0.000 000 008 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 008 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
0.000 000 008 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: 0.000 000 008 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) 0.000 000 008 1 × 2 = 0 + 0.000 000 016 2;
  • 2) 0.000 000 016 2 × 2 = 0 + 0.000 000 032 4;
  • 3) 0.000 000 032 4 × 2 = 0 + 0.000 000 064 8;
  • 4) 0.000 000 064 8 × 2 = 0 + 0.000 000 129 6;
  • 5) 0.000 000 129 6 × 2 = 0 + 0.000 000 259 2;
  • 6) 0.000 000 259 2 × 2 = 0 + 0.000 000 518 4;
  • 7) 0.000 000 518 4 × 2 = 0 + 0.000 001 036 8;
  • 8) 0.000 001 036 8 × 2 = 0 + 0.000 002 073 6;
  • 9) 0.000 002 073 6 × 2 = 0 + 0.000 004 147 2;
  • 10) 0.000 004 147 2 × 2 = 0 + 0.000 008 294 4;
  • 11) 0.000 008 294 4 × 2 = 0 + 0.000 016 588 8;
  • 12) 0.000 016 588 8 × 2 = 0 + 0.000 033 177 6;
  • 13) 0.000 033 177 6 × 2 = 0 + 0.000 066 355 2;
  • 14) 0.000 066 355 2 × 2 = 0 + 0.000 132 710 4;
  • 15) 0.000 132 710 4 × 2 = 0 + 0.000 265 420 8;
  • 16) 0.000 265 420 8 × 2 = 0 + 0.000 530 841 6;
  • 17) 0.000 530 841 6 × 2 = 0 + 0.001 061 683 2;
  • 18) 0.001 061 683 2 × 2 = 0 + 0.002 123 366 4;
  • 19) 0.002 123 366 4 × 2 = 0 + 0.004 246 732 8;
  • 20) 0.004 246 732 8 × 2 = 0 + 0.008 493 465 6;
  • 21) 0.008 493 465 6 × 2 = 0 + 0.016 986 931 2;
  • 22) 0.016 986 931 2 × 2 = 0 + 0.033 973 862 4;
  • 23) 0.033 973 862 4 × 2 = 0 + 0.067 947 724 8;
  • 24) 0.067 947 724 8 × 2 = 0 + 0.135 895 449 6;
  • 25) 0.135 895 449 6 × 2 = 0 + 0.271 790 899 2;
  • 26) 0.271 790 899 2 × 2 = 0 + 0.543 581 798 4;
  • 27) 0.543 581 798 4 × 2 = 1 + 0.087 163 596 8;
  • 28) 0.087 163 596 8 × 2 = 0 + 0.174 327 193 6;
  • 29) 0.174 327 193 6 × 2 = 0 + 0.348 654 387 2;
  • 30) 0.348 654 387 2 × 2 = 0 + 0.697 308 774 4;
  • 31) 0.697 308 774 4 × 2 = 1 + 0.394 617 548 8;
  • 32) 0.394 617 548 8 × 2 = 0 + 0.789 235 097 6;
  • 33) 0.789 235 097 6 × 2 = 1 + 0.578 470 195 2;
  • 34) 0.578 470 195 2 × 2 = 1 + 0.156 940 390 4;
  • 35) 0.156 940 390 4 × 2 = 0 + 0.313 880 780 8;
  • 36) 0.313 880 780 8 × 2 = 0 + 0.627 761 561 6;
  • 37) 0.627 761 561 6 × 2 = 1 + 0.255 523 123 2;
  • 38) 0.255 523 123 2 × 2 = 0 + 0.511 046 246 4;
  • 39) 0.511 046 246 4 × 2 = 1 + 0.022 092 492 8;
  • 40) 0.022 092 492 8 × 2 = 0 + 0.044 184 985 6;
  • 41) 0.044 184 985 6 × 2 = 0 + 0.088 369 971 2;
  • 42) 0.088 369 971 2 × 2 = 0 + 0.176 739 942 4;
  • 43) 0.176 739 942 4 × 2 = 0 + 0.353 479 884 8;
  • 44) 0.353 479 884 8 × 2 = 0 + 0.706 959 769 6;
  • 45) 0.706 959 769 6 × 2 = 1 + 0.413 919 539 2;
  • 46) 0.413 919 539 2 × 2 = 0 + 0.827 839 078 4;
  • 47) 0.827 839 078 4 × 2 = 1 + 0.655 678 156 8;
  • 48) 0.655 678 156 8 × 2 = 1 + 0.311 356 313 6;
  • 49) 0.311 356 313 6 × 2 = 0 + 0.622 712 627 2;
  • 50) 0.622 712 627 2 × 2 = 1 + 0.245 425 254 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:


0.000 000 008 1(10) =

0.0000 0000 0000 0000 0000 0000 0010 0010 1100 1010 0000 1011 01(2)

5. Positive number before normalization:

0.000 000 008 1(10) =

0.0000 0000 0000 0000 0000 0000 0010 0010 1100 1010 0000 1011 01(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:


0.000 000 008 1(10) =

0.0000 0000 0000 0000 0000 0000 0010 0010 1100 1010 0000 1011 01(2) =

0.0000 0000 0000 0000 0000 0000 0010 0010 1100 1010 0000 1011 01(2) × 20 =

1.0001 0110 0101 0000 0101 101(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):
1.0001 0110 0101 0000 0101 101


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. 000 1011 0010 1000 0010 1101 =

000 1011 0010 1000 0010 1101


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) =
000 1011 0010 1000 0010 1101


Decimal number 0.000 000 008 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 000 1011 0010 1000 0010 1101

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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