0.000 000 109 997 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 109 997 8(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
0.000 000 109 997 8(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: 0.000 000 109 997 8.

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) 0.000 000 109 997 8 × 2 = 0 + 0.000 000 219 995 6;
  • 2) 0.000 000 219 995 6 × 2 = 0 + 0.000 000 439 991 2;
  • 3) 0.000 000 439 991 2 × 2 = 0 + 0.000 000 879 982 4;
  • 4) 0.000 000 879 982 4 × 2 = 0 + 0.000 001 759 964 8;
  • 5) 0.000 001 759 964 8 × 2 = 0 + 0.000 003 519 929 6;
  • 6) 0.000 003 519 929 6 × 2 = 0 + 0.000 007 039 859 2;
  • 7) 0.000 007 039 859 2 × 2 = 0 + 0.000 014 079 718 4;
  • 8) 0.000 014 079 718 4 × 2 = 0 + 0.000 028 159 436 8;
  • 9) 0.000 028 159 436 8 × 2 = 0 + 0.000 056 318 873 6;
  • 10) 0.000 056 318 873 6 × 2 = 0 + 0.000 112 637 747 2;
  • 11) 0.000 112 637 747 2 × 2 = 0 + 0.000 225 275 494 4;
  • 12) 0.000 225 275 494 4 × 2 = 0 + 0.000 450 550 988 8;
  • 13) 0.000 450 550 988 8 × 2 = 0 + 0.000 901 101 977 6;
  • 14) 0.000 901 101 977 6 × 2 = 0 + 0.001 802 203 955 2;
  • 15) 0.001 802 203 955 2 × 2 = 0 + 0.003 604 407 910 4;
  • 16) 0.003 604 407 910 4 × 2 = 0 + 0.007 208 815 820 8;
  • 17) 0.007 208 815 820 8 × 2 = 0 + 0.014 417 631 641 6;
  • 18) 0.014 417 631 641 6 × 2 = 0 + 0.028 835 263 283 2;
  • 19) 0.028 835 263 283 2 × 2 = 0 + 0.057 670 526 566 4;
  • 20) 0.057 670 526 566 4 × 2 = 0 + 0.115 341 053 132 8;
  • 21) 0.115 341 053 132 8 × 2 = 0 + 0.230 682 106 265 6;
  • 22) 0.230 682 106 265 6 × 2 = 0 + 0.461 364 212 531 2;
  • 23) 0.461 364 212 531 2 × 2 = 0 + 0.922 728 425 062 4;
  • 24) 0.922 728 425 062 4 × 2 = 1 + 0.845 456 850 124 8;
  • 25) 0.845 456 850 124 8 × 2 = 1 + 0.690 913 700 249 6;
  • 26) 0.690 913 700 249 6 × 2 = 1 + 0.381 827 400 499 2;
  • 27) 0.381 827 400 499 2 × 2 = 0 + 0.763 654 800 998 4;
  • 28) 0.763 654 800 998 4 × 2 = 1 + 0.527 309 601 996 8;
  • 29) 0.527 309 601 996 8 × 2 = 1 + 0.054 619 203 993 6;
  • 30) 0.054 619 203 993 6 × 2 = 0 + 0.109 238 407 987 2;
  • 31) 0.109 238 407 987 2 × 2 = 0 + 0.218 476 815 974 4;
  • 32) 0.218 476 815 974 4 × 2 = 0 + 0.436 953 631 948 8;
  • 33) 0.436 953 631 948 8 × 2 = 0 + 0.873 907 263 897 6;
  • 34) 0.873 907 263 897 6 × 2 = 1 + 0.747 814 527 795 2;
  • 35) 0.747 814 527 795 2 × 2 = 1 + 0.495 629 055 590 4;
  • 36) 0.495 629 055 590 4 × 2 = 0 + 0.991 258 111 180 8;
  • 37) 0.991 258 111 180 8 × 2 = 1 + 0.982 516 222 361 6;
  • 38) 0.982 516 222 361 6 × 2 = 1 + 0.965 032 444 723 2;
  • 39) 0.965 032 444 723 2 × 2 = 1 + 0.930 064 889 446 4;
  • 40) 0.930 064 889 446 4 × 2 = 1 + 0.860 129 778 892 8;
  • 41) 0.860 129 778 892 8 × 2 = 1 + 0.720 259 557 785 6;
  • 42) 0.720 259 557 785 6 × 2 = 1 + 0.440 519 115 571 2;
  • 43) 0.440 519 115 571 2 × 2 = 0 + 0.881 038 231 142 4;
  • 44) 0.881 038 231 142 4 × 2 = 1 + 0.762 076 462 284 8;
  • 45) 0.762 076 462 284 8 × 2 = 1 + 0.524 152 924 569 6;
  • 46) 0.524 152 924 569 6 × 2 = 1 + 0.048 305 849 139 2;
  • 47) 0.048 305 849 139 2 × 2 = 0 + 0.096 611 698 278 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:


0.000 000 109 997 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1111 1101 110(2)

5. Positive number before normalization:

0.000 000 109 997 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1111 1101 110(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:


0.000 000 109 997 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1111 1101 110(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1111 1101 110(2) × 20 =

1.1101 1000 0110 1111 1101 110(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 0110 1111 1101 110


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 0011 0111 1110 1110 =

110 1100 0011 0111 1110 1110


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 0011 0111 1110 1110


Decimal number 0.000 000 109 997 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 0111 1110 1110

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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