-0.000 000 031 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 031 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
-0.000 000 031 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. Start with the positive version of the number:

|-0.000 000 031 3| = 0.000 000 031 3


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: 0.000 000 031 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) 0.000 000 031 3 × 2 = 0 + 0.000 000 062 6;
  • 2) 0.000 000 062 6 × 2 = 0 + 0.000 000 125 2;
  • 3) 0.000 000 125 2 × 2 = 0 + 0.000 000 250 4;
  • 4) 0.000 000 250 4 × 2 = 0 + 0.000 000 500 8;
  • 5) 0.000 000 500 8 × 2 = 0 + 0.000 001 001 6;
  • 6) 0.000 001 001 6 × 2 = 0 + 0.000 002 003 2;
  • 7) 0.000 002 003 2 × 2 = 0 + 0.000 004 006 4;
  • 8) 0.000 004 006 4 × 2 = 0 + 0.000 008 012 8;
  • 9) 0.000 008 012 8 × 2 = 0 + 0.000 016 025 6;
  • 10) 0.000 016 025 6 × 2 = 0 + 0.000 032 051 2;
  • 11) 0.000 032 051 2 × 2 = 0 + 0.000 064 102 4;
  • 12) 0.000 064 102 4 × 2 = 0 + 0.000 128 204 8;
  • 13) 0.000 128 204 8 × 2 = 0 + 0.000 256 409 6;
  • 14) 0.000 256 409 6 × 2 = 0 + 0.000 512 819 2;
  • 15) 0.000 512 819 2 × 2 = 0 + 0.001 025 638 4;
  • 16) 0.001 025 638 4 × 2 = 0 + 0.002 051 276 8;
  • 17) 0.002 051 276 8 × 2 = 0 + 0.004 102 553 6;
  • 18) 0.004 102 553 6 × 2 = 0 + 0.008 205 107 2;
  • 19) 0.008 205 107 2 × 2 = 0 + 0.016 410 214 4;
  • 20) 0.016 410 214 4 × 2 = 0 + 0.032 820 428 8;
  • 21) 0.032 820 428 8 × 2 = 0 + 0.065 640 857 6;
  • 22) 0.065 640 857 6 × 2 = 0 + 0.131 281 715 2;
  • 23) 0.131 281 715 2 × 2 = 0 + 0.262 563 430 4;
  • 24) 0.262 563 430 4 × 2 = 0 + 0.525 126 860 8;
  • 25) 0.525 126 860 8 × 2 = 1 + 0.050 253 721 6;
  • 26) 0.050 253 721 6 × 2 = 0 + 0.100 507 443 2;
  • 27) 0.100 507 443 2 × 2 = 0 + 0.201 014 886 4;
  • 28) 0.201 014 886 4 × 2 = 0 + 0.402 029 772 8;
  • 29) 0.402 029 772 8 × 2 = 0 + 0.804 059 545 6;
  • 30) 0.804 059 545 6 × 2 = 1 + 0.608 119 091 2;
  • 31) 0.608 119 091 2 × 2 = 1 + 0.216 238 182 4;
  • 32) 0.216 238 182 4 × 2 = 0 + 0.432 476 364 8;
  • 33) 0.432 476 364 8 × 2 = 0 + 0.864 952 729 6;
  • 34) 0.864 952 729 6 × 2 = 1 + 0.729 905 459 2;
  • 35) 0.729 905 459 2 × 2 = 1 + 0.459 810 918 4;
  • 36) 0.459 810 918 4 × 2 = 0 + 0.919 621 836 8;
  • 37) 0.919 621 836 8 × 2 = 1 + 0.839 243 673 6;
  • 38) 0.839 243 673 6 × 2 = 1 + 0.678 487 347 2;
  • 39) 0.678 487 347 2 × 2 = 1 + 0.356 974 694 4;
  • 40) 0.356 974 694 4 × 2 = 0 + 0.713 949 388 8;
  • 41) 0.713 949 388 8 × 2 = 1 + 0.427 898 777 6;
  • 42) 0.427 898 777 6 × 2 = 0 + 0.855 797 555 2;
  • 43) 0.855 797 555 2 × 2 = 1 + 0.711 595 110 4;
  • 44) 0.711 595 110 4 × 2 = 1 + 0.423 190 220 8;
  • 45) 0.423 190 220 8 × 2 = 0 + 0.846 380 441 6;
  • 46) 0.846 380 441 6 × 2 = 1 + 0.692 760 883 2;
  • 47) 0.692 760 883 2 × 2 = 1 + 0.385 521 766 4;
  • 48) 0.385 521 766 4 × 2 = 0 + 0.771 043 532 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).


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:


0.000 000 031 3(10) =

0.0000 0000 0000 0000 0000 0000 1000 0110 0110 1110 1011 0110(2)

6. Positive number before normalization:

0.000 000 031 3(10) =

0.0000 0000 0000 0000 0000 0000 1000 0110 0110 1110 1011 0110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 031 3(10) =

0.0000 0000 0000 0000 0000 0000 1000 0110 0110 1110 1011 0110(2) =

0.0000 0000 0000 0000 0000 0000 1000 0110 0110 1110 1011 0110(2) × 20 =

1.0000 1100 1101 1101 0110 110(2) × 2-25


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): -25

Mantissa (not normalized):
1.0000 1100 1101 1101 0110 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-25 + 2(8-1) - 1 =

(-25 + 127)(10) =

102(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;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 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) =

102(10) =

0110 0110(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. 000 0110 0110 1110 1011 0110 =

000 0110 0110 1110 1011 0110


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 0110

Mantissa (23 bits) =
000 0110 0110 1110 1011 0110


Decimal number -0.000 000 031 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0110 - 000 0110 0110 1110 1011 0110

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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