0.000 000 011 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 011(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
0.000 000 011(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: 0.000 000 011.

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) 0.000 000 011 × 2 = 0 + 0.000 000 022;
  • 2) 0.000 000 022 × 2 = 0 + 0.000 000 044;
  • 3) 0.000 000 044 × 2 = 0 + 0.000 000 088;
  • 4) 0.000 000 088 × 2 = 0 + 0.000 000 176;
  • 5) 0.000 000 176 × 2 = 0 + 0.000 000 352;
  • 6) 0.000 000 352 × 2 = 0 + 0.000 000 704;
  • 7) 0.000 000 704 × 2 = 0 + 0.000 001 408;
  • 8) 0.000 001 408 × 2 = 0 + 0.000 002 816;
  • 9) 0.000 002 816 × 2 = 0 + 0.000 005 632;
  • 10) 0.000 005 632 × 2 = 0 + 0.000 011 264;
  • 11) 0.000 011 264 × 2 = 0 + 0.000 022 528;
  • 12) 0.000 022 528 × 2 = 0 + 0.000 045 056;
  • 13) 0.000 045 056 × 2 = 0 + 0.000 090 112;
  • 14) 0.000 090 112 × 2 = 0 + 0.000 180 224;
  • 15) 0.000 180 224 × 2 = 0 + 0.000 360 448;
  • 16) 0.000 360 448 × 2 = 0 + 0.000 720 896;
  • 17) 0.000 720 896 × 2 = 0 + 0.001 441 792;
  • 18) 0.001 441 792 × 2 = 0 + 0.002 883 584;
  • 19) 0.002 883 584 × 2 = 0 + 0.005 767 168;
  • 20) 0.005 767 168 × 2 = 0 + 0.011 534 336;
  • 21) 0.011 534 336 × 2 = 0 + 0.023 068 672;
  • 22) 0.023 068 672 × 2 = 0 + 0.046 137 344;
  • 23) 0.046 137 344 × 2 = 0 + 0.092 274 688;
  • 24) 0.092 274 688 × 2 = 0 + 0.184 549 376;
  • 25) 0.184 549 376 × 2 = 0 + 0.369 098 752;
  • 26) 0.369 098 752 × 2 = 0 + 0.738 197 504;
  • 27) 0.738 197 504 × 2 = 1 + 0.476 395 008;
  • 28) 0.476 395 008 × 2 = 0 + 0.952 790 016;
  • 29) 0.952 790 016 × 2 = 1 + 0.905 580 032;
  • 30) 0.905 580 032 × 2 = 1 + 0.811 160 064;
  • 31) 0.811 160 064 × 2 = 1 + 0.622 320 128;
  • 32) 0.622 320 128 × 2 = 1 + 0.244 640 256;
  • 33) 0.244 640 256 × 2 = 0 + 0.489 280 512;
  • 34) 0.489 280 512 × 2 = 0 + 0.978 561 024;
  • 35) 0.978 561 024 × 2 = 1 + 0.957 122 048;
  • 36) 0.957 122 048 × 2 = 1 + 0.914 244 096;
  • 37) 0.914 244 096 × 2 = 1 + 0.828 488 192;
  • 38) 0.828 488 192 × 2 = 1 + 0.656 976 384;
  • 39) 0.656 976 384 × 2 = 1 + 0.313 952 768;
  • 40) 0.313 952 768 × 2 = 0 + 0.627 905 536;
  • 41) 0.627 905 536 × 2 = 1 + 0.255 811 072;
  • 42) 0.255 811 072 × 2 = 0 + 0.511 622 144;
  • 43) 0.511 622 144 × 2 = 1 + 0.023 244 288;
  • 44) 0.023 244 288 × 2 = 0 + 0.046 488 576;
  • 45) 0.046 488 576 × 2 = 0 + 0.092 977 152;
  • 46) 0.092 977 152 × 2 = 0 + 0.185 954 304;
  • 47) 0.185 954 304 × 2 = 0 + 0.371 908 608;
  • 48) 0.371 908 608 × 2 = 0 + 0.743 817 216;
  • 49) 0.743 817 216 × 2 = 1 + 0.487 634 432;
  • 50) 0.487 634 432 × 2 = 0 + 0.975 268 864;

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:


0.000 000 011(10) =

0.0000 0000 0000 0000 0000 0000 0010 1111 0011 1110 1010 0000 10(2)

5. Positive number before normalization:

0.000 000 011(10) =

0.0000 0000 0000 0000 0000 0000 0010 1111 0011 1110 1010 0000 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 011(10) =

0.0000 0000 0000 0000 0000 0000 0010 1111 0011 1110 1010 0000 10(2) =

0.0000 0000 0000 0000 0000 0000 0010 1111 0011 1110 1010 0000 10(2) × 20 =

1.0111 1001 1111 0101 0000 010(2) × 2-27


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): -27

Mantissa (not normalized):
1.0111 1001 1111 0101 0000 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-27 + 2(8-1) - 1 =

(-27 + 127)(10) =

100(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;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

100(10) =

0110 0100(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. 011 1100 1111 1010 1000 0010 =

011 1100 1111 1010 1000 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 0100

Mantissa (23 bits) =
011 1100 1111 1010 1000 0010


Decimal number 0.000 000 011 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 011 1100 1111 1010 1000 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