0.000 000 110 015 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 015 6(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
0.000 000 110 015 6(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: 0.000 000 110 015 6.

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) 0.000 000 110 015 6 × 2 = 0 + 0.000 000 220 031 2;
  • 2) 0.000 000 220 031 2 × 2 = 0 + 0.000 000 440 062 4;
  • 3) 0.000 000 440 062 4 × 2 = 0 + 0.000 000 880 124 8;
  • 4) 0.000 000 880 124 8 × 2 = 0 + 0.000 001 760 249 6;
  • 5) 0.000 001 760 249 6 × 2 = 0 + 0.000 003 520 499 2;
  • 6) 0.000 003 520 499 2 × 2 = 0 + 0.000 007 040 998 4;
  • 7) 0.000 007 040 998 4 × 2 = 0 + 0.000 014 081 996 8;
  • 8) 0.000 014 081 996 8 × 2 = 0 + 0.000 028 163 993 6;
  • 9) 0.000 028 163 993 6 × 2 = 0 + 0.000 056 327 987 2;
  • 10) 0.000 056 327 987 2 × 2 = 0 + 0.000 112 655 974 4;
  • 11) 0.000 112 655 974 4 × 2 = 0 + 0.000 225 311 948 8;
  • 12) 0.000 225 311 948 8 × 2 = 0 + 0.000 450 623 897 6;
  • 13) 0.000 450 623 897 6 × 2 = 0 + 0.000 901 247 795 2;
  • 14) 0.000 901 247 795 2 × 2 = 0 + 0.001 802 495 590 4;
  • 15) 0.001 802 495 590 4 × 2 = 0 + 0.003 604 991 180 8;
  • 16) 0.003 604 991 180 8 × 2 = 0 + 0.007 209 982 361 6;
  • 17) 0.007 209 982 361 6 × 2 = 0 + 0.014 419 964 723 2;
  • 18) 0.014 419 964 723 2 × 2 = 0 + 0.028 839 929 446 4;
  • 19) 0.028 839 929 446 4 × 2 = 0 + 0.057 679 858 892 8;
  • 20) 0.057 679 858 892 8 × 2 = 0 + 0.115 359 717 785 6;
  • 21) 0.115 359 717 785 6 × 2 = 0 + 0.230 719 435 571 2;
  • 22) 0.230 719 435 571 2 × 2 = 0 + 0.461 438 871 142 4;
  • 23) 0.461 438 871 142 4 × 2 = 0 + 0.922 877 742 284 8;
  • 24) 0.922 877 742 284 8 × 2 = 1 + 0.845 755 484 569 6;
  • 25) 0.845 755 484 569 6 × 2 = 1 + 0.691 510 969 139 2;
  • 26) 0.691 510 969 139 2 × 2 = 1 + 0.383 021 938 278 4;
  • 27) 0.383 021 938 278 4 × 2 = 0 + 0.766 043 876 556 8;
  • 28) 0.766 043 876 556 8 × 2 = 1 + 0.532 087 753 113 6;
  • 29) 0.532 087 753 113 6 × 2 = 1 + 0.064 175 506 227 2;
  • 30) 0.064 175 506 227 2 × 2 = 0 + 0.128 351 012 454 4;
  • 31) 0.128 351 012 454 4 × 2 = 0 + 0.256 702 024 908 8;
  • 32) 0.256 702 024 908 8 × 2 = 0 + 0.513 404 049 817 6;
  • 33) 0.513 404 049 817 6 × 2 = 1 + 0.026 808 099 635 2;
  • 34) 0.026 808 099 635 2 × 2 = 0 + 0.053 616 199 270 4;
  • 35) 0.053 616 199 270 4 × 2 = 0 + 0.107 232 398 540 8;
  • 36) 0.107 232 398 540 8 × 2 = 0 + 0.214 464 797 081 6;
  • 37) 0.214 464 797 081 6 × 2 = 0 + 0.428 929 594 163 2;
  • 38) 0.428 929 594 163 2 × 2 = 0 + 0.857 859 188 326 4;
  • 39) 0.857 859 188 326 4 × 2 = 1 + 0.715 718 376 652 8;
  • 40) 0.715 718 376 652 8 × 2 = 1 + 0.431 436 753 305 6;
  • 41) 0.431 436 753 305 6 × 2 = 0 + 0.862 873 506 611 2;
  • 42) 0.862 873 506 611 2 × 2 = 1 + 0.725 747 013 222 4;
  • 43) 0.725 747 013 222 4 × 2 = 1 + 0.451 494 026 444 8;
  • 44) 0.451 494 026 444 8 × 2 = 0 + 0.902 988 052 889 6;
  • 45) 0.902 988 052 889 6 × 2 = 1 + 0.805 976 105 779 2;
  • 46) 0.805 976 105 779 2 × 2 = 1 + 0.611 952 211 558 4;
  • 47) 0.611 952 211 558 4 × 2 = 1 + 0.223 904 423 116 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:


0.000 000 110 015 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0011 0110 111(2)

5. Positive number before normalization:

0.000 000 110 015 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0011 0110 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 110 015 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0011 0110 111(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0011 0110 111(2) × 20 =

1.1101 1000 1000 0011 0110 111(2) × 2-24


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): -24

Mantissa (not normalized):
1.1101 1000 1000 0011 0110 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 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) =

103(10) =

0110 0111(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. 110 1100 0100 0001 1011 0111 =

110 1100 0100 0001 1011 0111


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 0111

Mantissa (23 bits) =
110 1100 0100 0001 1011 0111


Decimal number 0.000 000 110 015 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0100 0001 1011 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