0.000 000 021 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 4(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 4(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 4.

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 4 × 2 = 0 + 0.000 000 042 8;
  • 2) 0.000 000 042 8 × 2 = 0 + 0.000 000 085 6;
  • 3) 0.000 000 085 6 × 2 = 0 + 0.000 000 171 2;
  • 4) 0.000 000 171 2 × 2 = 0 + 0.000 000 342 4;
  • 5) 0.000 000 342 4 × 2 = 0 + 0.000 000 684 8;
  • 6) 0.000 000 684 8 × 2 = 0 + 0.000 001 369 6;
  • 7) 0.000 001 369 6 × 2 = 0 + 0.000 002 739 2;
  • 8) 0.000 002 739 2 × 2 = 0 + 0.000 005 478 4;
  • 9) 0.000 005 478 4 × 2 = 0 + 0.000 010 956 8;
  • 10) 0.000 010 956 8 × 2 = 0 + 0.000 021 913 6;
  • 11) 0.000 021 913 6 × 2 = 0 + 0.000 043 827 2;
  • 12) 0.000 043 827 2 × 2 = 0 + 0.000 087 654 4;
  • 13) 0.000 087 654 4 × 2 = 0 + 0.000 175 308 8;
  • 14) 0.000 175 308 8 × 2 = 0 + 0.000 350 617 6;
  • 15) 0.000 350 617 6 × 2 = 0 + 0.000 701 235 2;
  • 16) 0.000 701 235 2 × 2 = 0 + 0.001 402 470 4;
  • 17) 0.001 402 470 4 × 2 = 0 + 0.002 804 940 8;
  • 18) 0.002 804 940 8 × 2 = 0 + 0.005 609 881 6;
  • 19) 0.005 609 881 6 × 2 = 0 + 0.011 219 763 2;
  • 20) 0.011 219 763 2 × 2 = 0 + 0.022 439 526 4;
  • 21) 0.022 439 526 4 × 2 = 0 + 0.044 879 052 8;
  • 22) 0.044 879 052 8 × 2 = 0 + 0.089 758 105 6;
  • 23) 0.089 758 105 6 × 2 = 0 + 0.179 516 211 2;
  • 24) 0.179 516 211 2 × 2 = 0 + 0.359 032 422 4;
  • 25) 0.359 032 422 4 × 2 = 0 + 0.718 064 844 8;
  • 26) 0.718 064 844 8 × 2 = 1 + 0.436 129 689 6;
  • 27) 0.436 129 689 6 × 2 = 0 + 0.872 259 379 2;
  • 28) 0.872 259 379 2 × 2 = 1 + 0.744 518 758 4;
  • 29) 0.744 518 758 4 × 2 = 1 + 0.489 037 516 8;
  • 30) 0.489 037 516 8 × 2 = 0 + 0.978 075 033 6;
  • 31) 0.978 075 033 6 × 2 = 1 + 0.956 150 067 2;
  • 32) 0.956 150 067 2 × 2 = 1 + 0.912 300 134 4;
  • 33) 0.912 300 134 4 × 2 = 1 + 0.824 600 268 8;
  • 34) 0.824 600 268 8 × 2 = 1 + 0.649 200 537 6;
  • 35) 0.649 200 537 6 × 2 = 1 + 0.298 401 075 2;
  • 36) 0.298 401 075 2 × 2 = 0 + 0.596 802 150 4;
  • 37) 0.596 802 150 4 × 2 = 1 + 0.193 604 300 8;
  • 38) 0.193 604 300 8 × 2 = 0 + 0.387 208 601 6;
  • 39) 0.387 208 601 6 × 2 = 0 + 0.774 417 203 2;
  • 40) 0.774 417 203 2 × 2 = 1 + 0.548 834 406 4;
  • 41) 0.548 834 406 4 × 2 = 1 + 0.097 668 812 8;
  • 42) 0.097 668 812 8 × 2 = 0 + 0.195 337 625 6;
  • 43) 0.195 337 625 6 × 2 = 0 + 0.390 675 251 2;
  • 44) 0.390 675 251 2 × 2 = 0 + 0.781 350 502 4;
  • 45) 0.781 350 502 4 × 2 = 1 + 0.562 701 004 8;
  • 46) 0.562 701 004 8 × 2 = 1 + 0.125 402 009 6;
  • 47) 0.125 402 009 6 × 2 = 0 + 0.250 804 019 2;
  • 48) 0.250 804 019 2 × 2 = 0 + 0.501 608 038 4;
  • 49) 0.501 608 038 4 × 2 = 1 + 0.003 216 076 8;

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 4(10) =

0.0000 0000 0000 0000 0000 0000 0101 1011 1110 1001 1000 1100 1(2)

5. Positive number before normalization:

0.000 000 021 4(10) =

0.0000 0000 0000 0000 0000 0000 0101 1011 1110 1001 1000 1100 1(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 4(10) =

0.0000 0000 0000 0000 0000 0000 0101 1011 1110 1001 1000 1100 1(2) =

0.0000 0000 0000 0000 0000 0000 0101 1011 1110 1001 1000 1100 1(2) × 20 =

1.0110 1111 1010 0110 0011 001(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 1111 1010 0110 0011 001


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 0111 1101 0011 0001 1001 =

011 0111 1101 0011 0001 1001


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 0111 1101 0011 0001 1001


Decimal number 0.000 000 021 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 011 0111 1101 0011 0001 1001

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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