0.000 000 538 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 538(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
0.000 000 538(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: 0.000 000 538.

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) 0.000 000 538 × 2 = 0 + 0.000 001 076;
  • 2) 0.000 001 076 × 2 = 0 + 0.000 002 152;
  • 3) 0.000 002 152 × 2 = 0 + 0.000 004 304;
  • 4) 0.000 004 304 × 2 = 0 + 0.000 008 608;
  • 5) 0.000 008 608 × 2 = 0 + 0.000 017 216;
  • 6) 0.000 017 216 × 2 = 0 + 0.000 034 432;
  • 7) 0.000 034 432 × 2 = 0 + 0.000 068 864;
  • 8) 0.000 068 864 × 2 = 0 + 0.000 137 728;
  • 9) 0.000 137 728 × 2 = 0 + 0.000 275 456;
  • 10) 0.000 275 456 × 2 = 0 + 0.000 550 912;
  • 11) 0.000 550 912 × 2 = 0 + 0.001 101 824;
  • 12) 0.001 101 824 × 2 = 0 + 0.002 203 648;
  • 13) 0.002 203 648 × 2 = 0 + 0.004 407 296;
  • 14) 0.004 407 296 × 2 = 0 + 0.008 814 592;
  • 15) 0.008 814 592 × 2 = 0 + 0.017 629 184;
  • 16) 0.017 629 184 × 2 = 0 + 0.035 258 368;
  • 17) 0.035 258 368 × 2 = 0 + 0.070 516 736;
  • 18) 0.070 516 736 × 2 = 0 + 0.141 033 472;
  • 19) 0.141 033 472 × 2 = 0 + 0.282 066 944;
  • 20) 0.282 066 944 × 2 = 0 + 0.564 133 888;
  • 21) 0.564 133 888 × 2 = 1 + 0.128 267 776;
  • 22) 0.128 267 776 × 2 = 0 + 0.256 535 552;
  • 23) 0.256 535 552 × 2 = 0 + 0.513 071 104;
  • 24) 0.513 071 104 × 2 = 1 + 0.026 142 208;
  • 25) 0.026 142 208 × 2 = 0 + 0.052 284 416;
  • 26) 0.052 284 416 × 2 = 0 + 0.104 568 832;
  • 27) 0.104 568 832 × 2 = 0 + 0.209 137 664;
  • 28) 0.209 137 664 × 2 = 0 + 0.418 275 328;
  • 29) 0.418 275 328 × 2 = 0 + 0.836 550 656;
  • 30) 0.836 550 656 × 2 = 1 + 0.673 101 312;
  • 31) 0.673 101 312 × 2 = 1 + 0.346 202 624;
  • 32) 0.346 202 624 × 2 = 0 + 0.692 405 248;
  • 33) 0.692 405 248 × 2 = 1 + 0.384 810 496;
  • 34) 0.384 810 496 × 2 = 0 + 0.769 620 992;
  • 35) 0.769 620 992 × 2 = 1 + 0.539 241 984;
  • 36) 0.539 241 984 × 2 = 1 + 0.078 483 968;
  • 37) 0.078 483 968 × 2 = 0 + 0.156 967 936;
  • 38) 0.156 967 936 × 2 = 0 + 0.313 935 872;
  • 39) 0.313 935 872 × 2 = 0 + 0.627 871 744;
  • 40) 0.627 871 744 × 2 = 1 + 0.255 743 488;
  • 41) 0.255 743 488 × 2 = 0 + 0.511 486 976;
  • 42) 0.511 486 976 × 2 = 1 + 0.022 973 952;
  • 43) 0.022 973 952 × 2 = 0 + 0.045 947 904;
  • 44) 0.045 947 904 × 2 = 0 + 0.091 895 808;

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:


0.000 000 538(10) =

0.0000 0000 0000 0000 0000 1001 0000 0110 1011 0001 0100(2)

5. Positive number before normalization:

0.000 000 538(10) =

0.0000 0000 0000 0000 0000 1001 0000 0110 1011 0001 0100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 538(10) =

0.0000 0000 0000 0000 0000 1001 0000 0110 1011 0001 0100(2) =

0.0000 0000 0000 0000 0000 1001 0000 0110 1011 0001 0100(2) × 20 =

1.0010 0000 1101 0110 0010 100(2) × 2-21


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): -21

Mantissa (not normalized):
1.0010 0000 1101 0110 0010 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 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) =

106(10) =

0110 1010(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. 001 0000 0110 1011 0001 0100 =

001 0000 0110 1011 0001 0100


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 1010

Mantissa (23 bits) =
001 0000 0110 1011 0001 0100


Decimal number 0.000 000 538 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 001 0000 0110 1011 0001 0100

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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