 = 0 + 0.554 051 584;
  • 34) 0.554 051 584 × 2 = 1 + 0.108 103 168;
  • 35) 0.108 103 168 × 2 = 0 + 0.216 206 336;
  • 36) 0.216 206 336 × 2 = 0 + 0.432 412 672;
  • 37) 0.432 412 672 × 2 = 0 + 0.864 825 344;
  • 38) 0.864 825 344 × 2 = 1 + 0.729 650 688;
  • 39) 0.729 650 688 × 2 = 1 + 0.459 301 376;
  • 40) 0.459 301 376 × 2 = 0 + 0.918 602 752;
  • 41) 0.918 602 752 × 2 = 1 + 0.837 205 504;
  • 42) 0.837 205 504 × 2 = 1 + 0.674 411 008;
  • 43) 0.674 411 008 × 2 = 1 + 0.348 822 016;
  • 44) 0.348 822 016 × 2 = 0 + 0.697 644 032;
  • 45) 0.697 644 032 × 2 = 1 + 0.395 288 064;
  • 46) 0.395 288 064 × 2 = 0 + 0.790 576 128;
  • 47) 0.790 576 128 × 2 = 1 + 0.581 152 256;
  • 48) 0.581 152 256 × 2 = 1 + 0.162 304 512;
  • 49) 0.162 304 512 × 2 = 0 + 0.324 609 024;
  • 50) 0.324 609 024 × 2 = 0 + 0.649 218 048;

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:


0.000 000 014 5(10) =

0.0000 0000 0000 0000 0000 0000 0011 1110 0100 0110 1110 1011 00(2)

5. Positive number before normalization:

0.000 000 014 5(10) =

0.0000 0000 0000 0000 0000 0000 0011 1110 0100 0110 1110 1011 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 014 5(10) =

0.0000 0000 0000 0000 0000 0000 0011 1110 0100 0110 1110 1011 00(2) =

0.0000 0000 0000 0000 0000 0000 0011 1110 0100 0110 1110 1011 00(2) × 20 =

1.1111 0010 0011 0111 0101 100(2) × 2-27


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): -27

Mantissa (not normalized):
1.1111 0010 0011 0111 0101 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-27 + 2(8-1) - 1 =

(-27 + 127)(10) =

100(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;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 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) =

100(10) =

0110 0100(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. 111 1001 0001 1011 1010 1100 =

111 1001 0001 1011 1010 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 0100

Mantissa (23 bits) =
111 1001 0001 1011 1010 1100


Decimal number 0.000 000 014 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 111 1001 0001 1011 1010 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