-0.000 000 004 23 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 004 23(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 23(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:

|-0.000 000 004 23| = 0.000 000 004 23


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: 0.000 000 004 23.

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 23 × 2 = 0 + 0.000 000 008 46;
  • 2) 0.000 000 008 46 × 2 = 0 + 0.000 000 016 92;
  • 3) 0.000 000 016 92 × 2 = 0 + 0.000 000 033 84;
  • 4) 0.000 000 033 84 × 2 = 0 + 0.000 000 067 68;
  • 5) 0.000 000 067 68 × 2 = 0 + 0.000 000 135 36;
  • 6) 0.000 000 135 36 × 2 = 0 + 0.000 000 270 72;
  • 7) 0.000 000 270 72 × 2 = 0 + 0.000 000 541 44;
  • 8) 0.000 000 541 44 × 2 = 0 + 0.000 001 082 88;
  • 9) 0.000 001 082 88 × 2 = 0 + 0.000 002 165 76;
  • 10) 0.000 002 165 76 × 2 = 0 + 0.000 004 331 52;
  • 11) 0.000 004 331 52 × 2 = 0 + 0.000 008 663 04;
  • 12) 0.000 008 663 04 × 2 = 0 + 0.000 017 326 08;
  • 13) 0.000 017 326 08 × 2 = 0 + 0.000 034 652 16;
  • 14) 0.000 034 652 16 × 2 = 0 + 0.000 069 304 32;
  • 15) 0.000 069 304 32 × 2 = 0 + 0.000 138 608 64;
  • 16) 0.000 138 608 64 × 2 = 0 + 0.000 277 217 28;
  • 17) 0.000 277 217 28 × 2 = 0 + 0.000 554 434 56;
  • 18) 0.000 554 434 56 × 2 = 0 + 0.001 108 869 12;
  • 19) 0.001 108 869 12 × 2 = 0 + 0.002 217 738 24;
  • 20) 0.002 217 738 24 × 2 = 0 + 0.004 435 476 48;
  • 21) 0.004 435 476 48 × 2 = 0 + 0.008 870 952 96;
  • 22) 0.008 870 952 96 × 2 = 0 + 0.017 741 905 92;
  • 23) 0.017 741 905 92 × 2 = 0 + 0.035 483 811 84;
  • 24) 0.035 483 811 84 × 2 = 0 + 0.070 967 623 68;
  • 25) 0.070 967 623 68 × 2 = 0 + 0.141 935 247 36;
  • 26) 0.141 935 247 36 × 2 = 0 + 0.283 870 494 72;
  • 27) 0.283 870 494 72 × 2 = 0 + 0.567 740 989 44;
  • 28) 0.567 740 989 44 × 2 = 1 + 0.135 481 978 88;
  • 29) 0.135 481 978 88 × 2 = 0 + 0.270 963 957 76;
  • 30) 0.270 963 957 76 × 2 = 0 + 0.541 927 915 52;
  • 31) 0.541 927 915 52 × 2 = 1 + 0.083 855 831 04;
  • 32) 0.083 855 831 04 × 2 = 0 + 0.167 711 662 08;
  • 33) 0.167 711 662 08 × 2 = 0 + 0.335 423 324 16;
  • 34) 0.335 423 324 16 × 2 = 0 + 0.670 846 648 32;
  • 35) 0.670 846 648 32 × 2 = 1 + 0.341 693 296 64;
  • 36) 0.341 693 296 64 × 2 = 0 + 0.683 386 593 28;
  • 37) 0.683 386 593 28 × 2 = 1 + 0.366 773 186 56;
  • 38) 0.366 773 186 56 × 2 = 0 + 0.733 546 373 12;
  • 39) 0.733 546 373 12 × 2 = 1 + 0.467 092 746 24;
  • 40) 0.467 092 746 24 × 2 = 0 + 0.934 185 492 48;
  • 41) 0.934 185 492 48 × 2 = 1 + 0.868 370 984 96;
  • 42) 0.868 370 984 96 × 2 = 1 + 0.736 741 969 92;
  • 43) 0.736 741 969 92 × 2 = 1 + 0.473 483 939 84;
  • 44) 0.473 483 939 84 × 2 = 0 + 0.946 967 879 68;
  • 45) 0.946 967 879 68 × 2 = 1 + 0.893 935 759 36;
  • 46) 0.893 935 759 36 × 2 = 1 + 0.787 871 518 72;
  • 47) 0.787 871 518 72 × 2 = 1 + 0.575 743 037 44;
  • 48) 0.575 743 037 44 × 2 = 1 + 0.151 486 074 88;
  • 49) 0.151 486 074 88 × 2 = 0 + 0.302 972 149 76;
  • 50) 0.302 972 149 76 × 2 = 0 + 0.605 944 299 52;
  • 51) 0.605 944 299 52 × 2 = 1 + 0.211 888 599 04;

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:


0.000 000 004 23(10) =

0.0000 0000 0000 0000 0000 0000 0001 0010 0010 1010 1110 1111 001(2)

6. Positive number before normalization:

0.000 000 004 23(10) =

0.0000 0000 0000 0000 0000 0000 0001 0010 0010 1010 1110 1111 001(2)

7. 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 23(10) =

0.0000 0000 0000 0000 0000 0000 0001 0010 0010 1010 1110 1111 001(2) =

0.0000 0000 0000 0000 0000 0000 0001 0010 0010 1010 1110 1111 001(2) × 20 =

1.0010 0010 1010 1110 1111 001(2) × 2-28


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): -28

Mantissa (not normalized):
1.0010 0010 1010 1110 1111 001


9. 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)


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;
  • 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;

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) =

99(10) =

0110 0011(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. 001 0001 0101 0111 0111 1001 =

001 0001 0101 0111 0111 1001


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 0011

Mantissa (23 bits) =
001 0001 0101 0111 0111 1001


Decimal number -0.000 000 004 23 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0011 - 001 0001 0101 0111 0111 1001

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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