0.000 000 110 012 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 012 6(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
0.000 000 110 012 6(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: 0.000 000 110 012 6.

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) 0.000 000 110 012 6 × 2 = 0 + 0.000 000 220 025 2;
  • 2) 0.000 000 220 025 2 × 2 = 0 + 0.000 000 440 050 4;
  • 3) 0.000 000 440 050 4 × 2 = 0 + 0.000 000 880 100 8;
  • 4) 0.000 000 880 100 8 × 2 = 0 + 0.000 001 760 201 6;
  • 5) 0.000 001 760 201 6 × 2 = 0 + 0.000 003 520 403 2;
  • 6) 0.000 003 520 403 2 × 2 = 0 + 0.000 007 040 806 4;
  • 7) 0.000 007 040 806 4 × 2 = 0 + 0.000 014 081 612 8;
  • 8) 0.000 014 081 612 8 × 2 = 0 + 0.000 028 163 225 6;
  • 9) 0.000 028 163 225 6 × 2 = 0 + 0.000 056 326 451 2;
  • 10) 0.000 056 326 451 2 × 2 = 0 + 0.000 112 652 902 4;
  • 11) 0.000 112 652 902 4 × 2 = 0 + 0.000 225 305 804 8;
  • 12) 0.000 225 305 804 8 × 2 = 0 + 0.000 450 611 609 6;
  • 13) 0.000 450 611 609 6 × 2 = 0 + 0.000 901 223 219 2;
  • 14) 0.000 901 223 219 2 × 2 = 0 + 0.001 802 446 438 4;
  • 15) 0.001 802 446 438 4 × 2 = 0 + 0.003 604 892 876 8;
  • 16) 0.003 604 892 876 8 × 2 = 0 + 0.007 209 785 753 6;
  • 17) 0.007 209 785 753 6 × 2 = 0 + 0.014 419 571 507 2;
  • 18) 0.014 419 571 507 2 × 2 = 0 + 0.028 839 143 014 4;
  • 19) 0.028 839 143 014 4 × 2 = 0 + 0.057 678 286 028 8;
  • 20) 0.057 678 286 028 8 × 2 = 0 + 0.115 356 572 057 6;
  • 21) 0.115 356 572 057 6 × 2 = 0 + 0.230 713 144 115 2;
  • 22) 0.230 713 144 115 2 × 2 = 0 + 0.461 426 288 230 4;
  • 23) 0.461 426 288 230 4 × 2 = 0 + 0.922 852 576 460 8;
  • 24) 0.922 852 576 460 8 × 2 = 1 + 0.845 705 152 921 6;
  • 25) 0.845 705 152 921 6 × 2 = 1 + 0.691 410 305 843 2;
  • 26) 0.691 410 305 843 2 × 2 = 1 + 0.382 820 611 686 4;
  • 27) 0.382 820 611 686 4 × 2 = 0 + 0.765 641 223 372 8;
  • 28) 0.765 641 223 372 8 × 2 = 1 + 0.531 282 446 745 6;
  • 29) 0.531 282 446 745 6 × 2 = 1 + 0.062 564 893 491 2;
  • 30) 0.062 564 893 491 2 × 2 = 0 + 0.125 129 786 982 4;
  • 31) 0.125 129 786 982 4 × 2 = 0 + 0.250 259 573 964 8;
  • 32) 0.250 259 573 964 8 × 2 = 0 + 0.500 519 147 929 6;
  • 33) 0.500 519 147 929 6 × 2 = 1 + 0.001 038 295 859 2;
  • 34) 0.001 038 295 859 2 × 2 = 0 + 0.002 076 591 718 4;
  • 35) 0.002 076 591 718 4 × 2 = 0 + 0.004 153 183 436 8;
  • 36) 0.004 153 183 436 8 × 2 = 0 + 0.008 306 366 873 6;
  • 37) 0.008 306 366 873 6 × 2 = 0 + 0.016 612 733 747 2;
  • 38) 0.016 612 733 747 2 × 2 = 0 + 0.033 225 467 494 4;
  • 39) 0.033 225 467 494 4 × 2 = 0 + 0.066 450 934 988 8;
  • 40) 0.066 450 934 988 8 × 2 = 0 + 0.132 901 869 977 6;
  • 41) 0.132 901 869 977 6 × 2 = 0 + 0.265 803 739 955 2;
  • 42) 0.265 803 739 955 2 × 2 = 0 + 0.531 607 479 910 4;
  • 43) 0.531 607 479 910 4 × 2 = 1 + 0.063 214 959 820 8;
  • 44) 0.063 214 959 820 8 × 2 = 0 + 0.126 429 919 641 6;
  • 45) 0.126 429 919 641 6 × 2 = 0 + 0.252 859 839 283 2;
  • 46) 0.252 859 839 283 2 × 2 = 0 + 0.505 719 678 566 4;
  • 47) 0.505 719 678 566 4 × 2 = 1 + 0.011 439 357 132 8;

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:


0.000 000 110 012 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0000 0010 001(2)

5. Positive number before normalization:

0.000 000 110 012 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0000 0010 001(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:


0.000 000 110 012 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0000 0010 001(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0000 0010 001(2) × 20 =

1.1101 1000 1000 0000 0010 001(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 0000 0010 001


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 0000 0001 0001 =

110 1100 0100 0000 0001 0001


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 0000 0001 0001


Decimal number 0.000 000 110 012 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0100 0000 0001 0001

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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