-0.000 000 025 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 025 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
-0.000 000 025 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. Start with the positive version of the number:

|-0.000 000 025 1| = 0.000 000 025 1


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: 0.000 000 025 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) 0.000 000 025 1 × 2 = 0 + 0.000 000 050 2;
  • 2) 0.000 000 050 2 × 2 = 0 + 0.000 000 100 4;
  • 3) 0.000 000 100 4 × 2 = 0 + 0.000 000 200 8;
  • 4) 0.000 000 200 8 × 2 = 0 + 0.000 000 401 6;
  • 5) 0.000 000 401 6 × 2 = 0 + 0.000 000 803 2;
  • 6) 0.000 000 803 2 × 2 = 0 + 0.000 001 606 4;
  • 7) 0.000 001 606 4 × 2 = 0 + 0.000 003 212 8;
  • 8) 0.000 003 212 8 × 2 = 0 + 0.000 006 425 6;
  • 9) 0.000 006 425 6 × 2 = 0 + 0.000 012 851 2;
  • 10) 0.000 012 851 2 × 2 = 0 + 0.000 025 702 4;
  • 11) 0.000 025 702 4 × 2 = 0 + 0.000 051 404 8;
  • 12) 0.000 051 404 8 × 2 = 0 + 0.000 102 809 6;
  • 13) 0.000 102 809 6 × 2 = 0 + 0.000 205 619 2;
  • 14) 0.000 205 619 2 × 2 = 0 + 0.000 411 238 4;
  • 15) 0.000 411 238 4 × 2 = 0 + 0.000 822 476 8;
  • 16) 0.000 822 476 8 × 2 = 0 + 0.001 644 953 6;
  • 17) 0.001 644 953 6 × 2 = 0 + 0.003 289 907 2;
  • 18) 0.003 289 907 2 × 2 = 0 + 0.006 579 814 4;
  • 19) 0.006 579 814 4 × 2 = 0 + 0.013 159 628 8;
  • 20) 0.013 159 628 8 × 2 = 0 + 0.026 319 257 6;
  • 21) 0.026 319 257 6 × 2 = 0 + 0.052 638 515 2;
  • 22) 0.052 638 515 2 × 2 = 0 + 0.105 277 030 4;
  • 23) 0.105 277 030 4 × 2 = 0 + 0.210 554 060 8;
  • 24) 0.210 554 060 8 × 2 = 0 + 0.421 108 121 6;
  • 25) 0.421 108 121 6 × 2 = 0 + 0.842 216 243 2;
  • 26) 0.842 216 243 2 × 2 = 1 + 0.684 432 486 4;
  • 27) 0.684 432 486 4 × 2 = 1 + 0.368 864 972 8;
  • 28) 0.368 864 972 8 × 2 = 0 + 0.737 729 945 6;
  • 29) 0.737 729 945 6 × 2 = 1 + 0.475 459 891 2;
  • 30) 0.475 459 891 2 × 2 = 0 + 0.950 919 782 4;
  • 31) 0.950 919 782 4 × 2 = 1 + 0.901 839 564 8;
  • 32) 0.901 839 564 8 × 2 = 1 + 0.803 679 129 6;
  • 33) 0.803 679 129 6 × 2 = 1 + 0.607 358 259 2;
  • 34) 0.607 358 259 2 × 2 = 1 + 0.214 716 518 4;
  • 35) 0.214 716 518 4 × 2 = 0 + 0.429 433 036 8;
  • 36) 0.429 433 036 8 × 2 = 0 + 0.858 866 073 6;
  • 37) 0.858 866 073 6 × 2 = 1 + 0.717 732 147 2;
  • 38) 0.717 732 147 2 × 2 = 1 + 0.435 464 294 4;
  • 39) 0.435 464 294 4 × 2 = 0 + 0.870 928 588 8;
  • 40) 0.870 928 588 8 × 2 = 1 + 0.741 857 177 6;
  • 41) 0.741 857 177 6 × 2 = 1 + 0.483 714 355 2;
  • 42) 0.483 714 355 2 × 2 = 0 + 0.967 428 710 4;
  • 43) 0.967 428 710 4 × 2 = 1 + 0.934 857 420 8;
  • 44) 0.934 857 420 8 × 2 = 1 + 0.869 714 841 6;
  • 45) 0.869 714 841 6 × 2 = 1 + 0.739 429 683 2;
  • 46) 0.739 429 683 2 × 2 = 1 + 0.478 859 366 4;
  • 47) 0.478 859 366 4 × 2 = 0 + 0.957 718 732 8;
  • 48) 0.957 718 732 8 × 2 = 1 + 0.915 437 465 6;
  • 49) 0.915 437 465 6 × 2 = 1 + 0.830 874 931 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).


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:


0.000 000 025 1(10) =

0.0000 0000 0000 0000 0000 0000 0110 1011 1100 1101 1011 1101 1(2)

6. Positive number before normalization:

0.000 000 025 1(10) =

0.0000 0000 0000 0000 0000 0000 0110 1011 1100 1101 1011 1101 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 025 1(10) =

0.0000 0000 0000 0000 0000 0000 0110 1011 1100 1101 1011 1101 1(2) =

0.0000 0000 0000 0000 0000 0000 0110 1011 1100 1101 1011 1101 1(2) × 20 =

1.1010 1111 0011 0110 1111 011(2) × 2-26


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): -26

Mantissa (not normalized):
1.1010 1111 0011 0110 1111 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(8-1) - 1 =

(-26 + 127)(10) =

101(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;
  • 101 ÷ 2 = 50 + 1;
  • 50 ÷ 2 = 25 + 0;
  • 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) =

101(10) =

0110 0101(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. 101 0111 1001 1011 0111 1011 =

101 0111 1001 1011 0111 1011


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 0101

Mantissa (23 bits) =
101 0111 1001 1011 0111 1011


Decimal number -0.000 000 025 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0101 - 101 0111 1001 1011 0111 1011

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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