0.000 000 324 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 324(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
0.000 000 324(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: 0.000 000 324.

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) 0.000 000 324 × 2 = 0 + 0.000 000 648;
  • 2) 0.000 000 648 × 2 = 0 + 0.000 001 296;
  • 3) 0.000 001 296 × 2 = 0 + 0.000 002 592;
  • 4) 0.000 002 592 × 2 = 0 + 0.000 005 184;
  • 5) 0.000 005 184 × 2 = 0 + 0.000 010 368;
  • 6) 0.000 010 368 × 2 = 0 + 0.000 020 736;
  • 7) 0.000 020 736 × 2 = 0 + 0.000 041 472;
  • 8) 0.000 041 472 × 2 = 0 + 0.000 082 944;
  • 9) 0.000 082 944 × 2 = 0 + 0.000 165 888;
  • 10) 0.000 165 888 × 2 = 0 + 0.000 331 776;
  • 11) 0.000 331 776 × 2 = 0 + 0.000 663 552;
  • 12) 0.000 663 552 × 2 = 0 + 0.001 327 104;
  • 13) 0.001 327 104 × 2 = 0 + 0.002 654 208;
  • 14) 0.002 654 208 × 2 = 0 + 0.005 308 416;
  • 15) 0.005 308 416 × 2 = 0 + 0.010 616 832;
  • 16) 0.010 616 832 × 2 = 0 + 0.021 233 664;
  • 17) 0.021 233 664 × 2 = 0 + 0.042 467 328;
  • 18) 0.042 467 328 × 2 = 0 + 0.084 934 656;
  • 19) 0.084 934 656 × 2 = 0 + 0.169 869 312;
  • 20) 0.169 869 312 × 2 = 0 + 0.339 738 624;
  • 21) 0.339 738 624 × 2 = 0 + 0.679 477 248;
  • 22) 0.679 477 248 × 2 = 1 + 0.358 954 496;
  • 23) 0.358 954 496 × 2 = 0 + 0.717 908 992;
  • 24) 0.717 908 992 × 2 = 1 + 0.435 817 984;
  • 25) 0.435 817 984 × 2 = 0 + 0.871 635 968;
  • 26) 0.871 635 968 × 2 = 1 + 0.743 271 936;
  • 27) 0.743 271 936 × 2 = 1 + 0.486 543 872;
  • 28) 0.486 543 872 × 2 = 0 + 0.973 087 744;
  • 29) 0.973 087 744 × 2 = 1 + 0.946 175 488;
  • 30) 0.946 175 488 × 2 = 1 + 0.892 350 976;
  • 31) 0.892 350 976 × 2 = 1 + 0.784 701 952;
  • 32) 0.784 701 952 × 2 = 1 + 0.569 403 904;
  • 33) 0.569 403 904 × 2 = 1 + 0.138 807 808;
  • 34) 0.138 807 808 × 2 = 0 + 0.277 615 616;
  • 35) 0.277 615 616 × 2 = 0 + 0.555 231 232;
  • 36) 0.555 231 232 × 2 = 1 + 0.110 462 464;
  • 37) 0.110 462 464 × 2 = 0 + 0.220 924 928;
  • 38) 0.220 924 928 × 2 = 0 + 0.441 849 856;
  • 39) 0.441 849 856 × 2 = 0 + 0.883 699 712;
  • 40) 0.883 699 712 × 2 = 1 + 0.767 399 424;
  • 41) 0.767 399 424 × 2 = 1 + 0.534 798 848;
  • 42) 0.534 798 848 × 2 = 1 + 0.069 597 696;
  • 43) 0.069 597 696 × 2 = 0 + 0.139 195 392;
  • 44) 0.139 195 392 × 2 = 0 + 0.278 390 784;
  • 45) 0.278 390 784 × 2 = 0 + 0.556 781 568;

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:


0.000 000 324(10) =

0.0000 0000 0000 0000 0000 0101 0110 1111 1001 0001 1100 0(2)

5. Positive number before normalization:

0.000 000 324(10) =

0.0000 0000 0000 0000 0000 0101 0110 1111 1001 0001 1100 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 324(10) =

0.0000 0000 0000 0000 0000 0101 0110 1111 1001 0001 1100 0(2) =

0.0000 0000 0000 0000 0000 0101 0110 1111 1001 0001 1100 0(2) × 20 =

1.0101 1011 1110 0100 0111 000(2) × 2-22


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): -22

Mantissa (not normalized):
1.0101 1011 1110 0100 0111 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(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;
  • 105 ÷ 2 = 52 + 1;
  • 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) =

105(10) =

0110 1001(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. 010 1101 1111 0010 0011 1000 =

010 1101 1111 0010 0011 1000


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 1001

Mantissa (23 bits) =
010 1101 1111 0010 0011 1000


Decimal number 0.000 000 324 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 010 1101 1111 0010 0011 1000

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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