0.000 000 497 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 497(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
0.000 000 497(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: 0.000 000 497.

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) 0.000 000 497 × 2 = 0 + 0.000 000 994;
  • 2) 0.000 000 994 × 2 = 0 + 0.000 001 988;
  • 3) 0.000 001 988 × 2 = 0 + 0.000 003 976;
  • 4) 0.000 003 976 × 2 = 0 + 0.000 007 952;
  • 5) 0.000 007 952 × 2 = 0 + 0.000 015 904;
  • 6) 0.000 015 904 × 2 = 0 + 0.000 031 808;
  • 7) 0.000 031 808 × 2 = 0 + 0.000 063 616;
  • 8) 0.000 063 616 × 2 = 0 + 0.000 127 232;
  • 9) 0.000 127 232 × 2 = 0 + 0.000 254 464;
  • 10) 0.000 254 464 × 2 = 0 + 0.000 508 928;
  • 11) 0.000 508 928 × 2 = 0 + 0.001 017 856;
  • 12) 0.001 017 856 × 2 = 0 + 0.002 035 712;
  • 13) 0.002 035 712 × 2 = 0 + 0.004 071 424;
  • 14) 0.004 071 424 × 2 = 0 + 0.008 142 848;
  • 15) 0.008 142 848 × 2 = 0 + 0.016 285 696;
  • 16) 0.016 285 696 × 2 = 0 + 0.032 571 392;
  • 17) 0.032 571 392 × 2 = 0 + 0.065 142 784;
  • 18) 0.065 142 784 × 2 = 0 + 0.130 285 568;
  • 19) 0.130 285 568 × 2 = 0 + 0.260 571 136;
  • 20) 0.260 571 136 × 2 = 0 + 0.521 142 272;
  • 21) 0.521 142 272 × 2 = 1 + 0.042 284 544;
  • 22) 0.042 284 544 × 2 = 0 + 0.084 569 088;
  • 23) 0.084 569 088 × 2 = 0 + 0.169 138 176;
  • 24) 0.169 138 176 × 2 = 0 + 0.338 276 352;
  • 25) 0.338 276 352 × 2 = 0 + 0.676 552 704;
  • 26) 0.676 552 704 × 2 = 1 + 0.353 105 408;
  • 27) 0.353 105 408 × 2 = 0 + 0.706 210 816;
  • 28) 0.706 210 816 × 2 = 1 + 0.412 421 632;
  • 29) 0.412 421 632 × 2 = 0 + 0.824 843 264;
  • 30) 0.824 843 264 × 2 = 1 + 0.649 686 528;
  • 31) 0.649 686 528 × 2 = 1 + 0.299 373 056;
  • 32) 0.299 373 056 × 2 = 0 + 0.598 746 112;
  • 33) 0.598 746 112 × 2 = 1 + 0.197 492 224;
  • 34) 0.197 492 224 × 2 = 0 + 0.394 984 448;
  • 35) 0.394 984 448 × 2 = 0 + 0.789 968 896;
  • 36) 0.789 968 896 × 2 = 1 + 0.579 937 792;
  • 37) 0.579 937 792 × 2 = 1 + 0.159 875 584;
  • 38) 0.159 875 584 × 2 = 0 + 0.319 751 168;
  • 39) 0.319 751 168 × 2 = 0 + 0.639 502 336;
  • 40) 0.639 502 336 × 2 = 1 + 0.279 004 672;
  • 41) 0.279 004 672 × 2 = 0 + 0.558 009 344;
  • 42) 0.558 009 344 × 2 = 1 + 0.116 018 688;
  • 43) 0.116 018 688 × 2 = 0 + 0.232 037 376;
  • 44) 0.232 037 376 × 2 = 0 + 0.464 074 752;

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:


0.000 000 497(10) =

0.0000 0000 0000 0000 0000 1000 0101 0110 1001 1001 0100(2)

5. Positive number before normalization:

0.000 000 497(10) =

0.0000 0000 0000 0000 0000 1000 0101 0110 1001 1001 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:


0.000 000 497(10) =

0.0000 0000 0000 0000 0000 1000 0101 0110 1001 1001 0100(2) =

0.0000 0000 0000 0000 0000 1000 0101 0110 1001 1001 0100(2) × 20 =

1.0000 1010 1101 0011 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):
1.0000 1010 1101 0011 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. 000 0101 0110 1001 1001 0100 =

000 0101 0110 1001 1001 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) =
000 0101 0110 1001 1001 0100


Decimal number 0.000 000 497 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 000 0101 0110 1001 1001 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