0.000 000 109 988 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 109 988 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
0.000 000 109 988 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: 0.000 000 109 988 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) 0.000 000 109 988 8 × 2 = 0 + 0.000 000 219 977 6;
  • 2) 0.000 000 219 977 6 × 2 = 0 + 0.000 000 439 955 2;
  • 3) 0.000 000 439 955 2 × 2 = 0 + 0.000 000 879 910 4;
  • 4) 0.000 000 879 910 4 × 2 = 0 + 0.000 001 759 820 8;
  • 5) 0.000 001 759 820 8 × 2 = 0 + 0.000 003 519 641 6;
  • 6) 0.000 003 519 641 6 × 2 = 0 + 0.000 007 039 283 2;
  • 7) 0.000 007 039 283 2 × 2 = 0 + 0.000 014 078 566 4;
  • 8) 0.000 014 078 566 4 × 2 = 0 + 0.000 028 157 132 8;
  • 9) 0.000 028 157 132 8 × 2 = 0 + 0.000 056 314 265 6;
  • 10) 0.000 056 314 265 6 × 2 = 0 + 0.000 112 628 531 2;
  • 11) 0.000 112 628 531 2 × 2 = 0 + 0.000 225 257 062 4;
  • 12) 0.000 225 257 062 4 × 2 = 0 + 0.000 450 514 124 8;
  • 13) 0.000 450 514 124 8 × 2 = 0 + 0.000 901 028 249 6;
  • 14) 0.000 901 028 249 6 × 2 = 0 + 0.001 802 056 499 2;
  • 15) 0.001 802 056 499 2 × 2 = 0 + 0.003 604 112 998 4;
  • 16) 0.003 604 112 998 4 × 2 = 0 + 0.007 208 225 996 8;
  • 17) 0.007 208 225 996 8 × 2 = 0 + 0.014 416 451 993 6;
  • 18) 0.014 416 451 993 6 × 2 = 0 + 0.028 832 903 987 2;
  • 19) 0.028 832 903 987 2 × 2 = 0 + 0.057 665 807 974 4;
  • 20) 0.057 665 807 974 4 × 2 = 0 + 0.115 331 615 948 8;
  • 21) 0.115 331 615 948 8 × 2 = 0 + 0.230 663 231 897 6;
  • 22) 0.230 663 231 897 6 × 2 = 0 + 0.461 326 463 795 2;
  • 23) 0.461 326 463 795 2 × 2 = 0 + 0.922 652 927 590 4;
  • 24) 0.922 652 927 590 4 × 2 = 1 + 0.845 305 855 180 8;
  • 25) 0.845 305 855 180 8 × 2 = 1 + 0.690 611 710 361 6;
  • 26) 0.690 611 710 361 6 × 2 = 1 + 0.381 223 420 723 2;
  • 27) 0.381 223 420 723 2 × 2 = 0 + 0.762 446 841 446 4;
  • 28) 0.762 446 841 446 4 × 2 = 1 + 0.524 893 682 892 8;
  • 29) 0.524 893 682 892 8 × 2 = 1 + 0.049 787 365 785 6;
  • 30) 0.049 787 365 785 6 × 2 = 0 + 0.099 574 731 571 2;
  • 31) 0.099 574 731 571 2 × 2 = 0 + 0.199 149 463 142 4;
  • 32) 0.199 149 463 142 4 × 2 = 0 + 0.398 298 926 284 8;
  • 33) 0.398 298 926 284 8 × 2 = 0 + 0.796 597 852 569 6;
  • 34) 0.796 597 852 569 6 × 2 = 1 + 0.593 195 705 139 2;
  • 35) 0.593 195 705 139 2 × 2 = 1 + 0.186 391 410 278 4;
  • 36) 0.186 391 410 278 4 × 2 = 0 + 0.372 782 820 556 8;
  • 37) 0.372 782 820 556 8 × 2 = 0 + 0.745 565 641 113 6;
  • 38) 0.745 565 641 113 6 × 2 = 1 + 0.491 131 282 227 2;
  • 39) 0.491 131 282 227 2 × 2 = 0 + 0.982 262 564 454 4;
  • 40) 0.982 262 564 454 4 × 2 = 1 + 0.964 525 128 908 8;
  • 41) 0.964 525 128 908 8 × 2 = 1 + 0.929 050 257 817 6;
  • 42) 0.929 050 257 817 6 × 2 = 1 + 0.858 100 515 635 2;
  • 43) 0.858 100 515 635 2 × 2 = 1 + 0.716 201 031 270 4;
  • 44) 0.716 201 031 270 4 × 2 = 1 + 0.432 402 062 540 8;
  • 45) 0.432 402 062 540 8 × 2 = 0 + 0.864 804 125 081 6;
  • 46) 0.864 804 125 081 6 × 2 = 1 + 0.729 608 250 163 2;
  • 47) 0.729 608 250 163 2 × 2 = 1 + 0.459 216 500 326 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:


0.000 000 109 988 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0101 1111 011(2)

5. Positive number before normalization:

0.000 000 109 988 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0101 1111 011(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:


0.000 000 109 988 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0101 1111 011(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0101 1111 011(2) × 20 =

1.1101 1000 0110 0101 1111 011(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 0101 1111 011


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 0010 1111 1011 =

110 1100 0011 0010 1111 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) =
0110 0111

Mantissa (23 bits) =
110 1100 0011 0010 1111 1011


Decimal number 0.000 000 109 988 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 0010 1111 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