0.000 000 287 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 287(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
0.000 000 287(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: 0.000 000 287.

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) 0.000 000 287 × 2 = 0 + 0.000 000 574;
  • 2) 0.000 000 574 × 2 = 0 + 0.000 001 148;
  • 3) 0.000 001 148 × 2 = 0 + 0.000 002 296;
  • 4) 0.000 002 296 × 2 = 0 + 0.000 004 592;
  • 5) 0.000 004 592 × 2 = 0 + 0.000 009 184;
  • 6) 0.000 009 184 × 2 = 0 + 0.000 018 368;
  • 7) 0.000 018 368 × 2 = 0 + 0.000 036 736;
  • 8) 0.000 036 736 × 2 = 0 + 0.000 073 472;
  • 9) 0.000 073 472 × 2 = 0 + 0.000 146 944;
  • 10) 0.000 146 944 × 2 = 0 + 0.000 293 888;
  • 11) 0.000 293 888 × 2 = 0 + 0.000 587 776;
  • 12) 0.000 587 776 × 2 = 0 + 0.001 175 552;
  • 13) 0.001 175 552 × 2 = 0 + 0.002 351 104;
  • 14) 0.002 351 104 × 2 = 0 + 0.004 702 208;
  • 15) 0.004 702 208 × 2 = 0 + 0.009 404 416;
  • 16) 0.009 404 416 × 2 = 0 + 0.018 808 832;
  • 17) 0.018 808 832 × 2 = 0 + 0.037 617 664;
  • 18) 0.037 617 664 × 2 = 0 + 0.075 235 328;
  • 19) 0.075 235 328 × 2 = 0 + 0.150 470 656;
  • 20) 0.150 470 656 × 2 = 0 + 0.300 941 312;
  • 21) 0.300 941 312 × 2 = 0 + 0.601 882 624;
  • 22) 0.601 882 624 × 2 = 1 + 0.203 765 248;
  • 23) 0.203 765 248 × 2 = 0 + 0.407 530 496;
  • 24) 0.407 530 496 × 2 = 0 + 0.815 060 992;
  • 25) 0.815 060 992 × 2 = 1 + 0.630 121 984;
  • 26) 0.630 121 984 × 2 = 1 + 0.260 243 968;
  • 27) 0.260 243 968 × 2 = 0 + 0.520 487 936;
  • 28) 0.520 487 936 × 2 = 1 + 0.040 975 872;
  • 29) 0.040 975 872 × 2 = 0 + 0.081 951 744;
  • 30) 0.081 951 744 × 2 = 0 + 0.163 903 488;
  • 31) 0.163 903 488 × 2 = 0 + 0.327 806 976;
  • 32) 0.327 806 976 × 2 = 0 + 0.655 613 952;
  • 33) 0.655 613 952 × 2 = 1 + 0.311 227 904;
  • 34) 0.311 227 904 × 2 = 0 + 0.622 455 808;
  • 35) 0.622 455 808 × 2 = 1 + 0.244 911 616;
  • 36) 0.244 911 616 × 2 = 0 + 0.489 823 232;
  • 37) 0.489 823 232 × 2 = 0 + 0.979 646 464;
  • 38) 0.979 646 464 × 2 = 1 + 0.959 292 928;
  • 39) 0.959 292 928 × 2 = 1 + 0.918 585 856;
  • 40) 0.918 585 856 × 2 = 1 + 0.837 171 712;
  • 41) 0.837 171 712 × 2 = 1 + 0.674 343 424;
  • 42) 0.674 343 424 × 2 = 1 + 0.348 686 848;
  • 43) 0.348 686 848 × 2 = 0 + 0.697 373 696;
  • 44) 0.697 373 696 × 2 = 1 + 0.394 747 392;
  • 45) 0.394 747 392 × 2 = 0 + 0.789 494 784;

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:


0.000 000 287(10) =

0.0000 0000 0000 0000 0000 0100 1101 0000 1010 0111 1101 0(2)

5. Positive number before normalization:

0.000 000 287(10) =

0.0000 0000 0000 0000 0000 0100 1101 0000 1010 0111 1101 0(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:


0.000 000 287(10) =

0.0000 0000 0000 0000 0000 0100 1101 0000 1010 0111 1101 0(2) =

0.0000 0000 0000 0000 0000 0100 1101 0000 1010 0111 1101 0(2) × 20 =

1.0011 0100 0010 1001 1111 010(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):
1.0011 0100 0010 1001 1111 010


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 1010 0001 0100 1111 1010 =

001 1010 0001 0100 1111 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 1001

Mantissa (23 bits) =
001 1010 0001 0100 1111 1010


Decimal number 0.000 000 287 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 001 1010 0001 0100 1111 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