-0.000 000 014 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 014 4(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
-0.000 000 014 4(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. Start with the positive version of the number:

|-0.000 000 014 4| = 0.000 000 014 4


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 014 4.

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) 0.000 000 014 4 × 2 = 0 + 0.000 000 028 8;
  • 2) 0.000 000 028 8 × 2 = 0 + 0.000 000 057 6;
  • 3) 0.000 000 057 6 × 2 = 0 + 0.000 000 115 2;
  • 4) 0.000 000 115 2 × 2 = 0 + 0.000 000 230 4;
  • 5) 0.000 000 230 4 × 2 = 0 + 0.000 000 460 8;
  • 6) 0.000 000 460 8 × 2 = 0 + 0.000 000 921 6;
  • 7) 0.000 000 921 6 × 2 = 0 + 0.000 001 843 2;
  • 8) 0.000 001 843 2 × 2 = 0 + 0.000 003 686 4;
  • 9) 0.000 003 686 4 × 2 = 0 + 0.000 007 372 8;
  • 10) 0.000 007 372 8 × 2 = 0 + 0.000 014 745 6;
  • 11) 0.000 014 745 6 × 2 = 0 + 0.000 029 491 2;
  • 12) 0.000 029 491 2 × 2 = 0 + 0.000 058 982 4;
  • 13) 0.000 058 982 4 × 2 = 0 + 0.000 117 964 8;
  • 14) 0.000 117 964 8 × 2 = 0 + 0.000 235 929 6;
  • 15) 0.000 235 929 6 × 2 = 0 + 0.000 471 859 2;
  • 16) 0.000 471 859 2 × 2 = 0 + 0.000 943 718 4;
  • 17) 0.000 943 718 4 × 2 = 0 + 0.001 887 436 8;
  • 18) 0.001 887 436 8 × 2 = 0 + 0.003 774 873 6;
  • 19) 0.003 774 873 6 × 2 = 0 + 0.007 549 747 2;
  • 20) 0.007 549 747 2 × 2 = 0 + 0.015 099 494 4;
  • 21) 0.015 099 494 4 × 2 = 0 + 0.030 198 988 8;
  • 22) 0.030 198 988 8 × 2 = 0 + 0.060 397 977 6;
  • 23) 0.060 397 977 6 × 2 = 0 + 0.120 795 955 2;
  • 24) 0.120 795 955 2 × 2 = 0 + 0.241 591 910 4;
  • 25) 0.241 591 910 4 × 2 = 0 + 0.483 183 820 8;
  • 26) 0.483 183 820 8 × 2 = 0 + 0.966 367 641 6;
  • 27) 0.966 367 641 6 × 2 = 1 + 0.932 735 283 2;
  • 28) 0.932 735 283 2 × 2 = 1 + 0.865 470 566 4;
  • 29) 0.865 470 566 4 × 2 = 1 + 0.730 941 132 8;
  • 30) 0.730 941 132 8 × 2 = 1 + 0.461 882 265 6;
  • 31) 0.461 882 265 6 × 2 = 0 + 0.923 764 531 2;
  • 32) 0.923 764 531 2 × 2 = 1 + 0.847 529 062 4;
  • 33) 0.847 529 062 4 × 2 = 1 + 0.695 058 124 8;
  • 34) 0.695 058 124 8 × 2 = 1 + 0.390 116 249 6;
  • 35) 0.390 116 249 6 × 2 = 0 + 0.780 232 499 2;
  • 36) 0.780 232 499 2 × 2 = 1 + 0.560 464 998 4;
  • 37) 0.560 464 998 4 × 2 = 1 + 0.120 929 996 8;
  • 38) 0.120 929 996 8 × 2 = 0 + 0.241 859 993 6;
  • 39) 0.241 859 993 6 × 2 = 0 + 0.483 719 987 2;
  • 40) 0.483 719 987 2 × 2 = 0 + 0.967 439 974 4;
  • 41) 0.967 439 974 4 × 2 = 1 + 0.934 879 948 8;
  • 42) 0.934 879 948 8 × 2 = 1 + 0.869 759 897 6;
  • 43) 0.869 759 897 6 × 2 = 1 + 0.739 519 795 2;
  • 44) 0.739 519 795 2 × 2 = 1 + 0.479 039 590 4;
  • 45) 0.479 039 590 4 × 2 = 0 + 0.958 079 180 8;
  • 46) 0.958 079 180 8 × 2 = 1 + 0.916 158 361 6;
  • 47) 0.916 158 361 6 × 2 = 1 + 0.832 316 723 2;
  • 48) 0.832 316 723 2 × 2 = 1 + 0.664 633 446 4;
  • 49) 0.664 633 446 4 × 2 = 1 + 0.329 266 892 8;
  • 50) 0.329 266 892 8 × 2 = 0 + 0.658 533 785 6;

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).


5. 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:


0.000 000 014 4(10) =

0.0000 0000 0000 0000 0000 0000 0011 1101 1101 1000 1111 0111 10(2)

6. Positive number before normalization:

0.000 000 014 4(10) =

0.0000 0000 0000 0000 0000 0000 0011 1101 1101 1000 1111 0111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 014 4(10) =

0.0000 0000 0000 0000 0000 0000 0011 1101 1101 1000 1111 0111 10(2) =

0.0000 0000 0000 0000 0000 0000 0011 1101 1101 1000 1111 0111 10(2) × 20 =

1.1110 1110 1100 0111 1011 110(2) × 2-27


8. 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 1 (a negative number)

Exponent (unadjusted): -27

Mantissa (not normalized):
1.1110 1110 1100 0111 1011 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-27 + 2(8-1) - 1 =

(-27 + 127)(10) =

100(10)


10. 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;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

100(10) =

0110 0100(2)


12. 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. 111 0111 0110 0011 1101 1110 =

111 0111 0110 0011 1101 1110


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0100

Mantissa (23 bits) =
111 0111 0110 0011 1101 1110


Decimal number -0.000 000 014 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0100 - 111 0111 0110 0011 1101 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