0.000 000 314 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 314(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
0.000 000 314(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: 0.000 000 314.

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) 0.000 000 314 × 2 = 0 + 0.000 000 628;
  • 2) 0.000 000 628 × 2 = 0 + 0.000 001 256;
  • 3) 0.000 001 256 × 2 = 0 + 0.000 002 512;
  • 4) 0.000 002 512 × 2 = 0 + 0.000 005 024;
  • 5) 0.000 005 024 × 2 = 0 + 0.000 010 048;
  • 6) 0.000 010 048 × 2 = 0 + 0.000 020 096;
  • 7) 0.000 020 096 × 2 = 0 + 0.000 040 192;
  • 8) 0.000 040 192 × 2 = 0 + 0.000 080 384;
  • 9) 0.000 080 384 × 2 = 0 + 0.000 160 768;
  • 10) 0.000 160 768 × 2 = 0 + 0.000 321 536;
  • 11) 0.000 321 536 × 2 = 0 + 0.000 643 072;
  • 12) 0.000 643 072 × 2 = 0 + 0.001 286 144;
  • 13) 0.001 286 144 × 2 = 0 + 0.002 572 288;
  • 14) 0.002 572 288 × 2 = 0 + 0.005 144 576;
  • 15) 0.005 144 576 × 2 = 0 + 0.010 289 152;
  • 16) 0.010 289 152 × 2 = 0 + 0.020 578 304;
  • 17) 0.020 578 304 × 2 = 0 + 0.041 156 608;
  • 18) 0.041 156 608 × 2 = 0 + 0.082 313 216;
  • 19) 0.082 313 216 × 2 = 0 + 0.164 626 432;
  • 20) 0.164 626 432 × 2 = 0 + 0.329 252 864;
  • 21) 0.329 252 864 × 2 = 0 + 0.658 505 728;
  • 22) 0.658 505 728 × 2 = 1 + 0.317 011 456;
  • 23) 0.317 011 456 × 2 = 0 + 0.634 022 912;
  • 24) 0.634 022 912 × 2 = 1 + 0.268 045 824;
  • 25) 0.268 045 824 × 2 = 0 + 0.536 091 648;
  • 26) 0.536 091 648 × 2 = 1 + 0.072 183 296;
  • 27) 0.072 183 296 × 2 = 0 + 0.144 366 592;
  • 28) 0.144 366 592 × 2 = 0 + 0.288 733 184;
  • 29) 0.288 733 184 × 2 = 0 + 0.577 466 368;
  • 30) 0.577 466 368 × 2 = 1 + 0.154 932 736;
  • 31) 0.154 932 736 × 2 = 0 + 0.309 865 472;
  • 32) 0.309 865 472 × 2 = 0 + 0.619 730 944;
  • 33) 0.619 730 944 × 2 = 1 + 0.239 461 888;
  • 34) 0.239 461 888 × 2 = 0 + 0.478 923 776;
  • 35) 0.478 923 776 × 2 = 0 + 0.957 847 552;
  • 36) 0.957 847 552 × 2 = 1 + 0.915 695 104;
  • 37) 0.915 695 104 × 2 = 1 + 0.831 390 208;
  • 38) 0.831 390 208 × 2 = 1 + 0.662 780 416;
  • 39) 0.662 780 416 × 2 = 1 + 0.325 560 832;
  • 40) 0.325 560 832 × 2 = 0 + 0.651 121 664;
  • 41) 0.651 121 664 × 2 = 1 + 0.302 243 328;
  • 42) 0.302 243 328 × 2 = 0 + 0.604 486 656;
  • 43) 0.604 486 656 × 2 = 1 + 0.208 973 312;
  • 44) 0.208 973 312 × 2 = 0 + 0.417 946 624;
  • 45) 0.417 946 624 × 2 = 0 + 0.835 893 248;

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:


0.000 000 314(10) =

0.0000 0000 0000 0000 0000 0101 0100 0100 1001 1110 1010 0(2)

5. Positive number before normalization:

0.000 000 314(10) =

0.0000 0000 0000 0000 0000 0101 0100 0100 1001 1110 1010 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:


0.000 000 314(10) =

0.0000 0000 0000 0000 0000 0101 0100 0100 1001 1110 1010 0(2) =

0.0000 0000 0000 0000 0000 0101 0100 0100 1001 1110 1010 0(2) × 20 =

1.0101 0001 0010 0111 1010 100(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):
1.0101 0001 0010 0111 1010 100


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 1000 1001 0011 1101 0100 =

010 1000 1001 0011 1101 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 1001

Mantissa (23 bits) =
010 1000 1001 0011 1101 0100


Decimal number 0.000 000 314 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 010 1000 1001 0011 1101 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