-0.000 001 54 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 001 54(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
-0.000 001 54(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:

|-0.000 001 54| = 0.000 001 54


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: 0.000 001 54.

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) 0.000 001 54 × 2 = 0 + 0.000 003 08;
  • 2) 0.000 003 08 × 2 = 0 + 0.000 006 16;
  • 3) 0.000 006 16 × 2 = 0 + 0.000 012 32;
  • 4) 0.000 012 32 × 2 = 0 + 0.000 024 64;
  • 5) 0.000 024 64 × 2 = 0 + 0.000 049 28;
  • 6) 0.000 049 28 × 2 = 0 + 0.000 098 56;
  • 7) 0.000 098 56 × 2 = 0 + 0.000 197 12;
  • 8) 0.000 197 12 × 2 = 0 + 0.000 394 24;
  • 9) 0.000 394 24 × 2 = 0 + 0.000 788 48;
  • 10) 0.000 788 48 × 2 = 0 + 0.001 576 96;
  • 11) 0.001 576 96 × 2 = 0 + 0.003 153 92;
  • 12) 0.003 153 92 × 2 = 0 + 0.006 307 84;
  • 13) 0.006 307 84 × 2 = 0 + 0.012 615 68;
  • 14) 0.012 615 68 × 2 = 0 + 0.025 231 36;
  • 15) 0.025 231 36 × 2 = 0 + 0.050 462 72;
  • 16) 0.050 462 72 × 2 = 0 + 0.100 925 44;
  • 17) 0.100 925 44 × 2 = 0 + 0.201 850 88;
  • 18) 0.201 850 88 × 2 = 0 + 0.403 701 76;
  • 19) 0.403 701 76 × 2 = 0 + 0.807 403 52;
  • 20) 0.807 403 52 × 2 = 1 + 0.614 807 04;
  • 21) 0.614 807 04 × 2 = 1 + 0.229 614 08;
  • 22) 0.229 614 08 × 2 = 0 + 0.459 228 16;
  • 23) 0.459 228 16 × 2 = 0 + 0.918 456 32;
  • 24) 0.918 456 32 × 2 = 1 + 0.836 912 64;
  • 25) 0.836 912 64 × 2 = 1 + 0.673 825 28;
  • 26) 0.673 825 28 × 2 = 1 + 0.347 650 56;
  • 27) 0.347 650 56 × 2 = 0 + 0.695 301 12;
  • 28) 0.695 301 12 × 2 = 1 + 0.390 602 24;
  • 29) 0.390 602 24 × 2 = 0 + 0.781 204 48;
  • 30) 0.781 204 48 × 2 = 1 + 0.562 408 96;
  • 31) 0.562 408 96 × 2 = 1 + 0.124 817 92;
  • 32) 0.124 817 92 × 2 = 0 + 0.249 635 84;
  • 33) 0.249 635 84 × 2 = 0 + 0.499 271 68;
  • 34) 0.499 271 68 × 2 = 0 + 0.998 543 36;
  • 35) 0.998 543 36 × 2 = 1 + 0.997 086 72;
  • 36) 0.997 086 72 × 2 = 1 + 0.994 173 44;
  • 37) 0.994 173 44 × 2 = 1 + 0.988 346 88;
  • 38) 0.988 346 88 × 2 = 1 + 0.976 693 76;
  • 39) 0.976 693 76 × 2 = 1 + 0.953 387 52;
  • 40) 0.953 387 52 × 2 = 1 + 0.906 775 04;
  • 41) 0.906 775 04 × 2 = 1 + 0.813 550 08;
  • 42) 0.813 550 08 × 2 = 1 + 0.627 100 16;
  • 43) 0.627 100 16 × 2 = 1 + 0.254 200 32;

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:


0.000 001 54(10) =

0.0000 0000 0000 0000 0001 1001 1101 0110 0011 1111 111(2)

6. Positive number before normalization:

0.000 001 54(10) =

0.0000 0000 0000 0000 0001 1001 1101 0110 0011 1111 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 54(10) =

0.0000 0000 0000 0000 0001 1001 1101 0110 0011 1111 111(2) =

0.0000 0000 0000 0000 0001 1001 1101 0110 0011 1111 111(2) × 20 =

1.1001 1101 0110 0011 1111 111(2) × 2-20


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): -20

Mantissa (not normalized):
1.1001 1101 0110 0011 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(8-1) - 1 =

(-20 + 127)(10) =

107(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;
  • 107 ÷ 2 = 53 + 1;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

107(10) =

0110 1011(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. 100 1110 1011 0001 1111 1111 =

100 1110 1011 0001 1111 1111


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 1011

Mantissa (23 bits) =
100 1110 1011 0001 1111 1111


Decimal number -0.000 001 54 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1011 - 100 1110 1011 0001 1111 1111

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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