0.000 000 132 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 132(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
0.000 000 132(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: 0.000 000 132.

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) 0.000 000 132 × 2 = 0 + 0.000 000 264;
  • 2) 0.000 000 264 × 2 = 0 + 0.000 000 528;
  • 3) 0.000 000 528 × 2 = 0 + 0.000 001 056;
  • 4) 0.000 001 056 × 2 = 0 + 0.000 002 112;
  • 5) 0.000 002 112 × 2 = 0 + 0.000 004 224;
  • 6) 0.000 004 224 × 2 = 0 + 0.000 008 448;
  • 7) 0.000 008 448 × 2 = 0 + 0.000 016 896;
  • 8) 0.000 016 896 × 2 = 0 + 0.000 033 792;
  • 9) 0.000 033 792 × 2 = 0 + 0.000 067 584;
  • 10) 0.000 067 584 × 2 = 0 + 0.000 135 168;
  • 11) 0.000 135 168 × 2 = 0 + 0.000 270 336;
  • 12) 0.000 270 336 × 2 = 0 + 0.000 540 672;
  • 13) 0.000 540 672 × 2 = 0 + 0.001 081 344;
  • 14) 0.001 081 344 × 2 = 0 + 0.002 162 688;
  • 15) 0.002 162 688 × 2 = 0 + 0.004 325 376;
  • 16) 0.004 325 376 × 2 = 0 + 0.008 650 752;
  • 17) 0.008 650 752 × 2 = 0 + 0.017 301 504;
  • 18) 0.017 301 504 × 2 = 0 + 0.034 603 008;
  • 19) 0.034 603 008 × 2 = 0 + 0.069 206 016;
  • 20) 0.069 206 016 × 2 = 0 + 0.138 412 032;
  • 21) 0.138 412 032 × 2 = 0 + 0.276 824 064;
  • 22) 0.276 824 064 × 2 = 0 + 0.553 648 128;
  • 23) 0.553 648 128 × 2 = 1 + 0.107 296 256;
  • 24) 0.107 296 256 × 2 = 0 + 0.214 592 512;
  • 25) 0.214 592 512 × 2 = 0 + 0.429 185 024;
  • 26) 0.429 185 024 × 2 = 0 + 0.858 370 048;
  • 27) 0.858 370 048 × 2 = 1 + 0.716 740 096;
  • 28) 0.716 740 096 × 2 = 1 + 0.433 480 192;
  • 29) 0.433 480 192 × 2 = 0 + 0.866 960 384;
  • 30) 0.866 960 384 × 2 = 1 + 0.733 920 768;
  • 31) 0.733 920 768 × 2 = 1 + 0.467 841 536;
  • 32) 0.467 841 536 × 2 = 0 + 0.935 683 072;
  • 33) 0.935 683 072 × 2 = 1 + 0.871 366 144;
  • 34) 0.871 366 144 × 2 = 1 + 0.742 732 288;
  • 35) 0.742 732 288 × 2 = 1 + 0.485 464 576;
  • 36) 0.485 464 576 × 2 = 0 + 0.970 929 152;
  • 37) 0.970 929 152 × 2 = 1 + 0.941 858 304;
  • 38) 0.941 858 304 × 2 = 1 + 0.883 716 608;
  • 39) 0.883 716 608 × 2 = 1 + 0.767 433 216;
  • 40) 0.767 433 216 × 2 = 1 + 0.534 866 432;
  • 41) 0.534 866 432 × 2 = 1 + 0.069 732 864;
  • 42) 0.069 732 864 × 2 = 0 + 0.139 465 728;
  • 43) 0.139 465 728 × 2 = 0 + 0.278 931 456;
  • 44) 0.278 931 456 × 2 = 0 + 0.557 862 912;
  • 45) 0.557 862 912 × 2 = 1 + 0.115 725 824;
  • 46) 0.115 725 824 × 2 = 0 + 0.231 451 648;

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:


0.000 000 132(10) =

0.0000 0000 0000 0000 0000 0010 0011 0110 1110 1111 1000 10(2)

5. Positive number before normalization:

0.000 000 132(10) =

0.0000 0000 0000 0000 0000 0010 0011 0110 1110 1111 1000 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 132(10) =

0.0000 0000 0000 0000 0000 0010 0011 0110 1110 1111 1000 10(2) =

0.0000 0000 0000 0000 0000 0010 0011 0110 1110 1111 1000 10(2) × 20 =

1.0001 1011 0111 0111 1100 010(2) × 2-23


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): -23

Mantissa (not normalized):
1.0001 1011 0111 0111 1100 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(8-1) - 1 =

(-23 + 127)(10) =

104(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;
  • 104 ÷ 2 = 52 + 0;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

104(10) =

0110 1000(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 1101 1011 1011 1110 0010 =

000 1101 1011 1011 1110 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 1000

Mantissa (23 bits) =
000 1101 1011 1011 1110 0010


Decimal number 0.000 000 132 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 000 1101 1011 1011 1110 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