0.000 000 022 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 022 9(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 9(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: 0.000 000 022 9.

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 9 × 2 = 0 + 0.000 000 045 8;
  • 2) 0.000 000 045 8 × 2 = 0 + 0.000 000 091 6;
  • 3) 0.000 000 091 6 × 2 = 0 + 0.000 000 183 2;
  • 4) 0.000 000 183 2 × 2 = 0 + 0.000 000 366 4;
  • 5) 0.000 000 366 4 × 2 = 0 + 0.000 000 732 8;
  • 6) 0.000 000 732 8 × 2 = 0 + 0.000 001 465 6;
  • 7) 0.000 001 465 6 × 2 = 0 + 0.000 002 931 2;
  • 8) 0.000 002 931 2 × 2 = 0 + 0.000 005 862 4;
  • 9) 0.000 005 862 4 × 2 = 0 + 0.000 011 724 8;
  • 10) 0.000 011 724 8 × 2 = 0 + 0.000 023 449 6;
  • 11) 0.000 023 449 6 × 2 = 0 + 0.000 046 899 2;
  • 12) 0.000 046 899 2 × 2 = 0 + 0.000 093 798 4;
  • 13) 0.000 093 798 4 × 2 = 0 + 0.000 187 596 8;
  • 14) 0.000 187 596 8 × 2 = 0 + 0.000 375 193 6;
  • 15) 0.000 375 193 6 × 2 = 0 + 0.000 750 387 2;
  • 16) 0.000 750 387 2 × 2 = 0 + 0.001 500 774 4;
  • 17) 0.001 500 774 4 × 2 = 0 + 0.003 001 548 8;
  • 18) 0.003 001 548 8 × 2 = 0 + 0.006 003 097 6;
  • 19) 0.006 003 097 6 × 2 = 0 + 0.012 006 195 2;
  • 20) 0.012 006 195 2 × 2 = 0 + 0.024 012 390 4;
  • 21) 0.024 012 390 4 × 2 = 0 + 0.048 024 780 8;
  • 22) 0.048 024 780 8 × 2 = 0 + 0.096 049 561 6;
  • 23) 0.096 049 561 6 × 2 = 0 + 0.192 099 123 2;
  • 24) 0.192 099 123 2 × 2 = 0 + 0.384 198 246 4;
  • 25) 0.384 198 246 4 × 2 = 0 + 0.768 396 492 8;
  • 26) 0.768 396 492 8 × 2 = 1 + 0.536 792 985 6;
  • 27) 0.536 792 985 6 × 2 = 1 + 0.073 585 971 2;
  • 28) 0.073 585 971 2 × 2 = 0 + 0.147 171 942 4;
  • 29) 0.147 171 942 4 × 2 = 0 + 0.294 343 884 8;
  • 30) 0.294 343 884 8 × 2 = 0 + 0.588 687 769 6;
  • 31) 0.588 687 769 6 × 2 = 1 + 0.177 375 539 2;
  • 32) 0.177 375 539 2 × 2 = 0 + 0.354 751 078 4;
  • 33) 0.354 751 078 4 × 2 = 0 + 0.709 502 156 8;
  • 34) 0.709 502 156 8 × 2 = 1 + 0.419 004 313 6;
  • 35) 0.419 004 313 6 × 2 = 0 + 0.838 008 627 2;
  • 36) 0.838 008 627 2 × 2 = 1 + 0.676 017 254 4;
  • 37) 0.676 017 254 4 × 2 = 1 + 0.352 034 508 8;
  • 38) 0.352 034 508 8 × 2 = 0 + 0.704 069 017 6;
  • 39) 0.704 069 017 6 × 2 = 1 + 0.408 138 035 2;
  • 40) 0.408 138 035 2 × 2 = 0 + 0.816 276 070 4;
  • 41) 0.816 276 070 4 × 2 = 1 + 0.632 552 140 8;
  • 42) 0.632 552 140 8 × 2 = 1 + 0.265 104 281 6;
  • 43) 0.265 104 281 6 × 2 = 0 + 0.530 208 563 2;
  • 44) 0.530 208 563 2 × 2 = 1 + 0.060 417 126 4;
  • 45) 0.060 417 126 4 × 2 = 0 + 0.120 834 252 8;
  • 46) 0.120 834 252 8 × 2 = 0 + 0.241 668 505 6;
  • 47) 0.241 668 505 6 × 2 = 0 + 0.483 337 011 2;
  • 48) 0.483 337 011 2 × 2 = 0 + 0.966 674 022 4;
  • 49) 0.966 674 022 4 × 2 = 1 + 0.933 348 044 8;

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:


0.000 000 022 9(10) =

0.0000 0000 0000 0000 0000 0000 0110 0010 0101 1010 1101 0000 1(2)

5. Positive number before normalization:

0.000 000 022 9(10) =

0.0000 0000 0000 0000 0000 0000 0110 0010 0101 1010 1101 0000 1(2)

6. 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 9(10) =

0.0000 0000 0000 0000 0000 0000 0110 0010 0101 1010 1101 0000 1(2) =

0.0000 0000 0000 0000 0000 0000 0110 0010 0101 1010 1101 0000 1(2) × 20 =

1.1000 1001 0110 1011 0100 001(2) × 2-26


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): -26

Mantissa (not normalized):
1.1000 1001 0110 1011 0100 001


8. 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)


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;
  • 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;

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) =

101(10) =

0110 0101(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. 100 0100 1011 0101 1010 0001 =

100 0100 1011 0101 1010 0001


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 0101

Mantissa (23 bits) =
100 0100 1011 0101 1010 0001


Decimal number 0.000 000 022 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 100 0100 1011 0101 1010 0001

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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