0.000 000 005 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 005 2(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
0.000 000 005 2(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: 0.000 000 005 2.

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) 0.000 000 005 2 × 2 = 0 + 0.000 000 010 4;
  • 2) 0.000 000 010 4 × 2 = 0 + 0.000 000 020 8;
  • 3) 0.000 000 020 8 × 2 = 0 + 0.000 000 041 6;
  • 4) 0.000 000 041 6 × 2 = 0 + 0.000 000 083 2;
  • 5) 0.000 000 083 2 × 2 = 0 + 0.000 000 166 4;
  • 6) 0.000 000 166 4 × 2 = 0 + 0.000 000 332 8;
  • 7) 0.000 000 332 8 × 2 = 0 + 0.000 000 665 6;
  • 8) 0.000 000 665 6 × 2 = 0 + 0.000 001 331 2;
  • 9) 0.000 001 331 2 × 2 = 0 + 0.000 002 662 4;
  • 10) 0.000 002 662 4 × 2 = 0 + 0.000 005 324 8;
  • 11) 0.000 005 324 8 × 2 = 0 + 0.000 010 649 6;
  • 12) 0.000 010 649 6 × 2 = 0 + 0.000 021 299 2;
  • 13) 0.000 021 299 2 × 2 = 0 + 0.000 042 598 4;
  • 14) 0.000 042 598 4 × 2 = 0 + 0.000 085 196 8;
  • 15) 0.000 085 196 8 × 2 = 0 + 0.000 170 393 6;
  • 16) 0.000 170 393 6 × 2 = 0 + 0.000 340 787 2;
  • 17) 0.000 340 787 2 × 2 = 0 + 0.000 681 574 4;
  • 18) 0.000 681 574 4 × 2 = 0 + 0.001 363 148 8;
  • 19) 0.001 363 148 8 × 2 = 0 + 0.002 726 297 6;
  • 20) 0.002 726 297 6 × 2 = 0 + 0.005 452 595 2;
  • 21) 0.005 452 595 2 × 2 = 0 + 0.010 905 190 4;
  • 22) 0.010 905 190 4 × 2 = 0 + 0.021 810 380 8;
  • 23) 0.021 810 380 8 × 2 = 0 + 0.043 620 761 6;
  • 24) 0.043 620 761 6 × 2 = 0 + 0.087 241 523 2;
  • 25) 0.087 241 523 2 × 2 = 0 + 0.174 483 046 4;
  • 26) 0.174 483 046 4 × 2 = 0 + 0.348 966 092 8;
  • 27) 0.348 966 092 8 × 2 = 0 + 0.697 932 185 6;
  • 28) 0.697 932 185 6 × 2 = 1 + 0.395 864 371 2;
  • 29) 0.395 864 371 2 × 2 = 0 + 0.791 728 742 4;
  • 30) 0.791 728 742 4 × 2 = 1 + 0.583 457 484 8;
  • 31) 0.583 457 484 8 × 2 = 1 + 0.166 914 969 6;
  • 32) 0.166 914 969 6 × 2 = 0 + 0.333 829 939 2;
  • 33) 0.333 829 939 2 × 2 = 0 + 0.667 659 878 4;
  • 34) 0.667 659 878 4 × 2 = 1 + 0.335 319 756 8;
  • 35) 0.335 319 756 8 × 2 = 0 + 0.670 639 513 6;
  • 36) 0.670 639 513 6 × 2 = 1 + 0.341 279 027 2;
  • 37) 0.341 279 027 2 × 2 = 0 + 0.682 558 054 4;
  • 38) 0.682 558 054 4 × 2 = 1 + 0.365 116 108 8;
  • 39) 0.365 116 108 8 × 2 = 0 + 0.730 232 217 6;
  • 40) 0.730 232 217 6 × 2 = 1 + 0.460 464 435 2;
  • 41) 0.460 464 435 2 × 2 = 0 + 0.920 928 870 4;
  • 42) 0.920 928 870 4 × 2 = 1 + 0.841 857 740 8;
  • 43) 0.841 857 740 8 × 2 = 1 + 0.683 715 481 6;
  • 44) 0.683 715 481 6 × 2 = 1 + 0.367 430 963 2;
  • 45) 0.367 430 963 2 × 2 = 0 + 0.734 861 926 4;
  • 46) 0.734 861 926 4 × 2 = 1 + 0.469 723 852 8;
  • 47) 0.469 723 852 8 × 2 = 0 + 0.939 447 705 6;
  • 48) 0.939 447 705 6 × 2 = 1 + 0.878 895 411 2;
  • 49) 0.878 895 411 2 × 2 = 1 + 0.757 790 822 4;
  • 50) 0.757 790 822 4 × 2 = 1 + 0.515 581 644 8;
  • 51) 0.515 581 644 8 × 2 = 1 + 0.031 163 289 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).


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:


0.000 000 005 2(10) =

0.0000 0000 0000 0000 0000 0000 0001 0110 0101 0101 0111 0101 111(2)

5. Positive number before normalization:

0.000 000 005 2(10) =

0.0000 0000 0000 0000 0000 0000 0001 0110 0101 0101 0111 0101 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 005 2(10) =

0.0000 0000 0000 0000 0000 0000 0001 0110 0101 0101 0111 0101 111(2) =

0.0000 0000 0000 0000 0000 0000 0001 0110 0101 0101 0111 0101 111(2) × 20 =

1.0110 0101 0101 0111 0101 111(2) × 2-28


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): -28

Mantissa (not normalized):
1.0110 0101 0101 0111 0101 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(8-1) - 1 =

(-28 + 127)(10) =

99(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;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

99(10) =

0110 0011(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 0010 1010 1011 1010 1111 =

011 0010 1010 1011 1010 1111


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 0011

Mantissa (23 bits) =
011 0010 1010 1011 1010 1111


Decimal number 0.000 000 005 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0011 - 011 0010 1010 1011 1010 1111

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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