0.000 000 503 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 503(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
0.000 000 503(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: 0.000 000 503.

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) 0.000 000 503 × 2 = 0 + 0.000 001 006;
  • 2) 0.000 001 006 × 2 = 0 + 0.000 002 012;
  • 3) 0.000 002 012 × 2 = 0 + 0.000 004 024;
  • 4) 0.000 004 024 × 2 = 0 + 0.000 008 048;
  • 5) 0.000 008 048 × 2 = 0 + 0.000 016 096;
  • 6) 0.000 016 096 × 2 = 0 + 0.000 032 192;
  • 7) 0.000 032 192 × 2 = 0 + 0.000 064 384;
  • 8) 0.000 064 384 × 2 = 0 + 0.000 128 768;
  • 9) 0.000 128 768 × 2 = 0 + 0.000 257 536;
  • 10) 0.000 257 536 × 2 = 0 + 0.000 515 072;
  • 11) 0.000 515 072 × 2 = 0 + 0.001 030 144;
  • 12) 0.001 030 144 × 2 = 0 + 0.002 060 288;
  • 13) 0.002 060 288 × 2 = 0 + 0.004 120 576;
  • 14) 0.004 120 576 × 2 = 0 + 0.008 241 152;
  • 15) 0.008 241 152 × 2 = 0 + 0.016 482 304;
  • 16) 0.016 482 304 × 2 = 0 + 0.032 964 608;
  • 17) 0.032 964 608 × 2 = 0 + 0.065 929 216;
  • 18) 0.065 929 216 × 2 = 0 + 0.131 858 432;
  • 19) 0.131 858 432 × 2 = 0 + 0.263 716 864;
  • 20) 0.263 716 864 × 2 = 0 + 0.527 433 728;
  • 21) 0.527 433 728 × 2 = 1 + 0.054 867 456;
  • 22) 0.054 867 456 × 2 = 0 + 0.109 734 912;
  • 23) 0.109 734 912 × 2 = 0 + 0.219 469 824;
  • 24) 0.219 469 824 × 2 = 0 + 0.438 939 648;
  • 25) 0.438 939 648 × 2 = 0 + 0.877 879 296;
  • 26) 0.877 879 296 × 2 = 1 + 0.755 758 592;
  • 27) 0.755 758 592 × 2 = 1 + 0.511 517 184;
  • 28) 0.511 517 184 × 2 = 1 + 0.023 034 368;
  • 29) 0.023 034 368 × 2 = 0 + 0.046 068 736;
  • 30) 0.046 068 736 × 2 = 0 + 0.092 137 472;
  • 31) 0.092 137 472 × 2 = 0 + 0.184 274 944;
  • 32) 0.184 274 944 × 2 = 0 + 0.368 549 888;
  • 33) 0.368 549 888 × 2 = 0 + 0.737 099 776;
  • 34) 0.737 099 776 × 2 = 1 + 0.474 199 552;
  • 35) 0.474 199 552 × 2 = 0 + 0.948 399 104;
  • 36) 0.948 399 104 × 2 = 1 + 0.896 798 208;
  • 37) 0.896 798 208 × 2 = 1 + 0.793 596 416;
  • 38) 0.793 596 416 × 2 = 1 + 0.587 192 832;
  • 39) 0.587 192 832 × 2 = 1 + 0.174 385 664;
  • 40) 0.174 385 664 × 2 = 0 + 0.348 771 328;
  • 41) 0.348 771 328 × 2 = 0 + 0.697 542 656;
  • 42) 0.697 542 656 × 2 = 1 + 0.395 085 312;
  • 43) 0.395 085 312 × 2 = 0 + 0.790 170 624;
  • 44) 0.790 170 624 × 2 = 1 + 0.580 341 248;

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:


0.000 000 503(10) =

0.0000 0000 0000 0000 0000 1000 0111 0000 0101 1110 0101(2)

5. Positive number before normalization:

0.000 000 503(10) =

0.0000 0000 0000 0000 0000 1000 0111 0000 0101 1110 0101(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:


0.000 000 503(10) =

0.0000 0000 0000 0000 0000 1000 0111 0000 0101 1110 0101(2) =

0.0000 0000 0000 0000 0000 1000 0111 0000 0101 1110 0101(2) × 20 =

1.0000 1110 0000 1011 1100 101(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):
1.0000 1110 0000 1011 1100 101


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 0111 0000 0101 1110 0101 =

000 0111 0000 0101 1110 0101


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 0111 0000 0101 1110 0101


Decimal number 0.000 000 503 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 000 0111 0000 0101 1110 0101

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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