111 111 111 111 111 104.95 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 111 111 111 111 111 104.95(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
111 111 111 111 111 104.95(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: 111 111 111 111 111 104.
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;
  • 111 111 111 111 111 104 ÷ 2 = 55 555 555 555 555 552 + 0;
  • 55 555 555 555 555 552 ÷ 2 = 27 777 777 777 777 776 + 0;
  • 27 777 777 777 777 776 ÷ 2 = 13 888 888 888 888 888 + 0;
  • 13 888 888 888 888 888 ÷ 2 = 6 944 444 444 444 444 + 0;
  • 6 944 444 444 444 444 ÷ 2 = 3 472 222 222 222 222 + 0;
  • 3 472 222 222 222 222 ÷ 2 = 1 736 111 111 111 111 + 0;
  • 1 736 111 111 111 111 ÷ 2 = 868 055 555 555 555 + 1;
  • 868 055 555 555 555 ÷ 2 = 434 027 777 777 777 + 1;
  • 434 027 777 777 777 ÷ 2 = 217 013 888 888 888 + 1;
  • 217 013 888 888 888 ÷ 2 = 108 506 944 444 444 + 0;
  • 108 506 944 444 444 ÷ 2 = 54 253 472 222 222 + 0;
  • 54 253 472 222 222 ÷ 2 = 27 126 736 111 111 + 0;
  • 27 126 736 111 111 ÷ 2 = 13 563 368 055 555 + 1;
  • 13 563 368 055 555 ÷ 2 = 6 781 684 027 777 + 1;
  • 6 781 684 027 777 ÷ 2 = 3 390 842 013 888 + 1;
  • 3 390 842 013 888 ÷ 2 = 1 695 421 006 944 + 0;
  • 1 695 421 006 944 ÷ 2 = 847 710 503 472 + 0;
  • 847 710 503 472 ÷ 2 = 423 855 251 736 + 0;
  • 423 855 251 736 ÷ 2 = 211 927 625 868 + 0;
  • 211 927 625 868 ÷ 2 = 105 963 812 934 + 0;
  • 105 963 812 934 ÷ 2 = 52 981 906 467 + 0;
  • 52 981 906 467 ÷ 2 = 26 490 953 233 + 1;
  • 26 490 953 233 ÷ 2 = 13 245 476 616 + 1;
  • 13 245 476 616 ÷ 2 = 6 622 738 308 + 0;
  • 6 622 738 308 ÷ 2 = 3 311 369 154 + 0;
  • 3 311 369 154 ÷ 2 = 1 655 684 577 + 0;
  • 1 655 684 577 ÷ 2 = 827 842 288 + 1;
  • 827 842 288 ÷ 2 = 413 921 144 + 0;
  • 413 921 144 ÷ 2 = 206 960 572 + 0;
  • 206 960 572 ÷ 2 = 103 480 286 + 0;
  • 103 480 286 ÷ 2 = 51 740 143 + 0;
  • 51 740 143 ÷ 2 = 25 870 071 + 1;
  • 25 870 071 ÷ 2 = 12 935 035 + 1;
  • 12 935 035 ÷ 2 = 6 467 517 + 1;
  • 6 467 517 ÷ 2 = 3 233 758 + 1;
  • 3 233 758 ÷ 2 = 1 616 879 + 0;
  • 1 616 879 ÷ 2 = 808 439 + 1;
  • 808 439 ÷ 2 = 404 219 + 1;
  • 404 219 ÷ 2 = 202 109 + 1;
  • 202 109 ÷ 2 = 101 054 + 1;
  • 101 054 ÷ 2 = 50 527 + 0;
  • 50 527 ÷ 2 = 25 263 + 1;
  • 25 263 ÷ 2 = 12 631 + 1;
  • 12 631 ÷ 2 = 6 315 + 1;
  • 6 315 ÷ 2 = 3 157 + 1;
  • 3 157 ÷ 2 = 1 578 + 1;
  • 1 578 ÷ 2 = 789 + 0;
  • 789 ÷ 2 = 394 + 1;
  • 394 ÷ 2 = 197 + 0;
  • 197 ÷ 2 = 98 + 1;
  • 98 ÷ 2 = 49 + 0;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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.

111 111 111 111 111 104(10) =

1 1000 1010 1011 1110 1111 0111 1000 0100 0110 0000 0111 0001 1100 0000(2)


3. Convert to binary (base 2) the fractional part: 0.95.

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) 0.95 × 2 = 1 + 0.9;
  • 2) 0.9 × 2 = 1 + 0.8;
  • 3) 0.8 × 2 = 1 + 0.6;
  • 4) 0.6 × 2 = 1 + 0.2;
  • 5) 0.2 × 2 = 0 + 0.4;
  • 6) 0.4 × 2 = 0 + 0.8;
  • 7) 0.8 × 2 = 1 + 0.6;
  • 8) 0.6 × 2 = 1 + 0.2;
  • 9) 0.2 × 2 = 0 + 0.4;
  • 10) 0.4 × 2 = 0 + 0.8;
  • 11) 0.8 × 2 = 1 + 0.6;
  • 12) 0.6 × 2 = 1 + 0.2;
  • 13) 0.2 × 2 = 0 + 0.4;
  • 14) 0.4 × 2 = 0 + 0.8;
  • 15) 0.8 × 2 = 1 + 0.6;
  • 16) 0.6 × 2 = 1 + 0.2;
  • 17) 0.2 × 2 = 0 + 0.4;
  • 18) 0.4 × 2 = 0 + 0.8;
  • 19) 0.8 × 2 = 1 + 0.6;
  • 20) 0.6 × 2 = 1 + 0.2;
  • 21) 0.2 × 2 = 0 + 0.4;
  • 22) 0.4 × 2 = 0 + 0.8;
  • 23) 0.8 × 2 = 1 + 0.6;
  • 24) 0.6 × 2 = 1 + 0.2;

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:


0.95(10) =

0.1111 0011 0011 0011 0011 0011(2)

5. Positive number before normalization:

111 111 111 111 111 104.95(10) =

1 1000 1010 1011 1110 1111 0111 1000 0100 0110 0000 0111 0001 1100 0000.1111 0011 0011 0011 0011 0011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 56 positions to the left, so that only one non zero digit remains to the left of it:


111 111 111 111 111 104.95(10) =

1 1000 1010 1011 1110 1111 0111 1000 0100 0110 0000 0111 0001 1100 0000.1111 0011 0011 0011 0011 0011(2) =

1 1000 1010 1011 1110 1111 0111 1000 0100 0110 0000 0111 0001 1100 0000.1111 0011 0011 0011 0011 0011(2) × 20 =

1.1000 1010 1011 1110 1111 0111 1000 0100 0110 0000 0111 0001 1100 0000 1111 0011 0011 0011 0011 0011(2) × 256


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): 56

Mantissa (not normalized):
1.1000 1010 1011 1110 1111 0111 1000 0100 0110 0000 0111 0001 1100 0000 1111 0011 0011 0011 0011 0011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

56 + 2(8-1) - 1 =

(56 + 127)(10) =

183(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;
  • 183 ÷ 2 = 91 + 1;
  • 91 ÷ 2 = 45 + 1;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 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) =

183(10) =

1011 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, 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. 100 0101 0101 1111 0111 1011 1 1000 0100 0110 0000 0111 0001 1100 0000 1111 0011 0011 0011 0011 0011 =

100 0101 0101 1111 0111 1011


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) =
1011 0111

Mantissa (23 bits) =
100 0101 0101 1111 0111 1011


Decimal number 111 111 111 111 111 104.95 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1011 0111 - 100 0101 0101 1111 0111 1011

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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