0.000 000 004 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 004 3(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
0.000 000 004 3(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: 0.000 000 004 3.

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) 0.000 000 004 3 × 2 = 0 + 0.000 000 008 6;
  • 2) 0.000 000 008 6 × 2 = 0 + 0.000 000 017 2;
  • 3) 0.000 000 017 2 × 2 = 0 + 0.000 000 034 4;
  • 4) 0.000 000 034 4 × 2 = 0 + 0.000 000 068 8;
  • 5) 0.000 000 068 8 × 2 = 0 + 0.000 000 137 6;
  • 6) 0.000 000 137 6 × 2 = 0 + 0.000 000 275 2;
  • 7) 0.000 000 275 2 × 2 = 0 + 0.000 000 550 4;
  • 8) 0.000 000 550 4 × 2 = 0 + 0.000 001 100 8;
  • 9) 0.000 001 100 8 × 2 = 0 + 0.000 002 201 6;
  • 10) 0.000 002 201 6 × 2 = 0 + 0.000 004 403 2;
  • 11) 0.000 004 403 2 × 2 = 0 + 0.000 008 806 4;
  • 12) 0.000 008 806 4 × 2 = 0 + 0.000 017 612 8;
  • 13) 0.000 017 612 8 × 2 = 0 + 0.000 035 225 6;
  • 14) 0.000 035 225 6 × 2 = 0 + 0.000 070 451 2;
  • 15) 0.000 070 451 2 × 2 = 0 + 0.000 140 902 4;
  • 16) 0.000 140 902 4 × 2 = 0 + 0.000 281 804 8;
  • 17) 0.000 281 804 8 × 2 = 0 + 0.000 563 609 6;
  • 18) 0.000 563 609 6 × 2 = 0 + 0.001 127 219 2;
  • 19) 0.001 127 219 2 × 2 = 0 + 0.002 254 438 4;
  • 20) 0.002 254 438 4 × 2 = 0 + 0.004 508 876 8;
  • 21) 0.004 508 876 8 × 2 = 0 + 0.009 017 753 6;
  • 22) 0.009 017 753 6 × 2 = 0 + 0.018 035 507 2;
  • 23) 0.018 035 507 2 × 2 = 0 + 0.036 071 014 4;
  • 24) 0.036 071 014 4 × 2 = 0 + 0.072 142 028 8;
  • 25) 0.072 142 028 8 × 2 = 0 + 0.144 284 057 6;
  • 26) 0.144 284 057 6 × 2 = 0 + 0.288 568 115 2;
  • 27) 0.288 568 115 2 × 2 = 0 + 0.577 136 230 4;
  • 28) 0.577 136 230 4 × 2 = 1 + 0.154 272 460 8;
  • 29) 0.154 272 460 8 × 2 = 0 + 0.308 544 921 6;
  • 30) 0.308 544 921 6 × 2 = 0 + 0.617 089 843 2;
  • 31) 0.617 089 843 2 × 2 = 1 + 0.234 179 686 4;
  • 32) 0.234 179 686 4 × 2 = 0 + 0.468 359 372 8;
  • 33) 0.468 359 372 8 × 2 = 0 + 0.936 718 745 6;
  • 34) 0.936 718 745 6 × 2 = 1 + 0.873 437 491 2;
  • 35) 0.873 437 491 2 × 2 = 1 + 0.746 874 982 4;
  • 36) 0.746 874 982 4 × 2 = 1 + 0.493 749 964 8;
  • 37) 0.493 749 964 8 × 2 = 0 + 0.987 499 929 6;
  • 38) 0.987 499 929 6 × 2 = 1 + 0.974 999 859 2;
  • 39) 0.974 999 859 2 × 2 = 1 + 0.949 999 718 4;
  • 40) 0.949 999 718 4 × 2 = 1 + 0.899 999 436 8;
  • 41) 0.899 999 436 8 × 2 = 1 + 0.799 998 873 6;
  • 42) 0.799 998 873 6 × 2 = 1 + 0.599 997 747 2;
  • 43) 0.599 997 747 2 × 2 = 1 + 0.199 995 494 4;
  • 44) 0.199 995 494 4 × 2 = 0 + 0.399 990 988 8;
  • 45) 0.399 990 988 8 × 2 = 0 + 0.799 981 977 6;
  • 46) 0.799 981 977 6 × 2 = 1 + 0.599 963 955 2;
  • 47) 0.599 963 955 2 × 2 = 1 + 0.199 927 910 4;
  • 48) 0.199 927 910 4 × 2 = 0 + 0.399 855 820 8;
  • 49) 0.399 855 820 8 × 2 = 0 + 0.799 711 641 6;
  • 50) 0.799 711 641 6 × 2 = 1 + 0.599 423 283 2;
  • 51) 0.599 423 283 2 × 2 = 1 + 0.198 846 566 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:


0.000 000 004 3(10) =

0.0000 0000 0000 0000 0000 0000 0001 0010 0111 0111 1110 0110 011(2)

5. Positive number before normalization:

0.000 000 004 3(10) =

0.0000 0000 0000 0000 0000 0000 0001 0010 0111 0111 1110 0110 011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 004 3(10) =

0.0000 0000 0000 0000 0000 0000 0001 0010 0111 0111 1110 0110 011(2) =

0.0000 0000 0000 0000 0000 0000 0001 0010 0111 0111 1110 0110 011(2) × 20 =

1.0010 0111 0111 1110 0110 011(2) × 2-28


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): -28

Mantissa (not normalized):
1.0010 0111 0111 1110 0110 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(8-1) - 1 =

(-28 + 127)(10) =

99(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;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

99(10) =

0110 0011(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. 001 0011 1011 1111 0011 0011 =

001 0011 1011 1111 0011 0011


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 0011

Mantissa (23 bits) =
001 0011 1011 1111 0011 0011


Decimal number 0.000 000 004 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0011 - 001 0011 1011 1111 0011 0011

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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