0.000 000 307 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 307(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
0.000 000 307(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: 0.000 000 307.

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) 0.000 000 307 × 2 = 0 + 0.000 000 614;
  • 2) 0.000 000 614 × 2 = 0 + 0.000 001 228;
  • 3) 0.000 001 228 × 2 = 0 + 0.000 002 456;
  • 4) 0.000 002 456 × 2 = 0 + 0.000 004 912;
  • 5) 0.000 004 912 × 2 = 0 + 0.000 009 824;
  • 6) 0.000 009 824 × 2 = 0 + 0.000 019 648;
  • 7) 0.000 019 648 × 2 = 0 + 0.000 039 296;
  • 8) 0.000 039 296 × 2 = 0 + 0.000 078 592;
  • 9) 0.000 078 592 × 2 = 0 + 0.000 157 184;
  • 10) 0.000 157 184 × 2 = 0 + 0.000 314 368;
  • 11) 0.000 314 368 × 2 = 0 + 0.000 628 736;
  • 12) 0.000 628 736 × 2 = 0 + 0.001 257 472;
  • 13) 0.001 257 472 × 2 = 0 + 0.002 514 944;
  • 14) 0.002 514 944 × 2 = 0 + 0.005 029 888;
  • 15) 0.005 029 888 × 2 = 0 + 0.010 059 776;
  • 16) 0.010 059 776 × 2 = 0 + 0.020 119 552;
  • 17) 0.020 119 552 × 2 = 0 + 0.040 239 104;
  • 18) 0.040 239 104 × 2 = 0 + 0.080 478 208;
  • 19) 0.080 478 208 × 2 = 0 + 0.160 956 416;
  • 20) 0.160 956 416 × 2 = 0 + 0.321 912 832;
  • 21) 0.321 912 832 × 2 = 0 + 0.643 825 664;
  • 22) 0.643 825 664 × 2 = 1 + 0.287 651 328;
  • 23) 0.287 651 328 × 2 = 0 + 0.575 302 656;
  • 24) 0.575 302 656 × 2 = 1 + 0.150 605 312;
  • 25) 0.150 605 312 × 2 = 0 + 0.301 210 624;
  • 26) 0.301 210 624 × 2 = 0 + 0.602 421 248;
  • 27) 0.602 421 248 × 2 = 1 + 0.204 842 496;
  • 28) 0.204 842 496 × 2 = 0 + 0.409 684 992;
  • 29) 0.409 684 992 × 2 = 0 + 0.819 369 984;
  • 30) 0.819 369 984 × 2 = 1 + 0.638 739 968;
  • 31) 0.638 739 968 × 2 = 1 + 0.277 479 936;
  • 32) 0.277 479 936 × 2 = 0 + 0.554 959 872;
  • 33) 0.554 959 872 × 2 = 1 + 0.109 919 744;
  • 34) 0.109 919 744 × 2 = 0 + 0.219 839 488;
  • 35) 0.219 839 488 × 2 = 0 + 0.439 678 976;
  • 36) 0.439 678 976 × 2 = 0 + 0.879 357 952;
  • 37) 0.879 357 952 × 2 = 1 + 0.758 715 904;
  • 38) 0.758 715 904 × 2 = 1 + 0.517 431 808;
  • 39) 0.517 431 808 × 2 = 1 + 0.034 863 616;
  • 40) 0.034 863 616 × 2 = 0 + 0.069 727 232;
  • 41) 0.069 727 232 × 2 = 0 + 0.139 454 464;
  • 42) 0.139 454 464 × 2 = 0 + 0.278 908 928;
  • 43) 0.278 908 928 × 2 = 0 + 0.557 817 856;
  • 44) 0.557 817 856 × 2 = 1 + 0.115 635 712;
  • 45) 0.115 635 712 × 2 = 0 + 0.231 271 424;

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:


0.000 000 307(10) =

0.0000 0000 0000 0000 0000 0101 0010 0110 1000 1110 0001 0(2)

5. Positive number before normalization:

0.000 000 307(10) =

0.0000 0000 0000 0000 0000 0101 0010 0110 1000 1110 0001 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:


0.000 000 307(10) =

0.0000 0000 0000 0000 0000 0101 0010 0110 1000 1110 0001 0(2) =

0.0000 0000 0000 0000 0000 0101 0010 0110 1000 1110 0001 0(2) × 20 =

1.0100 1001 1010 0011 1000 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):
1.0100 1001 1010 0011 1000 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. 010 0100 1101 0001 1100 0010 =

010 0100 1101 0001 1100 0010


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) =
010 0100 1101 0001 1100 0010


Decimal number 0.000 000 307 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 010 0100 1101 0001 1100 0010

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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