0.000 000 099 95 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 099 95(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
0.000 000 099 95(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: 0.000 000 099 95.

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) 0.000 000 099 95 × 2 = 0 + 0.000 000 199 9;
  • 2) 0.000 000 199 9 × 2 = 0 + 0.000 000 399 8;
  • 3) 0.000 000 399 8 × 2 = 0 + 0.000 000 799 6;
  • 4) 0.000 000 799 6 × 2 = 0 + 0.000 001 599 2;
  • 5) 0.000 001 599 2 × 2 = 0 + 0.000 003 198 4;
  • 6) 0.000 003 198 4 × 2 = 0 + 0.000 006 396 8;
  • 7) 0.000 006 396 8 × 2 = 0 + 0.000 012 793 6;
  • 8) 0.000 012 793 6 × 2 = 0 + 0.000 025 587 2;
  • 9) 0.000 025 587 2 × 2 = 0 + 0.000 051 174 4;
  • 10) 0.000 051 174 4 × 2 = 0 + 0.000 102 348 8;
  • 11) 0.000 102 348 8 × 2 = 0 + 0.000 204 697 6;
  • 12) 0.000 204 697 6 × 2 = 0 + 0.000 409 395 2;
  • 13) 0.000 409 395 2 × 2 = 0 + 0.000 818 790 4;
  • 14) 0.000 818 790 4 × 2 = 0 + 0.001 637 580 8;
  • 15) 0.001 637 580 8 × 2 = 0 + 0.003 275 161 6;
  • 16) 0.003 275 161 6 × 2 = 0 + 0.006 550 323 2;
  • 17) 0.006 550 323 2 × 2 = 0 + 0.013 100 646 4;
  • 18) 0.013 100 646 4 × 2 = 0 + 0.026 201 292 8;
  • 19) 0.026 201 292 8 × 2 = 0 + 0.052 402 585 6;
  • 20) 0.052 402 585 6 × 2 = 0 + 0.104 805 171 2;
  • 21) 0.104 805 171 2 × 2 = 0 + 0.209 610 342 4;
  • 22) 0.209 610 342 4 × 2 = 0 + 0.419 220 684 8;
  • 23) 0.419 220 684 8 × 2 = 0 + 0.838 441 369 6;
  • 24) 0.838 441 369 6 × 2 = 1 + 0.676 882 739 2;
  • 25) 0.676 882 739 2 × 2 = 1 + 0.353 765 478 4;
  • 26) 0.353 765 478 4 × 2 = 0 + 0.707 530 956 8;
  • 27) 0.707 530 956 8 × 2 = 1 + 0.415 061 913 6;
  • 28) 0.415 061 913 6 × 2 = 0 + 0.830 123 827 2;
  • 29) 0.830 123 827 2 × 2 = 1 + 0.660 247 654 4;
  • 30) 0.660 247 654 4 × 2 = 1 + 0.320 495 308 8;
  • 31) 0.320 495 308 8 × 2 = 0 + 0.640 990 617 6;
  • 32) 0.640 990 617 6 × 2 = 1 + 0.281 981 235 2;
  • 33) 0.281 981 235 2 × 2 = 0 + 0.563 962 470 4;
  • 34) 0.563 962 470 4 × 2 = 1 + 0.127 924 940 8;
  • 35) 0.127 924 940 8 × 2 = 0 + 0.255 849 881 6;
  • 36) 0.255 849 881 6 × 2 = 0 + 0.511 699 763 2;
  • 37) 0.511 699 763 2 × 2 = 1 + 0.023 399 526 4;
  • 38) 0.023 399 526 4 × 2 = 0 + 0.046 799 052 8;
  • 39) 0.046 799 052 8 × 2 = 0 + 0.093 598 105 6;
  • 40) 0.093 598 105 6 × 2 = 0 + 0.187 196 211 2;
  • 41) 0.187 196 211 2 × 2 = 0 + 0.374 392 422 4;
  • 42) 0.374 392 422 4 × 2 = 0 + 0.748 784 844 8;
  • 43) 0.748 784 844 8 × 2 = 1 + 0.497 569 689 6;
  • 44) 0.497 569 689 6 × 2 = 0 + 0.995 139 379 2;
  • 45) 0.995 139 379 2 × 2 = 1 + 0.990 278 758 4;
  • 46) 0.990 278 758 4 × 2 = 1 + 0.980 557 516 8;
  • 47) 0.980 557 516 8 × 2 = 1 + 0.961 115 033 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:


0.000 000 099 95(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0100 1000 0010 111(2)

5. Positive number before normalization:

0.000 000 099 95(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0100 1000 0010 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 099 95(10) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0100 1000 0010 111(2) =

0.0000 0000 0000 0000 0000 0001 1010 1101 0100 1000 0010 111(2) × 20 =

1.1010 1101 0100 1000 0010 111(2) × 2-24


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): -24

Mantissa (not normalized):
1.1010 1101 0100 1000 0010 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 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) =

103(10) =

0110 0111(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. 101 0110 1010 0100 0001 0111 =

101 0110 1010 0100 0001 0111


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 0111

Mantissa (23 bits) =
101 0110 1010 0100 0001 0111


Decimal number 0.000 000 099 95 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 101 0110 1010 0100 0001 0111

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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