0.000 000 010 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 010 6(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
0.000 000 010 6(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: 0.000 000 010 6.

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) 0.000 000 010 6 × 2 = 0 + 0.000 000 021 2;
  • 2) 0.000 000 021 2 × 2 = 0 + 0.000 000 042 4;
  • 3) 0.000 000 042 4 × 2 = 0 + 0.000 000 084 8;
  • 4) 0.000 000 084 8 × 2 = 0 + 0.000 000 169 6;
  • 5) 0.000 000 169 6 × 2 = 0 + 0.000 000 339 2;
  • 6) 0.000 000 339 2 × 2 = 0 + 0.000 000 678 4;
  • 7) 0.000 000 678 4 × 2 = 0 + 0.000 001 356 8;
  • 8) 0.000 001 356 8 × 2 = 0 + 0.000 002 713 6;
  • 9) 0.000 002 713 6 × 2 = 0 + 0.000 005 427 2;
  • 10) 0.000 005 427 2 × 2 = 0 + 0.000 010 854 4;
  • 11) 0.000 010 854 4 × 2 = 0 + 0.000 021 708 8;
  • 12) 0.000 021 708 8 × 2 = 0 + 0.000 043 417 6;
  • 13) 0.000 043 417 6 × 2 = 0 + 0.000 086 835 2;
  • 14) 0.000 086 835 2 × 2 = 0 + 0.000 173 670 4;
  • 15) 0.000 173 670 4 × 2 = 0 + 0.000 347 340 8;
  • 16) 0.000 347 340 8 × 2 = 0 + 0.000 694 681 6;
  • 17) 0.000 694 681 6 × 2 = 0 + 0.001 389 363 2;
  • 18) 0.001 389 363 2 × 2 = 0 + 0.002 778 726 4;
  • 19) 0.002 778 726 4 × 2 = 0 + 0.005 557 452 8;
  • 20) 0.005 557 452 8 × 2 = 0 + 0.011 114 905 6;
  • 21) 0.011 114 905 6 × 2 = 0 + 0.022 229 811 2;
  • 22) 0.022 229 811 2 × 2 = 0 + 0.044 459 622 4;
  • 23) 0.044 459 622 4 × 2 = 0 + 0.088 919 244 8;
  • 24) 0.088 919 244 8 × 2 = 0 + 0.177 838 489 6;
  • 25) 0.177 838 489 6 × 2 = 0 + 0.355 676 979 2;
  • 26) 0.355 676 979 2 × 2 = 0 + 0.711 353 958 4;
  • 27) 0.711 353 958 4 × 2 = 1 + 0.422 707 916 8;
  • 28) 0.422 707 916 8 × 2 = 0 + 0.845 415 833 6;
  • 29) 0.845 415 833 6 × 2 = 1 + 0.690 831 667 2;
  • 30) 0.690 831 667 2 × 2 = 1 + 0.381 663 334 4;
  • 31) 0.381 663 334 4 × 2 = 0 + 0.763 326 668 8;
  • 32) 0.763 326 668 8 × 2 = 1 + 0.526 653 337 6;
  • 33) 0.526 653 337 6 × 2 = 1 + 0.053 306 675 2;
  • 34) 0.053 306 675 2 × 2 = 0 + 0.106 613 350 4;
  • 35) 0.106 613 350 4 × 2 = 0 + 0.213 226 700 8;
  • 36) 0.213 226 700 8 × 2 = 0 + 0.426 453 401 6;
  • 37) 0.426 453 401 6 × 2 = 0 + 0.852 906 803 2;
  • 38) 0.852 906 803 2 × 2 = 1 + 0.705 813 606 4;
  • 39) 0.705 813 606 4 × 2 = 1 + 0.411 627 212 8;
  • 40) 0.411 627 212 8 × 2 = 0 + 0.823 254 425 6;
  • 41) 0.823 254 425 6 × 2 = 1 + 0.646 508 851 2;
  • 42) 0.646 508 851 2 × 2 = 1 + 0.293 017 702 4;
  • 43) 0.293 017 702 4 × 2 = 0 + 0.586 035 404 8;
  • 44) 0.586 035 404 8 × 2 = 1 + 0.172 070 809 6;
  • 45) 0.172 070 809 6 × 2 = 0 + 0.344 141 619 2;
  • 46) 0.344 141 619 2 × 2 = 0 + 0.688 283 238 4;
  • 47) 0.688 283 238 4 × 2 = 1 + 0.376 566 476 8;
  • 48) 0.376 566 476 8 × 2 = 0 + 0.753 132 953 6;
  • 49) 0.753 132 953 6 × 2 = 1 + 0.506 265 907 2;
  • 50) 0.506 265 907 2 × 2 = 1 + 0.012 531 814 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).


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:


0.000 000 010 6(10) =

0.0000 0000 0000 0000 0000 0000 0010 1101 1000 0110 1101 0010 11(2)

5. Positive number before normalization:

0.000 000 010 6(10) =

0.0000 0000 0000 0000 0000 0000 0010 1101 1000 0110 1101 0010 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 010 6(10) =

0.0000 0000 0000 0000 0000 0000 0010 1101 1000 0110 1101 0010 11(2) =

0.0000 0000 0000 0000 0000 0000 0010 1101 1000 0110 1101 0010 11(2) × 20 =

1.0110 1100 0011 0110 1001 011(2) × 2-27


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): -27

Mantissa (not normalized):
1.0110 1100 0011 0110 1001 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-27 + 2(8-1) - 1 =

(-27 + 127)(10) =

100(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;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

100(10) =

0110 0100(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. 011 0110 0001 1011 0100 1011 =

011 0110 0001 1011 0100 1011


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) =
0110 0100

Mantissa (23 bits) =
011 0110 0001 1011 0100 1011


Decimal number 0.000 000 010 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 011 0110 0001 1011 0100 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