-0.000 000 113 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 113(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
-0.000 000 113(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:

|-0.000 000 113| = 0.000 000 113


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: 0.000 000 113.

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) 0.000 000 113 × 2 = 0 + 0.000 000 226;
  • 2) 0.000 000 226 × 2 = 0 + 0.000 000 452;
  • 3) 0.000 000 452 × 2 = 0 + 0.000 000 904;
  • 4) 0.000 000 904 × 2 = 0 + 0.000 001 808;
  • 5) 0.000 001 808 × 2 = 0 + 0.000 003 616;
  • 6) 0.000 003 616 × 2 = 0 + 0.000 007 232;
  • 7) 0.000 007 232 × 2 = 0 + 0.000 014 464;
  • 8) 0.000 014 464 × 2 = 0 + 0.000 028 928;
  • 9) 0.000 028 928 × 2 = 0 + 0.000 057 856;
  • 10) 0.000 057 856 × 2 = 0 + 0.000 115 712;
  • 11) 0.000 115 712 × 2 = 0 + 0.000 231 424;
  • 12) 0.000 231 424 × 2 = 0 + 0.000 462 848;
  • 13) 0.000 462 848 × 2 = 0 + 0.000 925 696;
  • 14) 0.000 925 696 × 2 = 0 + 0.001 851 392;
  • 15) 0.001 851 392 × 2 = 0 + 0.003 702 784;
  • 16) 0.003 702 784 × 2 = 0 + 0.007 405 568;
  • 17) 0.007 405 568 × 2 = 0 + 0.014 811 136;
  • 18) 0.014 811 136 × 2 = 0 + 0.029 622 272;
  • 19) 0.029 622 272 × 2 = 0 + 0.059 244 544;
  • 20) 0.059 244 544 × 2 = 0 + 0.118 489 088;
  • 21) 0.118 489 088 × 2 = 0 + 0.236 978 176;
  • 22) 0.236 978 176 × 2 = 0 + 0.473 956 352;
  • 23) 0.473 956 352 × 2 = 0 + 0.947 912 704;
  • 24) 0.947 912 704 × 2 = 1 + 0.895 825 408;
  • 25) 0.895 825 408 × 2 = 1 + 0.791 650 816;
  • 26) 0.791 650 816 × 2 = 1 + 0.583 301 632;
  • 27) 0.583 301 632 × 2 = 1 + 0.166 603 264;
  • 28) 0.166 603 264 × 2 = 0 + 0.333 206 528;
  • 29) 0.333 206 528 × 2 = 0 + 0.666 413 056;
  • 30) 0.666 413 056 × 2 = 1 + 0.332 826 112;
  • 31) 0.332 826 112 × 2 = 0 + 0.665 652 224;
  • 32) 0.665 652 224 × 2 = 1 + 0.331 304 448;
  • 33) 0.331 304 448 × 2 = 0 + 0.662 608 896;
  • 34) 0.662 608 896 × 2 = 1 + 0.325 217 792;
  • 35) 0.325 217 792 × 2 = 0 + 0.650 435 584;
  • 36) 0.650 435 584 × 2 = 1 + 0.300 871 168;
  • 37) 0.300 871 168 × 2 = 0 + 0.601 742 336;
  • 38) 0.601 742 336 × 2 = 1 + 0.203 484 672;
  • 39) 0.203 484 672 × 2 = 0 + 0.406 969 344;
  • 40) 0.406 969 344 × 2 = 0 + 0.813 938 688;
  • 41) 0.813 938 688 × 2 = 1 + 0.627 877 376;
  • 42) 0.627 877 376 × 2 = 1 + 0.255 754 752;
  • 43) 0.255 754 752 × 2 = 0 + 0.511 509 504;
  • 44) 0.511 509 504 × 2 = 1 + 0.023 019 008;
  • 45) 0.023 019 008 × 2 = 0 + 0.046 038 016;
  • 46) 0.046 038 016 × 2 = 0 + 0.092 076 032;
  • 47) 0.092 076 032 × 2 = 0 + 0.184 152 064;

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:


0.000 000 113(10) =

0.0000 0000 0000 0000 0000 0001 1110 0101 0101 0100 1101 000(2)

6. Positive number before normalization:

0.000 000 113(10) =

0.0000 0000 0000 0000 0000 0001 1110 0101 0101 0100 1101 000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 113(10) =

0.0000 0000 0000 0000 0000 0001 1110 0101 0101 0100 1101 000(2) =

0.0000 0000 0000 0000 0000 0001 1110 0101 0101 0100 1101 000(2) × 20 =

1.1110 0101 0101 0100 1101 000(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1110 0101 0101 0100 1101 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 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) =

103(10) =

0110 0111(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 0010 1010 1010 0110 1000 =

111 0010 1010 1010 0110 1000


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 0111

Mantissa (23 bits) =
111 0010 1010 1010 0110 1000


Decimal number -0.000 000 113 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 111 0010 1010 1010 0110 1000

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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