0.000 000 279 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 279(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
0.000 000 279(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: 0.000 000 279.

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) 0.000 000 279 × 2 = 0 + 0.000 000 558;
  • 2) 0.000 000 558 × 2 = 0 + 0.000 001 116;
  • 3) 0.000 001 116 × 2 = 0 + 0.000 002 232;
  • 4) 0.000 002 232 × 2 = 0 + 0.000 004 464;
  • 5) 0.000 004 464 × 2 = 0 + 0.000 008 928;
  • 6) 0.000 008 928 × 2 = 0 + 0.000 017 856;
  • 7) 0.000 017 856 × 2 = 0 + 0.000 035 712;
  • 8) 0.000 035 712 × 2 = 0 + 0.000 071 424;
  • 9) 0.000 071 424 × 2 = 0 + 0.000 142 848;
  • 10) 0.000 142 848 × 2 = 0 + 0.000 285 696;
  • 11) 0.000 285 696 × 2 = 0 + 0.000 571 392;
  • 12) 0.000 571 392 × 2 = 0 + 0.001 142 784;
  • 13) 0.001 142 784 × 2 = 0 + 0.002 285 568;
  • 14) 0.002 285 568 × 2 = 0 + 0.004 571 136;
  • 15) 0.004 571 136 × 2 = 0 + 0.009 142 272;
  • 16) 0.009 142 272 × 2 = 0 + 0.018 284 544;
  • 17) 0.018 284 544 × 2 = 0 + 0.036 569 088;
  • 18) 0.036 569 088 × 2 = 0 + 0.073 138 176;
  • 19) 0.073 138 176 × 2 = 0 + 0.146 276 352;
  • 20) 0.146 276 352 × 2 = 0 + 0.292 552 704;
  • 21) 0.292 552 704 × 2 = 0 + 0.585 105 408;
  • 22) 0.585 105 408 × 2 = 1 + 0.170 210 816;
  • 23) 0.170 210 816 × 2 = 0 + 0.340 421 632;
  • 24) 0.340 421 632 × 2 = 0 + 0.680 843 264;
  • 25) 0.680 843 264 × 2 = 1 + 0.361 686 528;
  • 26) 0.361 686 528 × 2 = 0 + 0.723 373 056;
  • 27) 0.723 373 056 × 2 = 1 + 0.446 746 112;
  • 28) 0.446 746 112 × 2 = 0 + 0.893 492 224;
  • 29) 0.893 492 224 × 2 = 1 + 0.786 984 448;
  • 30) 0.786 984 448 × 2 = 1 + 0.573 968 896;
  • 31) 0.573 968 896 × 2 = 1 + 0.147 937 792;
  • 32) 0.147 937 792 × 2 = 0 + 0.295 875 584;
  • 33) 0.295 875 584 × 2 = 0 + 0.591 751 168;
  • 34) 0.591 751 168 × 2 = 1 + 0.183 502 336;
  • 35) 0.183 502 336 × 2 = 0 + 0.367 004 672;
  • 36) 0.367 004 672 × 2 = 0 + 0.734 009 344;
  • 37) 0.734 009 344 × 2 = 1 + 0.468 018 688;
  • 38) 0.468 018 688 × 2 = 0 + 0.936 037 376;
  • 39) 0.936 037 376 × 2 = 1 + 0.872 074 752;
  • 40) 0.872 074 752 × 2 = 1 + 0.744 149 504;
  • 41) 0.744 149 504 × 2 = 1 + 0.488 299 008;
  • 42) 0.488 299 008 × 2 = 0 + 0.976 598 016;
  • 43) 0.976 598 016 × 2 = 1 + 0.953 196 032;
  • 44) 0.953 196 032 × 2 = 1 + 0.906 392 064;
  • 45) 0.906 392 064 × 2 = 1 + 0.812 784 128;

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:


0.000 000 279(10) =

0.0000 0000 0000 0000 0000 0100 1010 1110 0100 1011 1011 1(2)

5. Positive number before normalization:

0.000 000 279(10) =

0.0000 0000 0000 0000 0000 0100 1010 1110 0100 1011 1011 1(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:


0.000 000 279(10) =

0.0000 0000 0000 0000 0000 0100 1010 1110 0100 1011 1011 1(2) =

0.0000 0000 0000 0000 0000 0100 1010 1110 0100 1011 1011 1(2) × 20 =

1.0010 1011 1001 0010 1110 111(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):
1.0010 1011 1001 0010 1110 111


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. 001 0101 1100 1001 0111 0111 =

001 0101 1100 1001 0111 0111


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) =
001 0101 1100 1001 0111 0111


Decimal number 0.000 000 279 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 001 0101 1100 1001 0111 0111

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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