11 110 000 111 100.001 111 000 009 65 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 11 110 000 111 100.001 111 000 009 65(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
11 110 000 111 100.001 111 000 009 65(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: 11 110 000 111 100.
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;
  • 11 110 000 111 100 ÷ 2 = 5 555 000 055 550 + 0;
  • 5 555 000 055 550 ÷ 2 = 2 777 500 027 775 + 0;
  • 2 777 500 027 775 ÷ 2 = 1 388 750 013 887 + 1;
  • 1 388 750 013 887 ÷ 2 = 694 375 006 943 + 1;
  • 694 375 006 943 ÷ 2 = 347 187 503 471 + 1;
  • 347 187 503 471 ÷ 2 = 173 593 751 735 + 1;
  • 173 593 751 735 ÷ 2 = 86 796 875 867 + 1;
  • 86 796 875 867 ÷ 2 = 43 398 437 933 + 1;
  • 43 398 437 933 ÷ 2 = 21 699 218 966 + 1;
  • 21 699 218 966 ÷ 2 = 10 849 609 483 + 0;
  • 10 849 609 483 ÷ 2 = 5 424 804 741 + 1;
  • 5 424 804 741 ÷ 2 = 2 712 402 370 + 1;
  • 2 712 402 370 ÷ 2 = 1 356 201 185 + 0;
  • 1 356 201 185 ÷ 2 = 678 100 592 + 1;
  • 678 100 592 ÷ 2 = 339 050 296 + 0;
  • 339 050 296 ÷ 2 = 169 525 148 + 0;
  • 169 525 148 ÷ 2 = 84 762 574 + 0;
  • 84 762 574 ÷ 2 = 42 381 287 + 0;
  • 42 381 287 ÷ 2 = 21 190 643 + 1;
  • 21 190 643 ÷ 2 = 10 595 321 + 1;
  • 10 595 321 ÷ 2 = 5 297 660 + 1;
  • 5 297 660 ÷ 2 = 2 648 830 + 0;
  • 2 648 830 ÷ 2 = 1 324 415 + 0;
  • 1 324 415 ÷ 2 = 662 207 + 1;
  • 662 207 ÷ 2 = 331 103 + 1;
  • 331 103 ÷ 2 = 165 551 + 1;
  • 165 551 ÷ 2 = 82 775 + 1;
  • 82 775 ÷ 2 = 41 387 + 1;
  • 41 387 ÷ 2 = 20 693 + 1;
  • 20 693 ÷ 2 = 10 346 + 1;
  • 10 346 ÷ 2 = 5 173 + 0;
  • 5 173 ÷ 2 = 2 586 + 1;
  • 2 586 ÷ 2 = 1 293 + 0;
  • 1 293 ÷ 2 = 646 + 1;
  • 646 ÷ 2 = 323 + 0;
  • 323 ÷ 2 = 161 + 1;
  • 161 ÷ 2 = 80 + 1;
  • 80 ÷ 2 = 40 + 0;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

11 110 000 111 100(10) =

1010 0001 1010 1011 1111 1001 1100 0010 1101 1111 1100(2)


3. Convert to binary (base 2) the fractional part: 0.001 111 000 009 65.

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) 0.001 111 000 009 65 × 2 = 0 + 0.002 222 000 019 3;
  • 2) 0.002 222 000 019 3 × 2 = 0 + 0.004 444 000 038 6;
  • 3) 0.004 444 000 038 6 × 2 = 0 + 0.008 888 000 077 2;
  • 4) 0.008 888 000 077 2 × 2 = 0 + 0.017 776 000 154 4;
  • 5) 0.017 776 000 154 4 × 2 = 0 + 0.035 552 000 308 8;
  • 6) 0.035 552 000 308 8 × 2 = 0 + 0.071 104 000 617 6;
  • 7) 0.071 104 000 617 6 × 2 = 0 + 0.142 208 001 235 2;
  • 8) 0.142 208 001 235 2 × 2 = 0 + 0.284 416 002 470 4;
  • 9) 0.284 416 002 470 4 × 2 = 0 + 0.568 832 004 940 8;
  • 10) 0.568 832 004 940 8 × 2 = 1 + 0.137 664 009 881 6;
  • 11) 0.137 664 009 881 6 × 2 = 0 + 0.275 328 019 763 2;
  • 12) 0.275 328 019 763 2 × 2 = 0 + 0.550 656 039 526 4;
  • 13) 0.550 656 039 526 4 × 2 = 1 + 0.101 312 079 052 8;
  • 14) 0.101 312 079 052 8 × 2 = 0 + 0.202 624 158 105 6;
  • 15) 0.202 624 158 105 6 × 2 = 0 + 0.405 248 316 211 2;
  • 16) 0.405 248 316 211 2 × 2 = 0 + 0.810 496 632 422 4;
  • 17) 0.810 496 632 422 4 × 2 = 1 + 0.620 993 264 844 8;
  • 18) 0.620 993 264 844 8 × 2 = 1 + 0.241 986 529 689 6;
  • 19) 0.241 986 529 689 6 × 2 = 0 + 0.483 973 059 379 2;
  • 20) 0.483 973 059 379 2 × 2 = 0 + 0.967 946 118 758 4;
  • 21) 0.967 946 118 758 4 × 2 = 1 + 0.935 892 237 516 8;
  • 22) 0.935 892 237 516 8 × 2 = 1 + 0.871 784 475 033 6;
  • 23) 0.871 784 475 033 6 × 2 = 1 + 0.743 568 950 067 2;
  • 24) 0.743 568 950 067 2 × 2 = 1 + 0.487 137 900 134 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:


0.001 111 000 009 65(10) =

0.0000 0000 0100 1000 1100 1111(2)

5. Positive number before normalization:

11 110 000 111 100.001 111 000 009 65(10) =

1010 0001 1010 1011 1111 1001 1100 0010 1101 1111 1100.0000 0000 0100 1000 1100 1111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the left, so that only one non zero digit remains to the left of it:


11 110 000 111 100.001 111 000 009 65(10) =

1010 0001 1010 1011 1111 1001 1100 0010 1101 1111 1100.0000 0000 0100 1000 1100 1111(2) =

1010 0001 1010 1011 1111 1001 1100 0010 1101 1111 1100.0000 0000 0100 1000 1100 1111(2) × 20 =

1.0100 0011 0101 0111 1111 0011 1000 0101 1011 1111 1000 0000 0000 1001 0001 1001 111(2) × 243


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): 43

Mantissa (not normalized):
1.0100 0011 0101 0111 1111 0011 1000 0101 1011 1111 1000 0000 0000 1001 0001 1001 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

43 + 2(8-1) - 1 =

(43 + 127)(10) =

170(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;
  • 170 ÷ 2 = 85 + 0;
  • 85 ÷ 2 = 42 + 1;
  • 42 ÷ 2 = 21 + 0;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

170(10) =

1010 1010(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 010 0001 1010 1011 1111 1001 1100 0010 1101 1111 1100 0000 0000 0100 1000 1100 1111 =

010 0001 1010 1011 1111 1001


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) =
1010 1010

Mantissa (23 bits) =
010 0001 1010 1011 1111 1001


Decimal number 11 110 000 111 100.001 111 000 009 65 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1010 1010 - 010 0001 1010 1011 1111 1001

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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