0.000 000 021 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 3(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
0.000 000 021 3(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: 0.000 000 021 3.

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) 0.000 000 021 3 × 2 = 0 + 0.000 000 042 6;
  • 2) 0.000 000 042 6 × 2 = 0 + 0.000 000 085 2;
  • 3) 0.000 000 085 2 × 2 = 0 + 0.000 000 170 4;
  • 4) 0.000 000 170 4 × 2 = 0 + 0.000 000 340 8;
  • 5) 0.000 000 340 8 × 2 = 0 + 0.000 000 681 6;
  • 6) 0.000 000 681 6 × 2 = 0 + 0.000 001 363 2;
  • 7) 0.000 001 363 2 × 2 = 0 + 0.000 002 726 4;
  • 8) 0.000 002 726 4 × 2 = 0 + 0.000 005 452 8;
  • 9) 0.000 005 452 8 × 2 = 0 + 0.000 010 905 6;
  • 10) 0.000 010 905 6 × 2 = 0 + 0.000 021 811 2;
  • 11) 0.000 021 811 2 × 2 = 0 + 0.000 043 622 4;
  • 12) 0.000 043 622 4 × 2 = 0 + 0.000 087 244 8;
  • 13) 0.000 087 244 8 × 2 = 0 + 0.000 174 489 6;
  • 14) 0.000 174 489 6 × 2 = 0 + 0.000 348 979 2;
  • 15) 0.000 348 979 2 × 2 = 0 + 0.000 697 958 4;
  • 16) 0.000 697 958 4 × 2 = 0 + 0.001 395 916 8;
  • 17) 0.001 395 916 8 × 2 = 0 + 0.002 791 833 6;
  • 18) 0.002 791 833 6 × 2 = 0 + 0.005 583 667 2;
  • 19) 0.005 583 667 2 × 2 = 0 + 0.011 167 334 4;
  • 20) 0.011 167 334 4 × 2 = 0 + 0.022 334 668 8;
  • 21) 0.022 334 668 8 × 2 = 0 + 0.044 669 337 6;
  • 22) 0.044 669 337 6 × 2 = 0 + 0.089 338 675 2;
  • 23) 0.089 338 675 2 × 2 = 0 + 0.178 677 350 4;
  • 24) 0.178 677 350 4 × 2 = 0 + 0.357 354 700 8;
  • 25) 0.357 354 700 8 × 2 = 0 + 0.714 709 401 6;
  • 26) 0.714 709 401 6 × 2 = 1 + 0.429 418 803 2;
  • 27) 0.429 418 803 2 × 2 = 0 + 0.858 837 606 4;
  • 28) 0.858 837 606 4 × 2 = 1 + 0.717 675 212 8;
  • 29) 0.717 675 212 8 × 2 = 1 + 0.435 350 425 6;
  • 30) 0.435 350 425 6 × 2 = 0 + 0.870 700 851 2;
  • 31) 0.870 700 851 2 × 2 = 1 + 0.741 401 702 4;
  • 32) 0.741 401 702 4 × 2 = 1 + 0.482 803 404 8;
  • 33) 0.482 803 404 8 × 2 = 0 + 0.965 606 809 6;
  • 34) 0.965 606 809 6 × 2 = 1 + 0.931 213 619 2;
  • 35) 0.931 213 619 2 × 2 = 1 + 0.862 427 238 4;
  • 36) 0.862 427 238 4 × 2 = 1 + 0.724 854 476 8;
  • 37) 0.724 854 476 8 × 2 = 1 + 0.449 708 953 6;
  • 38) 0.449 708 953 6 × 2 = 0 + 0.899 417 907 2;
  • 39) 0.899 417 907 2 × 2 = 1 + 0.798 835 814 4;
  • 40) 0.798 835 814 4 × 2 = 1 + 0.597 671 628 8;
  • 41) 0.597 671 628 8 × 2 = 1 + 0.195 343 257 6;
  • 42) 0.195 343 257 6 × 2 = 0 + 0.390 686 515 2;
  • 43) 0.390 686 515 2 × 2 = 0 + 0.781 373 030 4;
  • 44) 0.781 373 030 4 × 2 = 1 + 0.562 746 060 8;
  • 45) 0.562 746 060 8 × 2 = 1 + 0.125 492 121 6;
  • 46) 0.125 492 121 6 × 2 = 0 + 0.250 984 243 2;
  • 47) 0.250 984 243 2 × 2 = 0 + 0.501 968 486 4;
  • 48) 0.501 968 486 4 × 2 = 1 + 0.003 936 972 8;
  • 49) 0.003 936 972 8 × 2 = 0 + 0.007 873 945 6;

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:


0.000 000 021 3(10) =

0.0000 0000 0000 0000 0000 0000 0101 1011 0111 1011 1001 1001 0(2)

5. Positive number before normalization:

0.000 000 021 3(10) =

0.0000 0000 0000 0000 0000 0000 0101 1011 0111 1011 1001 1001 0(2)

6. 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:


0.000 000 021 3(10) =

0.0000 0000 0000 0000 0000 0000 0101 1011 0111 1011 1001 1001 0(2) =

0.0000 0000 0000 0000 0000 0000 0101 1011 0111 1011 1001 1001 0(2) × 20 =

1.0110 1101 1110 1110 0110 010(2) × 2-26


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): -26

Mantissa (not normalized):
1.0110 1101 1110 1110 0110 010


8. 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)


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;
  • 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;

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) =

101(10) =

0110 0101(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 1111 0111 0011 0010 =

011 0110 1111 0111 0011 0010


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 0101

Mantissa (23 bits) =
011 0110 1111 0111 0011 0010


Decimal number 0.000 000 021 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 011 0110 1111 0111 0011 0010

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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