0.000 000 110 021 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 021 4(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
0.000 000 110 021 4(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: 0.000 000 110 021 4.

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) 0.000 000 110 021 4 × 2 = 0 + 0.000 000 220 042 8;
  • 2) 0.000 000 220 042 8 × 2 = 0 + 0.000 000 440 085 6;
  • 3) 0.000 000 440 085 6 × 2 = 0 + 0.000 000 880 171 2;
  • 4) 0.000 000 880 171 2 × 2 = 0 + 0.000 001 760 342 4;
  • 5) 0.000 001 760 342 4 × 2 = 0 + 0.000 003 520 684 8;
  • 6) 0.000 003 520 684 8 × 2 = 0 + 0.000 007 041 369 6;
  • 7) 0.000 007 041 369 6 × 2 = 0 + 0.000 014 082 739 2;
  • 8) 0.000 014 082 739 2 × 2 = 0 + 0.000 028 165 478 4;
  • 9) 0.000 028 165 478 4 × 2 = 0 + 0.000 056 330 956 8;
  • 10) 0.000 056 330 956 8 × 2 = 0 + 0.000 112 661 913 6;
  • 11) 0.000 112 661 913 6 × 2 = 0 + 0.000 225 323 827 2;
  • 12) 0.000 225 323 827 2 × 2 = 0 + 0.000 450 647 654 4;
  • 13) 0.000 450 647 654 4 × 2 = 0 + 0.000 901 295 308 8;
  • 14) 0.000 901 295 308 8 × 2 = 0 + 0.001 802 590 617 6;
  • 15) 0.001 802 590 617 6 × 2 = 0 + 0.003 605 181 235 2;
  • 16) 0.003 605 181 235 2 × 2 = 0 + 0.007 210 362 470 4;
  • 17) 0.007 210 362 470 4 × 2 = 0 + 0.014 420 724 940 8;
  • 18) 0.014 420 724 940 8 × 2 = 0 + 0.028 841 449 881 6;
  • 19) 0.028 841 449 881 6 × 2 = 0 + 0.057 682 899 763 2;
  • 20) 0.057 682 899 763 2 × 2 = 0 + 0.115 365 799 526 4;
  • 21) 0.115 365 799 526 4 × 2 = 0 + 0.230 731 599 052 8;
  • 22) 0.230 731 599 052 8 × 2 = 0 + 0.461 463 198 105 6;
  • 23) 0.461 463 198 105 6 × 2 = 0 + 0.922 926 396 211 2;
  • 24) 0.922 926 396 211 2 × 2 = 1 + 0.845 852 792 422 4;
  • 25) 0.845 852 792 422 4 × 2 = 1 + 0.691 705 584 844 8;
  • 26) 0.691 705 584 844 8 × 2 = 1 + 0.383 411 169 689 6;
  • 27) 0.383 411 169 689 6 × 2 = 0 + 0.766 822 339 379 2;
  • 28) 0.766 822 339 379 2 × 2 = 1 + 0.533 644 678 758 4;
  • 29) 0.533 644 678 758 4 × 2 = 1 + 0.067 289 357 516 8;
  • 30) 0.067 289 357 516 8 × 2 = 0 + 0.134 578 715 033 6;
  • 31) 0.134 578 715 033 6 × 2 = 0 + 0.269 157 430 067 2;
  • 32) 0.269 157 430 067 2 × 2 = 0 + 0.538 314 860 134 4;
  • 33) 0.538 314 860 134 4 × 2 = 1 + 0.076 629 720 268 8;
  • 34) 0.076 629 720 268 8 × 2 = 0 + 0.153 259 440 537 6;
  • 35) 0.153 259 440 537 6 × 2 = 0 + 0.306 518 881 075 2;
  • 36) 0.306 518 881 075 2 × 2 = 0 + 0.613 037 762 150 4;
  • 37) 0.613 037 762 150 4 × 2 = 1 + 0.226 075 524 300 8;
  • 38) 0.226 075 524 300 8 × 2 = 0 + 0.452 151 048 601 6;
  • 39) 0.452 151 048 601 6 × 2 = 0 + 0.904 302 097 203 2;
  • 40) 0.904 302 097 203 2 × 2 = 1 + 0.808 604 194 406 4;
  • 41) 0.808 604 194 406 4 × 2 = 1 + 0.617 208 388 812 8;
  • 42) 0.617 208 388 812 8 × 2 = 1 + 0.234 416 777 625 6;
  • 43) 0.234 416 777 625 6 × 2 = 0 + 0.468 833 555 251 2;
  • 44) 0.468 833 555 251 2 × 2 = 0 + 0.937 667 110 502 4;
  • 45) 0.937 667 110 502 4 × 2 = 1 + 0.875 334 221 004 8;
  • 46) 0.875 334 221 004 8 × 2 = 1 + 0.750 668 442 009 6;
  • 47) 0.750 668 442 009 6 × 2 = 1 + 0.501 336 884 019 2;

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:


0.000 000 110 021 4(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 1001 1100 111(2)

5. Positive number before normalization:

0.000 000 110 021 4(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 1001 1100 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:


0.000 000 110 021 4(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 1001 1100 111(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 1001 1100 111(2) × 20 =

1.1101 1000 1000 1001 1100 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 1001 1100 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 0100 1110 0111 =

110 1100 0100 0100 1110 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 0100 1110 0111


Decimal number 0.000 000 110 021 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0100 0100 1110 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