0.000 000 013 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 013 7(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
0.000 000 013 7(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: 0.000 000 013 7.

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) 0.000 000 013 7 × 2 = 0 + 0.000 000 027 4;
  • 2) 0.000 000 027 4 × 2 = 0 + 0.000 000 054 8;
  • 3) 0.000 000 054 8 × 2 = 0 + 0.000 000 109 6;
  • 4) 0.000 000 109 6 × 2 = 0 + 0.000 000 219 2;
  • 5) 0.000 000 219 2 × 2 = 0 + 0.000 000 438 4;
  • 6) 0.000 000 438 4 × 2 = 0 + 0.000 000 876 8;
  • 7) 0.000 000 876 8 × 2 = 0 + 0.000 001 753 6;
  • 8) 0.000 001 753 6 × 2 = 0 + 0.000 003 507 2;
  • 9) 0.000 003 507 2 × 2 = 0 + 0.000 007 014 4;
  • 10) 0.000 007 014 4 × 2 = 0 + 0.000 014 028 8;
  • 11) 0.000 014 028 8 × 2 = 0 + 0.000 028 057 6;
  • 12) 0.000 028 057 6 × 2 = 0 + 0.000 056 115 2;
  • 13) 0.000 056 115 2 × 2 = 0 + 0.000 112 230 4;
  • 14) 0.000 112 230 4 × 2 = 0 + 0.000 224 460 8;
  • 15) 0.000 224 460 8 × 2 = 0 + 0.000 448 921 6;
  • 16) 0.000 448 921 6 × 2 = 0 + 0.000 897 843 2;
  • 17) 0.000 897 843 2 × 2 = 0 + 0.001 795 686 4;
  • 18) 0.001 795 686 4 × 2 = 0 + 0.003 591 372 8;
  • 19) 0.003 591 372 8 × 2 = 0 + 0.007 182 745 6;
  • 20) 0.007 182 745 6 × 2 = 0 + 0.014 365 491 2;
  • 21) 0.014 365 491 2 × 2 = 0 + 0.028 730 982 4;
  • 22) 0.028 730 982 4 × 2 = 0 + 0.057 461 964 8;
  • 23) 0.057 461 964 8 × 2 = 0 + 0.114 923 929 6;
  • 24) 0.114 923 929 6 × 2 = 0 + 0.229 847 859 2;
  • 25) 0.229 847 859 2 × 2 = 0 + 0.459 695 718 4;
  • 26) 0.459 695 718 4 × 2 = 0 + 0.919 391 436 8;
  • 27) 0.919 391 436 8 × 2 = 1 + 0.838 782 873 6;
  • 28) 0.838 782 873 6 × 2 = 1 + 0.677 565 747 2;
  • 29) 0.677 565 747 2 × 2 = 1 + 0.355 131 494 4;
  • 30) 0.355 131 494 4 × 2 = 0 + 0.710 262 988 8;
  • 31) 0.710 262 988 8 × 2 = 1 + 0.420 525 977 6;
  • 32) 0.420 525 977 6 × 2 = 0 + 0.841 051 955 2;
  • 33) 0.841 051 955 2 × 2 = 1 + 0.682 103 910 4;
  • 34) 0.682 103 910 4 × 2 = 1 + 0.364 207 820 8;
  • 35) 0.364 207 820 8 × 2 = 0 + 0.728 415 641 6;
  • 36) 0.728 415 641 6 × 2 = 1 + 0.456 831 283 2;
  • 37) 0.456 831 283 2 × 2 = 0 + 0.913 662 566 4;
  • 38) 0.913 662 566 4 × 2 = 1 + 0.827 325 132 8;
  • 39) 0.827 325 132 8 × 2 = 1 + 0.654 650 265 6;
  • 40) 0.654 650 265 6 × 2 = 1 + 0.309 300 531 2;
  • 41) 0.309 300 531 2 × 2 = 0 + 0.618 601 062 4;
  • 42) 0.618 601 062 4 × 2 = 1 + 0.237 202 124 8;
  • 43) 0.237 202 124 8 × 2 = 0 + 0.474 404 249 6;
  • 44) 0.474 404 249 6 × 2 = 0 + 0.948 808 499 2;
  • 45) 0.948 808 499 2 × 2 = 1 + 0.897 616 998 4;
  • 46) 0.897 616 998 4 × 2 = 1 + 0.795 233 996 8;
  • 47) 0.795 233 996 8 × 2 = 1 + 0.590 467 993 6;
  • 48) 0.590 467 993 6 × 2 = 1 + 0.180 935 987 2;
  • 49) 0.180 935 987 2 × 2 = 0 + 0.361 871 974 4;
  • 50) 0.361 871 974 4 × 2 = 0 + 0.723 743 948 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 - 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:


0.000 000 013 7(10) =

0.0000 0000 0000 0000 0000 0000 0011 1010 1101 0111 0100 1111 00(2)

5. Positive number before normalization:

0.000 000 013 7(10) =

0.0000 0000 0000 0000 0000 0000 0011 1010 1101 0111 0100 1111 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:


0.000 000 013 7(10) =

0.0000 0000 0000 0000 0000 0000 0011 1010 1101 0111 0100 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0011 1010 1101 0111 0100 1111 00(2) × 20 =

1.1101 0110 1011 1010 0111 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):
1.1101 0110 1011 1010 0111 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. 110 1011 0101 1101 0011 1100 =

110 1011 0101 1101 0011 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) =
110 1011 0101 1101 0011 1100


Decimal number 0.000 000 013 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 110 1011 0101 1101 0011 1100

Share

» Convert decimal 0.000 000 019 1 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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