0.000 000 110 017 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 017 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
0.000 000 110 017 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: 0.000 000 110 017 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) 0.000 000 110 017 3 × 2 = 0 + 0.000 000 220 034 6;
  • 2) 0.000 000 220 034 6 × 2 = 0 + 0.000 000 440 069 2;
  • 3) 0.000 000 440 069 2 × 2 = 0 + 0.000 000 880 138 4;
  • 4) 0.000 000 880 138 4 × 2 = 0 + 0.000 001 760 276 8;
  • 5) 0.000 001 760 276 8 × 2 = 0 + 0.000 003 520 553 6;
  • 6) 0.000 003 520 553 6 × 2 = 0 + 0.000 007 041 107 2;
  • 7) 0.000 007 041 107 2 × 2 = 0 + 0.000 014 082 214 4;
  • 8) 0.000 014 082 214 4 × 2 = 0 + 0.000 028 164 428 8;
  • 9) 0.000 028 164 428 8 × 2 = 0 + 0.000 056 328 857 6;
  • 10) 0.000 056 328 857 6 × 2 = 0 + 0.000 112 657 715 2;
  • 11) 0.000 112 657 715 2 × 2 = 0 + 0.000 225 315 430 4;
  • 12) 0.000 225 315 430 4 × 2 = 0 + 0.000 450 630 860 8;
  • 13) 0.000 450 630 860 8 × 2 = 0 + 0.000 901 261 721 6;
  • 14) 0.000 901 261 721 6 × 2 = 0 + 0.001 802 523 443 2;
  • 15) 0.001 802 523 443 2 × 2 = 0 + 0.003 605 046 886 4;
  • 16) 0.003 605 046 886 4 × 2 = 0 + 0.007 210 093 772 8;
  • 17) 0.007 210 093 772 8 × 2 = 0 + 0.014 420 187 545 6;
  • 18) 0.014 420 187 545 6 × 2 = 0 + 0.028 840 375 091 2;
  • 19) 0.028 840 375 091 2 × 2 = 0 + 0.057 680 750 182 4;
  • 20) 0.057 680 750 182 4 × 2 = 0 + 0.115 361 500 364 8;
  • 21) 0.115 361 500 364 8 × 2 = 0 + 0.230 723 000 729 6;
  • 22) 0.230 723 000 729 6 × 2 = 0 + 0.461 446 001 459 2;
  • 23) 0.461 446 001 459 2 × 2 = 0 + 0.922 892 002 918 4;
  • 24) 0.922 892 002 918 4 × 2 = 1 + 0.845 784 005 836 8;
  • 25) 0.845 784 005 836 8 × 2 = 1 + 0.691 568 011 673 6;
  • 26) 0.691 568 011 673 6 × 2 = 1 + 0.383 136 023 347 2;
  • 27) 0.383 136 023 347 2 × 2 = 0 + 0.766 272 046 694 4;
  • 28) 0.766 272 046 694 4 × 2 = 1 + 0.532 544 093 388 8;
  • 29) 0.532 544 093 388 8 × 2 = 1 + 0.065 088 186 777 6;
  • 30) 0.065 088 186 777 6 × 2 = 0 + 0.130 176 373 555 2;
  • 31) 0.130 176 373 555 2 × 2 = 0 + 0.260 352 747 110 4;
  • 32) 0.260 352 747 110 4 × 2 = 0 + 0.520 705 494 220 8;
  • 33) 0.520 705 494 220 8 × 2 = 1 + 0.041 410 988 441 6;
  • 34) 0.041 410 988 441 6 × 2 = 0 + 0.082 821 976 883 2;
  • 35) 0.082 821 976 883 2 × 2 = 0 + 0.165 643 953 766 4;
  • 36) 0.165 643 953 766 4 × 2 = 0 + 0.331 287 907 532 8;
  • 37) 0.331 287 907 532 8 × 2 = 0 + 0.662 575 815 065 6;
  • 38) 0.662 575 815 065 6 × 2 = 1 + 0.325 151 630 131 2;
  • 39) 0.325 151 630 131 2 × 2 = 0 + 0.650 303 260 262 4;
  • 40) 0.650 303 260 262 4 × 2 = 1 + 0.300 606 520 524 8;
  • 41) 0.300 606 520 524 8 × 2 = 0 + 0.601 213 041 049 6;
  • 42) 0.601 213 041 049 6 × 2 = 1 + 0.202 426 082 099 2;
  • 43) 0.202 426 082 099 2 × 2 = 0 + 0.404 852 164 198 4;
  • 44) 0.404 852 164 198 4 × 2 = 0 + 0.809 704 328 396 8;
  • 45) 0.809 704 328 396 8 × 2 = 1 + 0.619 408 656 793 6;
  • 46) 0.619 408 656 793 6 × 2 = 1 + 0.238 817 313 587 2;
  • 47) 0.238 817 313 587 2 × 2 = 0 + 0.477 634 627 174 4;

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:


0.000 000 110 017 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0101 0100 110(2)

5. Positive number before normalization:

0.000 000 110 017 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0101 0100 110(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:


0.000 000 110 017 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0101 0100 110(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0101 0100 110(2) × 20 =

1.1101 1000 1000 0101 0100 110(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 0101 0100 110


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 0010 1010 0110 =

110 1100 0100 0010 1010 0110


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 0010 1010 0110


Decimal number 0.000 000 110 017 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0100 0010 1010 0110

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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