0.000 000 316 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 316(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
0.000 000 316(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: 0.000 000 316.

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) 0.000 000 316 × 2 = 0 + 0.000 000 632;
  • 2) 0.000 000 632 × 2 = 0 + 0.000 001 264;
  • 3) 0.000 001 264 × 2 = 0 + 0.000 002 528;
  • 4) 0.000 002 528 × 2 = 0 + 0.000 005 056;
  • 5) 0.000 005 056 × 2 = 0 + 0.000 010 112;
  • 6) 0.000 010 112 × 2 = 0 + 0.000 020 224;
  • 7) 0.000 020 224 × 2 = 0 + 0.000 040 448;
  • 8) 0.000 040 448 × 2 = 0 + 0.000 080 896;
  • 9) 0.000 080 896 × 2 = 0 + 0.000 161 792;
  • 10) 0.000 161 792 × 2 = 0 + 0.000 323 584;
  • 11) 0.000 323 584 × 2 = 0 + 0.000 647 168;
  • 12) 0.000 647 168 × 2 = 0 + 0.001 294 336;
  • 13) 0.001 294 336 × 2 = 0 + 0.002 588 672;
  • 14) 0.002 588 672 × 2 = 0 + 0.005 177 344;
  • 15) 0.005 177 344 × 2 = 0 + 0.010 354 688;
  • 16) 0.010 354 688 × 2 = 0 + 0.020 709 376;
  • 17) 0.020 709 376 × 2 = 0 + 0.041 418 752;
  • 18) 0.041 418 752 × 2 = 0 + 0.082 837 504;
  • 19) 0.082 837 504 × 2 = 0 + 0.165 675 008;
  • 20) 0.165 675 008 × 2 = 0 + 0.331 350 016;
  • 21) 0.331 350 016 × 2 = 0 + 0.662 700 032;
  • 22) 0.662 700 032 × 2 = 1 + 0.325 400 064;
  • 23) 0.325 400 064 × 2 = 0 + 0.650 800 128;
  • 24) 0.650 800 128 × 2 = 1 + 0.301 600 256;
  • 25) 0.301 600 256 × 2 = 0 + 0.603 200 512;
  • 26) 0.603 200 512 × 2 = 1 + 0.206 401 024;
  • 27) 0.206 401 024 × 2 = 0 + 0.412 802 048;
  • 28) 0.412 802 048 × 2 = 0 + 0.825 604 096;
  • 29) 0.825 604 096 × 2 = 1 + 0.651 208 192;
  • 30) 0.651 208 192 × 2 = 1 + 0.302 416 384;
  • 31) 0.302 416 384 × 2 = 0 + 0.604 832 768;
  • 32) 0.604 832 768 × 2 = 1 + 0.209 665 536;
  • 33) 0.209 665 536 × 2 = 0 + 0.419 331 072;
  • 34) 0.419 331 072 × 2 = 0 + 0.838 662 144;
  • 35) 0.838 662 144 × 2 = 1 + 0.677 324 288;
  • 36) 0.677 324 288 × 2 = 1 + 0.354 648 576;
  • 37) 0.354 648 576 × 2 = 0 + 0.709 297 152;
  • 38) 0.709 297 152 × 2 = 1 + 0.418 594 304;
  • 39) 0.418 594 304 × 2 = 0 + 0.837 188 608;
  • 40) 0.837 188 608 × 2 = 1 + 0.674 377 216;
  • 41) 0.674 377 216 × 2 = 1 + 0.348 754 432;
  • 42) 0.348 754 432 × 2 = 0 + 0.697 508 864;
  • 43) 0.697 508 864 × 2 = 1 + 0.395 017 728;
  • 44) 0.395 017 728 × 2 = 0 + 0.790 035 456;
  • 45) 0.790 035 456 × 2 = 1 + 0.580 070 912;

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:


0.000 000 316(10) =

0.0000 0000 0000 0000 0000 0101 0100 1101 0011 0101 1010 1(2)

5. Positive number before normalization:

0.000 000 316(10) =

0.0000 0000 0000 0000 0000 0101 0100 1101 0011 0101 1010 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:


0.000 000 316(10) =

0.0000 0000 0000 0000 0000 0101 0100 1101 0011 0101 1010 1(2) =

0.0000 0000 0000 0000 0000 0101 0100 1101 0011 0101 1010 1(2) × 20 =

1.0101 0011 0100 1101 0110 101(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 0011 0100 1101 0110 101


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 1001 1010 0110 1011 0101 =

010 1001 1010 0110 1011 0101


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 1001 1010 0110 1011 0101


Decimal number 0.000 000 316 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 010 1001 1010 0110 1011 0101

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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