-0.000 000 022 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 022 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
-0.000 000 022 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. Start with the positive version of the number:

|-0.000 000 022 3| = 0.000 000 022 3


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 022 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) 0.000 000 022 3 × 2 = 0 + 0.000 000 044 6;
  • 2) 0.000 000 044 6 × 2 = 0 + 0.000 000 089 2;
  • 3) 0.000 000 089 2 × 2 = 0 + 0.000 000 178 4;
  • 4) 0.000 000 178 4 × 2 = 0 + 0.000 000 356 8;
  • 5) 0.000 000 356 8 × 2 = 0 + 0.000 000 713 6;
  • 6) 0.000 000 713 6 × 2 = 0 + 0.000 001 427 2;
  • 7) 0.000 001 427 2 × 2 = 0 + 0.000 002 854 4;
  • 8) 0.000 002 854 4 × 2 = 0 + 0.000 005 708 8;
  • 9) 0.000 005 708 8 × 2 = 0 + 0.000 011 417 6;
  • 10) 0.000 011 417 6 × 2 = 0 + 0.000 022 835 2;
  • 11) 0.000 022 835 2 × 2 = 0 + 0.000 045 670 4;
  • 12) 0.000 045 670 4 × 2 = 0 + 0.000 091 340 8;
  • 13) 0.000 091 340 8 × 2 = 0 + 0.000 182 681 6;
  • 14) 0.000 182 681 6 × 2 = 0 + 0.000 365 363 2;
  • 15) 0.000 365 363 2 × 2 = 0 + 0.000 730 726 4;
  • 16) 0.000 730 726 4 × 2 = 0 + 0.001 461 452 8;
  • 17) 0.001 461 452 8 × 2 = 0 + 0.002 922 905 6;
  • 18) 0.002 922 905 6 × 2 = 0 + 0.005 845 811 2;
  • 19) 0.005 845 811 2 × 2 = 0 + 0.011 691 622 4;
  • 20) 0.011 691 622 4 × 2 = 0 + 0.023 383 244 8;
  • 21) 0.023 383 244 8 × 2 = 0 + 0.046 766 489 6;
  • 22) 0.046 766 489 6 × 2 = 0 + 0.093 532 979 2;
  • 23) 0.093 532 979 2 × 2 = 0 + 0.187 065 958 4;
  • 24) 0.187 065 958 4 × 2 = 0 + 0.374 131 916 8;
  • 25) 0.374 131 916 8 × 2 = 0 + 0.748 263 833 6;
  • 26) 0.748 263 833 6 × 2 = 1 + 0.496 527 667 2;
  • 27) 0.496 527 667 2 × 2 = 0 + 0.993 055 334 4;
  • 28) 0.993 055 334 4 × 2 = 1 + 0.986 110 668 8;
  • 29) 0.986 110 668 8 × 2 = 1 + 0.972 221 337 6;
  • 30) 0.972 221 337 6 × 2 = 1 + 0.944 442 675 2;
  • 31) 0.944 442 675 2 × 2 = 1 + 0.888 885 350 4;
  • 32) 0.888 885 350 4 × 2 = 1 + 0.777 770 700 8;
  • 33) 0.777 770 700 8 × 2 = 1 + 0.555 541 401 6;
  • 34) 0.555 541 401 6 × 2 = 1 + 0.111 082 803 2;
  • 35) 0.111 082 803 2 × 2 = 0 + 0.222 165 606 4;
  • 36) 0.222 165 606 4 × 2 = 0 + 0.444 331 212 8;
  • 37) 0.444 331 212 8 × 2 = 0 + 0.888 662 425 6;
  • 38) 0.888 662 425 6 × 2 = 1 + 0.777 324 851 2;
  • 39) 0.777 324 851 2 × 2 = 1 + 0.554 649 702 4;
  • 40) 0.554 649 702 4 × 2 = 1 + 0.109 299 404 8;
  • 41) 0.109 299 404 8 × 2 = 0 + 0.218 598 809 6;
  • 42) 0.218 598 809 6 × 2 = 0 + 0.437 197 619 2;
  • 43) 0.437 197 619 2 × 2 = 0 + 0.874 395 238 4;
  • 44) 0.874 395 238 4 × 2 = 1 + 0.748 790 476 8;
  • 45) 0.748 790 476 8 × 2 = 1 + 0.497 580 953 6;
  • 46) 0.497 580 953 6 × 2 = 0 + 0.995 161 907 2;
  • 47) 0.995 161 907 2 × 2 = 1 + 0.990 323 814 4;
  • 48) 0.990 323 814 4 × 2 = 1 + 0.980 647 628 8;
  • 49) 0.980 647 628 8 × 2 = 1 + 0.961 295 257 6;

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).


5. 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:


0.000 000 022 3(10) =

0.0000 0000 0000 0000 0000 0000 0101 1111 1100 0111 0001 1011 1(2)

6. Positive number before normalization:

0.000 000 022 3(10) =

0.0000 0000 0000 0000 0000 0000 0101 1111 1100 0111 0001 1011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 022 3(10) =

0.0000 0000 0000 0000 0000 0000 0101 1111 1100 0111 0001 1011 1(2) =

0.0000 0000 0000 0000 0000 0000 0101 1111 1100 0111 0001 1011 1(2) × 20 =

1.0111 1111 0001 1100 0110 111(2) × 2-26


8. 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 1 (a negative number)

Exponent (unadjusted): -26

Mantissa (not normalized):
1.0111 1111 0001 1100 0110 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(8-1) - 1 =

(-26 + 127)(10) =

101(10)


10. 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;
  • 101 ÷ 2 = 50 + 1;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

101(10) =

0110 0101(2)


12. 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 1111 1000 1110 0011 0111 =

011 1111 1000 1110 0011 0111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0101

Mantissa (23 bits) =
011 1111 1000 1110 0011 0111


Decimal number -0.000 000 022 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0101 - 011 1111 1000 1110 0011 0111

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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