0.000 000 527 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 527(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
0.000 000 527(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: 0.000 000 527.

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) 0.000 000 527 × 2 = 0 + 0.000 001 054;
  • 2) 0.000 001 054 × 2 = 0 + 0.000 002 108;
  • 3) 0.000 002 108 × 2 = 0 + 0.000 004 216;
  • 4) 0.000 004 216 × 2 = 0 + 0.000 008 432;
  • 5) 0.000 008 432 × 2 = 0 + 0.000 016 864;
  • 6) 0.000 016 864 × 2 = 0 + 0.000 033 728;
  • 7) 0.000 033 728 × 2 = 0 + 0.000 067 456;
  • 8) 0.000 067 456 × 2 = 0 + 0.000 134 912;
  • 9) 0.000 134 912 × 2 = 0 + 0.000 269 824;
  • 10) 0.000 269 824 × 2 = 0 + 0.000 539 648;
  • 11) 0.000 539 648 × 2 = 0 + 0.001 079 296;
  • 12) 0.001 079 296 × 2 = 0 + 0.002 158 592;
  • 13) 0.002 158 592 × 2 = 0 + 0.004 317 184;
  • 14) 0.004 317 184 × 2 = 0 + 0.008 634 368;
  • 15) 0.008 634 368 × 2 = 0 + 0.017 268 736;
  • 16) 0.017 268 736 × 2 = 0 + 0.034 537 472;
  • 17) 0.034 537 472 × 2 = 0 + 0.069 074 944;
  • 18) 0.069 074 944 × 2 = 0 + 0.138 149 888;
  • 19) 0.138 149 888 × 2 = 0 + 0.276 299 776;
  • 20) 0.276 299 776 × 2 = 0 + 0.552 599 552;
  • 21) 0.552 599 552 × 2 = 1 + 0.105 199 104;
  • 22) 0.105 199 104 × 2 = 0 + 0.210 398 208;
  • 23) 0.210 398 208 × 2 = 0 + 0.420 796 416;
  • 24) 0.420 796 416 × 2 = 0 + 0.841 592 832;
  • 25) 0.841 592 832 × 2 = 1 + 0.683 185 664;
  • 26) 0.683 185 664 × 2 = 1 + 0.366 371 328;
  • 27) 0.366 371 328 × 2 = 0 + 0.732 742 656;
  • 28) 0.732 742 656 × 2 = 1 + 0.465 485 312;
  • 29) 0.465 485 312 × 2 = 0 + 0.930 970 624;
  • 30) 0.930 970 624 × 2 = 1 + 0.861 941 248;
  • 31) 0.861 941 248 × 2 = 1 + 0.723 882 496;
  • 32) 0.723 882 496 × 2 = 1 + 0.447 764 992;
  • 33) 0.447 764 992 × 2 = 0 + 0.895 529 984;
  • 34) 0.895 529 984 × 2 = 1 + 0.791 059 968;
  • 35) 0.791 059 968 × 2 = 1 + 0.582 119 936;
  • 36) 0.582 119 936 × 2 = 1 + 0.164 239 872;
  • 37) 0.164 239 872 × 2 = 0 + 0.328 479 744;
  • 38) 0.328 479 744 × 2 = 0 + 0.656 959 488;
  • 39) 0.656 959 488 × 2 = 1 + 0.313 918 976;
  • 40) 0.313 918 976 × 2 = 0 + 0.627 837 952;
  • 41) 0.627 837 952 × 2 = 1 + 0.255 675 904;
  • 42) 0.255 675 904 × 2 = 0 + 0.511 351 808;
  • 43) 0.511 351 808 × 2 = 1 + 0.022 703 616;
  • 44) 0.022 703 616 × 2 = 0 + 0.045 407 232;

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:


0.000 000 527(10) =

0.0000 0000 0000 0000 0000 1000 1101 0111 0111 0010 1010(2)

5. Positive number before normalization:

0.000 000 527(10) =

0.0000 0000 0000 0000 0000 1000 1101 0111 0111 0010 1010(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:


0.000 000 527(10) =

0.0000 0000 0000 0000 0000 1000 1101 0111 0111 0010 1010(2) =

0.0000 0000 0000 0000 0000 1000 1101 0111 0111 0010 1010(2) × 20 =

1.0001 1010 1110 1110 0101 010(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):
1.0001 1010 1110 1110 0101 010


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. 000 1101 0111 0111 0010 1010 =

000 1101 0111 0111 0010 1010


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) =
000 1101 0111 0111 0010 1010


Decimal number 0.000 000 527 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 000 1101 0111 0111 0010 1010

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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