-0.000 000 032 421 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 032 421 9(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
-0.000 000 032 421 9(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:

|-0.000 000 032 421 9| = 0.000 000 032 421 9


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: 0.000 000 032 421 9.

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) 0.000 000 032 421 9 × 2 = 0 + 0.000 000 064 843 8;
  • 2) 0.000 000 064 843 8 × 2 = 0 + 0.000 000 129 687 6;
  • 3) 0.000 000 129 687 6 × 2 = 0 + 0.000 000 259 375 2;
  • 4) 0.000 000 259 375 2 × 2 = 0 + 0.000 000 518 750 4;
  • 5) 0.000 000 518 750 4 × 2 = 0 + 0.000 001 037 500 8;
  • 6) 0.000 001 037 500 8 × 2 = 0 + 0.000 002 075 001 6;
  • 7) 0.000 002 075 001 6 × 2 = 0 + 0.000 004 150 003 2;
  • 8) 0.000 004 150 003 2 × 2 = 0 + 0.000 008 300 006 4;
  • 9) 0.000 008 300 006 4 × 2 = 0 + 0.000 016 600 012 8;
  • 10) 0.000 016 600 012 8 × 2 = 0 + 0.000 033 200 025 6;
  • 11) 0.000 033 200 025 6 × 2 = 0 + 0.000 066 400 051 2;
  • 12) 0.000 066 400 051 2 × 2 = 0 + 0.000 132 800 102 4;
  • 13) 0.000 132 800 102 4 × 2 = 0 + 0.000 265 600 204 8;
  • 14) 0.000 265 600 204 8 × 2 = 0 + 0.000 531 200 409 6;
  • 15) 0.000 531 200 409 6 × 2 = 0 + 0.001 062 400 819 2;
  • 16) 0.001 062 400 819 2 × 2 = 0 + 0.002 124 801 638 4;
  • 17) 0.002 124 801 638 4 × 2 = 0 + 0.004 249 603 276 8;
  • 18) 0.004 249 603 276 8 × 2 = 0 + 0.008 499 206 553 6;
  • 19) 0.008 499 206 553 6 × 2 = 0 + 0.016 998 413 107 2;
  • 20) 0.016 998 413 107 2 × 2 = 0 + 0.033 996 826 214 4;
  • 21) 0.033 996 826 214 4 × 2 = 0 + 0.067 993 652 428 8;
  • 22) 0.067 993 652 428 8 × 2 = 0 + 0.135 987 304 857 6;
  • 23) 0.135 987 304 857 6 × 2 = 0 + 0.271 974 609 715 2;
  • 24) 0.271 974 609 715 2 × 2 = 0 + 0.543 949 219 430 4;
  • 25) 0.543 949 219 430 4 × 2 = 1 + 0.087 898 438 860 8;
  • 26) 0.087 898 438 860 8 × 2 = 0 + 0.175 796 877 721 6;
  • 27) 0.175 796 877 721 6 × 2 = 0 + 0.351 593 755 443 2;
  • 28) 0.351 593 755 443 2 × 2 = 0 + 0.703 187 510 886 4;
  • 29) 0.703 187 510 886 4 × 2 = 1 + 0.406 375 021 772 8;
  • 30) 0.406 375 021 772 8 × 2 = 0 + 0.812 750 043 545 6;
  • 31) 0.812 750 043 545 6 × 2 = 1 + 0.625 500 087 091 2;
  • 32) 0.625 500 087 091 2 × 2 = 1 + 0.251 000 174 182 4;
  • 33) 0.251 000 174 182 4 × 2 = 0 + 0.502 000 348 364 8;
  • 34) 0.502 000 348 364 8 × 2 = 1 + 0.004 000 696 729 6;
  • 35) 0.004 000 696 729 6 × 2 = 0 + 0.008 001 393 459 2;
  • 36) 0.008 001 393 459 2 × 2 = 0 + 0.016 002 786 918 4;
  • 37) 0.016 002 786 918 4 × 2 = 0 + 0.032 005 573 836 8;
  • 38) 0.032 005 573 836 8 × 2 = 0 + 0.064 011 147 673 6;
  • 39) 0.064 011 147 673 6 × 2 = 0 + 0.128 022 295 347 2;
  • 40) 0.128 022 295 347 2 × 2 = 0 + 0.256 044 590 694 4;
  • 41) 0.256 044 590 694 4 × 2 = 0 + 0.512 089 181 388 8;
  • 42) 0.512 089 181 388 8 × 2 = 1 + 0.024 178 362 777 6;
  • 43) 0.024 178 362 777 6 × 2 = 0 + 0.048 356 725 555 2;
  • 44) 0.048 356 725 555 2 × 2 = 0 + 0.096 713 451 110 4;
  • 45) 0.096 713 451 110 4 × 2 = 0 + 0.193 426 902 220 8;
  • 46) 0.193 426 902 220 8 × 2 = 0 + 0.386 853 804 441 6;
  • 47) 0.386 853 804 441 6 × 2 = 0 + 0.773 707 608 883 2;
  • 48) 0.773 707 608 883 2 × 2 = 1 + 0.547 415 217 766 4;

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:


0.000 000 032 421 9(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0000 0100 0001(2)

6. Positive number before normalization:

0.000 000 032 421 9(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0000 0100 0001(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:


0.000 000 032 421 9(10) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0000 0100 0001(2) =

0.0000 0000 0000 0000 0000 0000 1000 1011 0100 0000 0100 0001(2) × 20 =

1.0001 0110 1000 0000 1000 001(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):
1.0001 0110 1000 0000 1000 001


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 1011 0100 0000 0100 0001 =

000 1011 0100 0000 0100 0001


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 1011 0100 0000 0100 0001


Decimal number -0.000 000 032 421 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0110 - 000 1011 0100 0000 0100 0001

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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