0.000 000 009 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 009 1(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
0.000 000 009 1(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: 0.000 000 009 1.

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) 0.000 000 009 1 × 2 = 0 + 0.000 000 018 2;
  • 2) 0.000 000 018 2 × 2 = 0 + 0.000 000 036 4;
  • 3) 0.000 000 036 4 × 2 = 0 + 0.000 000 072 8;
  • 4) 0.000 000 072 8 × 2 = 0 + 0.000 000 145 6;
  • 5) 0.000 000 145 6 × 2 = 0 + 0.000 000 291 2;
  • 6) 0.000 000 291 2 × 2 = 0 + 0.000 000 582 4;
  • 7) 0.000 000 582 4 × 2 = 0 + 0.000 001 164 8;
  • 8) 0.000 001 164 8 × 2 = 0 + 0.000 002 329 6;
  • 9) 0.000 002 329 6 × 2 = 0 + 0.000 004 659 2;
  • 10) 0.000 004 659 2 × 2 = 0 + 0.000 009 318 4;
  • 11) 0.000 009 318 4 × 2 = 0 + 0.000 018 636 8;
  • 12) 0.000 018 636 8 × 2 = 0 + 0.000 037 273 6;
  • 13) 0.000 037 273 6 × 2 = 0 + 0.000 074 547 2;
  • 14) 0.000 074 547 2 × 2 = 0 + 0.000 149 094 4;
  • 15) 0.000 149 094 4 × 2 = 0 + 0.000 298 188 8;
  • 16) 0.000 298 188 8 × 2 = 0 + 0.000 596 377 6;
  • 17) 0.000 596 377 6 × 2 = 0 + 0.001 192 755 2;
  • 18) 0.001 192 755 2 × 2 = 0 + 0.002 385 510 4;
  • 19) 0.002 385 510 4 × 2 = 0 + 0.004 771 020 8;
  • 20) 0.004 771 020 8 × 2 = 0 + 0.009 542 041 6;
  • 21) 0.009 542 041 6 × 2 = 0 + 0.019 084 083 2;
  • 22) 0.019 084 083 2 × 2 = 0 + 0.038 168 166 4;
  • 23) 0.038 168 166 4 × 2 = 0 + 0.076 336 332 8;
  • 24) 0.076 336 332 8 × 2 = 0 + 0.152 672 665 6;
  • 25) 0.152 672 665 6 × 2 = 0 + 0.305 345 331 2;
  • 26) 0.305 345 331 2 × 2 = 0 + 0.610 690 662 4;
  • 27) 0.610 690 662 4 × 2 = 1 + 0.221 381 324 8;
  • 28) 0.221 381 324 8 × 2 = 0 + 0.442 762 649 6;
  • 29) 0.442 762 649 6 × 2 = 0 + 0.885 525 299 2;
  • 30) 0.885 525 299 2 × 2 = 1 + 0.771 050 598 4;
  • 31) 0.771 050 598 4 × 2 = 1 + 0.542 101 196 8;
  • 32) 0.542 101 196 8 × 2 = 1 + 0.084 202 393 6;
  • 33) 0.084 202 393 6 × 2 = 0 + 0.168 404 787 2;
  • 34) 0.168 404 787 2 × 2 = 0 + 0.336 809 574 4;
  • 35) 0.336 809 574 4 × 2 = 0 + 0.673 619 148 8;
  • 36) 0.673 619 148 8 × 2 = 1 + 0.347 238 297 6;
  • 37) 0.347 238 297 6 × 2 = 0 + 0.694 476 595 2;
  • 38) 0.694 476 595 2 × 2 = 1 + 0.388 953 190 4;
  • 39) 0.388 953 190 4 × 2 = 0 + 0.777 906 380 8;
  • 40) 0.777 906 380 8 × 2 = 1 + 0.555 812 761 6;
  • 41) 0.555 812 761 6 × 2 = 1 + 0.111 625 523 2;
  • 42) 0.111 625 523 2 × 2 = 0 + 0.223 251 046 4;
  • 43) 0.223 251 046 4 × 2 = 0 + 0.446 502 092 8;
  • 44) 0.446 502 092 8 × 2 = 0 + 0.893 004 185 6;
  • 45) 0.893 004 185 6 × 2 = 1 + 0.786 008 371 2;
  • 46) 0.786 008 371 2 × 2 = 1 + 0.572 016 742 4;
  • 47) 0.572 016 742 4 × 2 = 1 + 0.144 033 484 8;
  • 48) 0.144 033 484 8 × 2 = 0 + 0.288 066 969 6;
  • 49) 0.288 066 969 6 × 2 = 0 + 0.576 133 939 2;
  • 50) 0.576 133 939 2 × 2 = 1 + 0.152 267 878 4;

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:


0.000 000 009 1(10) =

0.0000 0000 0000 0000 0000 0000 0010 0111 0001 0101 1000 1110 01(2)

5. Positive number before normalization:

0.000 000 009 1(10) =

0.0000 0000 0000 0000 0000 0000 0010 0111 0001 0101 1000 1110 01(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:


0.000 000 009 1(10) =

0.0000 0000 0000 0000 0000 0000 0010 0111 0001 0101 1000 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0010 0111 0001 0101 1000 1110 01(2) × 20 =

1.0011 1000 1010 1100 0111 001(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):
1.0011 1000 1010 1100 0111 001


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. 001 1100 0101 0110 0011 1001 =

001 1100 0101 0110 0011 1001


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) =
001 1100 0101 0110 0011 1001


Decimal number 0.000 000 009 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 001 1100 0101 0110 0011 1001

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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