0.000 000 009 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 009 3(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 3(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 3.

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 3 × 2 = 0 + 0.000 000 018 6;
  • 2) 0.000 000 018 6 × 2 = 0 + 0.000 000 037 2;
  • 3) 0.000 000 037 2 × 2 = 0 + 0.000 000 074 4;
  • 4) 0.000 000 074 4 × 2 = 0 + 0.000 000 148 8;
  • 5) 0.000 000 148 8 × 2 = 0 + 0.000 000 297 6;
  • 6) 0.000 000 297 6 × 2 = 0 + 0.000 000 595 2;
  • 7) 0.000 000 595 2 × 2 = 0 + 0.000 001 190 4;
  • 8) 0.000 001 190 4 × 2 = 0 + 0.000 002 380 8;
  • 9) 0.000 002 380 8 × 2 = 0 + 0.000 004 761 6;
  • 10) 0.000 004 761 6 × 2 = 0 + 0.000 009 523 2;
  • 11) 0.000 009 523 2 × 2 = 0 + 0.000 019 046 4;
  • 12) 0.000 019 046 4 × 2 = 0 + 0.000 038 092 8;
  • 13) 0.000 038 092 8 × 2 = 0 + 0.000 076 185 6;
  • 14) 0.000 076 185 6 × 2 = 0 + 0.000 152 371 2;
  • 15) 0.000 152 371 2 × 2 = 0 + 0.000 304 742 4;
  • 16) 0.000 304 742 4 × 2 = 0 + 0.000 609 484 8;
  • 17) 0.000 609 484 8 × 2 = 0 + 0.001 218 969 6;
  • 18) 0.001 218 969 6 × 2 = 0 + 0.002 437 939 2;
  • 19) 0.002 437 939 2 × 2 = 0 + 0.004 875 878 4;
  • 20) 0.004 875 878 4 × 2 = 0 + 0.009 751 756 8;
  • 21) 0.009 751 756 8 × 2 = 0 + 0.019 503 513 6;
  • 22) 0.019 503 513 6 × 2 = 0 + 0.039 007 027 2;
  • 23) 0.039 007 027 2 × 2 = 0 + 0.078 014 054 4;
  • 24) 0.078 014 054 4 × 2 = 0 + 0.156 028 108 8;
  • 25) 0.156 028 108 8 × 2 = 0 + 0.312 056 217 6;
  • 26) 0.312 056 217 6 × 2 = 0 + 0.624 112 435 2;
  • 27) 0.624 112 435 2 × 2 = 1 + 0.248 224 870 4;
  • 28) 0.248 224 870 4 × 2 = 0 + 0.496 449 740 8;
  • 29) 0.496 449 740 8 × 2 = 0 + 0.992 899 481 6;
  • 30) 0.992 899 481 6 × 2 = 1 + 0.985 798 963 2;
  • 31) 0.985 798 963 2 × 2 = 1 + 0.971 597 926 4;
  • 32) 0.971 597 926 4 × 2 = 1 + 0.943 195 852 8;
  • 33) 0.943 195 852 8 × 2 = 1 + 0.886 391 705 6;
  • 34) 0.886 391 705 6 × 2 = 1 + 0.772 783 411 2;
  • 35) 0.772 783 411 2 × 2 = 1 + 0.545 566 822 4;
  • 36) 0.545 566 822 4 × 2 = 1 + 0.091 133 644 8;
  • 37) 0.091 133 644 8 × 2 = 0 + 0.182 267 289 6;
  • 38) 0.182 267 289 6 × 2 = 0 + 0.364 534 579 2;
  • 39) 0.364 534 579 2 × 2 = 0 + 0.729 069 158 4;
  • 40) 0.729 069 158 4 × 2 = 1 + 0.458 138 316 8;
  • 41) 0.458 138 316 8 × 2 = 0 + 0.916 276 633 6;
  • 42) 0.916 276 633 6 × 2 = 1 + 0.832 553 267 2;
  • 43) 0.832 553 267 2 × 2 = 1 + 0.665 106 534 4;
  • 44) 0.665 106 534 4 × 2 = 1 + 0.330 213 068 8;
  • 45) 0.330 213 068 8 × 2 = 0 + 0.660 426 137 6;
  • 46) 0.660 426 137 6 × 2 = 1 + 0.320 852 275 2;
  • 47) 0.320 852 275 2 × 2 = 0 + 0.641 704 550 4;
  • 48) 0.641 704 550 4 × 2 = 1 + 0.283 409 100 8;
  • 49) 0.283 409 100 8 × 2 = 0 + 0.566 818 201 6;
  • 50) 0.566 818 201 6 × 2 = 1 + 0.133 636 403 2;

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 3(10) =

0.0000 0000 0000 0000 0000 0000 0010 0111 1111 0001 0111 0101 01(2)

5. Positive number before normalization:

0.000 000 009 3(10) =

0.0000 0000 0000 0000 0000 0000 0010 0111 1111 0001 0111 0101 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 3(10) =

0.0000 0000 0000 0000 0000 0000 0010 0111 1111 0001 0111 0101 01(2) =

0.0000 0000 0000 0000 0000 0000 0010 0111 1111 0001 0111 0101 01(2) × 20 =

1.0011 1111 1000 1011 1010 101(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 1111 1000 1011 1010 101


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 1111 1100 0101 1101 0101 =

001 1111 1100 0101 1101 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 0100

Mantissa (23 bits) =
001 1111 1100 0101 1101 0101


Decimal number 0.000 000 009 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 001 1111 1100 0101 1101 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