0.000 000 000 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 1(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
0.000 000 000 1(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 1.

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) 0.000 000 000 1 × 2 = 0 + 0.000 000 000 2;
  • 2) 0.000 000 000 2 × 2 = 0 + 0.000 000 000 4;
  • 3) 0.000 000 000 4 × 2 = 0 + 0.000 000 000 8;
  • 4) 0.000 000 000 8 × 2 = 0 + 0.000 000 001 6;
  • 5) 0.000 000 001 6 × 2 = 0 + 0.000 000 003 2;
  • 6) 0.000 000 003 2 × 2 = 0 + 0.000 000 006 4;
  • 7) 0.000 000 006 4 × 2 = 0 + 0.000 000 012 8;
  • 8) 0.000 000 012 8 × 2 = 0 + 0.000 000 025 6;
  • 9) 0.000 000 025 6 × 2 = 0 + 0.000 000 051 2;
  • 10) 0.000 000 051 2 × 2 = 0 + 0.000 000 102 4;
  • 11) 0.000 000 102 4 × 2 = 0 + 0.000 000 204 8;
  • 12) 0.000 000 204 8 × 2 = 0 + 0.000 000 409 6;
  • 13) 0.000 000 409 6 × 2 = 0 + 0.000 000 819 2;
  • 14) 0.000 000 819 2 × 2 = 0 + 0.000 001 638 4;
  • 15) 0.000 001 638 4 × 2 = 0 + 0.000 003 276 8;
  • 16) 0.000 003 276 8 × 2 = 0 + 0.000 006 553 6;
  • 17) 0.000 006 553 6 × 2 = 0 + 0.000 013 107 2;
  • 18) 0.000 013 107 2 × 2 = 0 + 0.000 026 214 4;
  • 19) 0.000 026 214 4 × 2 = 0 + 0.000 052 428 8;
  • 20) 0.000 052 428 8 × 2 = 0 + 0.000 104 857 6;
  • 21) 0.000 104 857 6 × 2 = 0 + 0.000 209 715 2;
  • 22) 0.000 209 715 2 × 2 = 0 + 0.000 419 430 4;
  • 23) 0.000 419 430 4 × 2 = 0 + 0.000 838 860 8;
  • 24) 0.000 838 860 8 × 2 = 0 + 0.001 677 721 6;
  • 25) 0.001 677 721 6 × 2 = 0 + 0.003 355 443 2;
  • 26) 0.003 355 443 2 × 2 = 0 + 0.006 710 886 4;
  • 27) 0.006 710 886 4 × 2 = 0 + 0.013 421 772 8;
  • 28) 0.013 421 772 8 × 2 = 0 + 0.026 843 545 6;
  • 29) 0.026 843 545 6 × 2 = 0 + 0.053 687 091 2;
  • 30) 0.053 687 091 2 × 2 = 0 + 0.107 374 182 4;
  • 31) 0.107 374 182 4 × 2 = 0 + 0.214 748 364 8;
  • 32) 0.214 748 364 8 × 2 = 0 + 0.429 496 729 6;
  • 33) 0.429 496 729 6 × 2 = 0 + 0.858 993 459 2;
  • 34) 0.858 993 459 2 × 2 = 1 + 0.717 986 918 4;
  • 35) 0.717 986 918 4 × 2 = 1 + 0.435 973 836 8;
  • 36) 0.435 973 836 8 × 2 = 0 + 0.871 947 673 6;
  • 37) 0.871 947 673 6 × 2 = 1 + 0.743 895 347 2;
  • 38) 0.743 895 347 2 × 2 = 1 + 0.487 790 694 4;
  • 39) 0.487 790 694 4 × 2 = 0 + 0.975 581 388 8;
  • 40) 0.975 581 388 8 × 2 = 1 + 0.951 162 777 6;
  • 41) 0.951 162 777 6 × 2 = 1 + 0.902 325 555 2;
  • 42) 0.902 325 555 2 × 2 = 1 + 0.804 651 110 4;
  • 43) 0.804 651 110 4 × 2 = 1 + 0.609 302 220 8;
  • 44) 0.609 302 220 8 × 2 = 1 + 0.218 604 441 6;
  • 45) 0.218 604 441 6 × 2 = 0 + 0.437 208 883 2;
  • 46) 0.437 208 883 2 × 2 = 0 + 0.874 417 766 4;
  • 47) 0.874 417 766 4 × 2 = 1 + 0.748 835 532 8;
  • 48) 0.748 835 532 8 × 2 = 1 + 0.497 671 065 6;
  • 49) 0.497 671 065 6 × 2 = 0 + 0.995 342 131 2;
  • 50) 0.995 342 131 2 × 2 = 1 + 0.990 684 262 4;
  • 51) 0.990 684 262 4 × 2 = 1 + 0.981 368 524 8;
  • 52) 0.981 368 524 8 × 2 = 1 + 0.962 737 049 6;
  • 53) 0.962 737 049 6 × 2 = 1 + 0.925 474 099 2;
  • 54) 0.925 474 099 2 × 2 = 1 + 0.850 948 198 4;
  • 55) 0.850 948 198 4 × 2 = 1 + 0.701 896 396 8;
  • 56) 0.701 896 396 8 × 2 = 1 + 0.403 792 793 6;
  • 57) 0.403 792 793 6 × 2 = 0 + 0.807 585 587 2;

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).


4. 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:


0.000 000 000 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0110 1101 1111 0011 0111 1111 0(2)

5. Positive number before normalization:

0.000 000 000 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0110 1101 1111 0011 0111 1111 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 34 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0110 1101 1111 0011 0111 1111 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0110 1101 1111 0011 0111 1111 0(2) × 20 =

1.1011 0111 1100 1101 1111 110(2) × 2-34


7. 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 0 (a positive number)

Exponent (unadjusted): -34

Mantissa (not normalized):
1.1011 0111 1100 1101 1111 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-34 + 2(8-1) - 1 =

(-34 + 127)(10) =

93(10)


9. 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;
  • 93 ÷ 2 = 46 + 1;
  • 46 ÷ 2 = 23 + 0;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

93(10) =

0101 1101(2)


11. 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. 101 1011 1110 0110 1111 1110 =

101 1011 1110 0110 1111 1110


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0101 1101

Mantissa (23 bits) =
101 1011 1110 0110 1111 1110


Decimal number 0.000 000 000 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0101 1101 - 101 1011 1110 0110 1111 1110

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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