56.654 654 654 654 654 654 654 654 654 654 654 678 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 56.654 654 654 654 654 654 654 654 654 654 654 678 7(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
56.654 654 654 654 654 654 654 654 654 654 654 678 7(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: 56.
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;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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.

56(10) =

11 1000(2)


3. Convert to binary (base 2) the fractional part: 0.654 654 654 654 654 654 654 654 654 654 654 678 7.

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) 0.654 654 654 654 654 654 654 654 654 654 654 678 7 × 2 = 1 + 0.309 309 309 309 309 309 309 309 309 309 309 357 4;
  • 2) 0.309 309 309 309 309 309 309 309 309 309 309 357 4 × 2 = 0 + 0.618 618 618 618 618 618 618 618 618 618 618 714 8;
  • 3) 0.618 618 618 618 618 618 618 618 618 618 618 714 8 × 2 = 1 + 0.237 237 237 237 237 237 237 237 237 237 237 429 6;
  • 4) 0.237 237 237 237 237 237 237 237 237 237 237 429 6 × 2 = 0 + 0.474 474 474 474 474 474 474 474 474 474 474 859 2;
  • 5) 0.474 474 474 474 474 474 474 474 474 474 474 859 2 × 2 = 0 + 0.948 948 948 948 948 948 948 948 948 948 949 718 4;
  • 6) 0.948 948 948 948 948 948 948 948 948 948 949 718 4 × 2 = 1 + 0.897 897 897 897 897 897 897 897 897 897 899 436 8;
  • 7) 0.897 897 897 897 897 897 897 897 897 897 899 436 8 × 2 = 1 + 0.795 795 795 795 795 795 795 795 795 795 798 873 6;
  • 8) 0.795 795 795 795 795 795 795 795 795 795 798 873 6 × 2 = 1 + 0.591 591 591 591 591 591 591 591 591 591 597 747 2;
  • 9) 0.591 591 591 591 591 591 591 591 591 591 597 747 2 × 2 = 1 + 0.183 183 183 183 183 183 183 183 183 183 195 494 4;
  • 10) 0.183 183 183 183 183 183 183 183 183 183 195 494 4 × 2 = 0 + 0.366 366 366 366 366 366 366 366 366 366 390 988 8;
  • 11) 0.366 366 366 366 366 366 366 366 366 366 390 988 8 × 2 = 0 + 0.732 732 732 732 732 732 732 732 732 732 781 977 6;
  • 12) 0.732 732 732 732 732 732 732 732 732 732 781 977 6 × 2 = 1 + 0.465 465 465 465 465 465 465 465 465 465 563 955 2;
  • 13) 0.465 465 465 465 465 465 465 465 465 465 563 955 2 × 2 = 0 + 0.930 930 930 930 930 930 930 930 930 931 127 910 4;
  • 14) 0.930 930 930 930 930 930 930 930 930 931 127 910 4 × 2 = 1 + 0.861 861 861 861 861 861 861 861 861 862 255 820 8;
  • 15) 0.861 861 861 861 861 861 861 861 861 862 255 820 8 × 2 = 1 + 0.723 723 723 723 723 723 723 723 723 724 511 641 6;
  • 16) 0.723 723 723 723 723 723 723 723 723 724 511 641 6 × 2 = 1 + 0.447 447 447 447 447 447 447 447 447 449 023 283 2;
  • 17) 0.447 447 447 447 447 447 447 447 447 449 023 283 2 × 2 = 0 + 0.894 894 894 894 894 894 894 894 894 898 046 566 4;
  • 18) 0.894 894 894 894 894 894 894 894 894 898 046 566 4 × 2 = 1 + 0.789 789 789 789 789 789 789 789 789 796 093 132 8;
  • 19) 0.789 789 789 789 789 789 789 789 789 796 093 132 8 × 2 = 1 + 0.579 579 579 579 579 579 579 579 579 592 186 265 6;
  • 20) 0.579 579 579 579 579 579 579 579 579 592 186 265 6 × 2 = 1 + 0.159 159 159 159 159 159 159 159 159 184 372 531 2;
  • 21) 0.159 159 159 159 159 159 159 159 159 184 372 531 2 × 2 = 0 + 0.318 318 318 318 318 318 318 318 318 368 745 062 4;
  • 22) 0.318 318 318 318 318 318 318 318 318 368 745 062 4 × 2 = 0 + 0.636 636 636 636 636 636 636 636 636 737 490 124 8;
  • 23) 0.636 636 636 636 636 636 636 636 636 737 490 124 8 × 2 = 1 + 0.273 273 273 273 273 273 273 273 273 474 980 249 6;
  • 24) 0.273 273 273 273 273 273 273 273 273 474 980 249 6 × 2 = 0 + 0.546 546 546 546 546 546 546 546 546 949 960 499 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:


0.654 654 654 654 654 654 654 654 654 654 654 678 7(10) =

0.1010 0111 1001 0111 0111 0010(2)

5. Positive number before normalization:

56.654 654 654 654 654 654 654 654 654 654 654 678 7(10) =

11 1000.1010 0111 1001 0111 0111 0010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left, so that only one non zero digit remains to the left of it:


56.654 654 654 654 654 654 654 654 654 654 654 678 7(10) =

11 1000.1010 0111 1001 0111 0111 0010(2) =

11 1000.1010 0111 1001 0111 0111 0010(2) × 20 =

1.1100 0101 0011 1100 1011 1011 1001 0(2) × 25


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): 5

Mantissa (not normalized):
1.1100 0101 0011 1100 1011 1011 1001 0


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

5 + 2(8-1) - 1 =

(5 + 127)(10) =

132(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;
  • 132 ÷ 2 = 66 + 0;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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) =

132(10) =

1000 0100(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. 110 0010 1001 1110 0101 1101 11 0010 =

110 0010 1001 1110 0101 1101


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) =
1000 0100

Mantissa (23 bits) =
110 0010 1001 1110 0101 1101


Decimal number 56.654 654 654 654 654 654 654 654 654 654 654 678 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0100 - 110 0010 1001 1110 0101 1101

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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