0.000 000 011 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 011 7(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
0.000 000 011 7(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: 0.000 000 011 7.

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) 0.000 000 011 7 × 2 = 0 + 0.000 000 023 4;
  • 2) 0.000 000 023 4 × 2 = 0 + 0.000 000 046 8;
  • 3) 0.000 000 046 8 × 2 = 0 + 0.000 000 093 6;
  • 4) 0.000 000 093 6 × 2 = 0 + 0.000 000 187 2;
  • 5) 0.000 000 187 2 × 2 = 0 + 0.000 000 374 4;
  • 6) 0.000 000 374 4 × 2 = 0 + 0.000 000 748 8;
  • 7) 0.000 000 748 8 × 2 = 0 + 0.000 001 497 6;
  • 8) 0.000 001 497 6 × 2 = 0 + 0.000 002 995 2;
  • 9) 0.000 002 995 2 × 2 = 0 + 0.000 005 990 4;
  • 10) 0.000 005 990 4 × 2 = 0 + 0.000 011 980 8;
  • 11) 0.000 011 980 8 × 2 = 0 + 0.000 023 961 6;
  • 12) 0.000 023 961 6 × 2 = 0 + 0.000 047 923 2;
  • 13) 0.000 047 923 2 × 2 = 0 + 0.000 095 846 4;
  • 14) 0.000 095 846 4 × 2 = 0 + 0.000 191 692 8;
  • 15) 0.000 191 692 8 × 2 = 0 + 0.000 383 385 6;
  • 16) 0.000 383 385 6 × 2 = 0 + 0.000 766 771 2;
  • 17) 0.000 766 771 2 × 2 = 0 + 0.001 533 542 4;
  • 18) 0.001 533 542 4 × 2 = 0 + 0.003 067 084 8;
  • 19) 0.003 067 084 8 × 2 = 0 + 0.006 134 169 6;
  • 20) 0.006 134 169 6 × 2 = 0 + 0.012 268 339 2;
  • 21) 0.012 268 339 2 × 2 = 0 + 0.024 536 678 4;
  • 22) 0.024 536 678 4 × 2 = 0 + 0.049 073 356 8;
  • 23) 0.049 073 356 8 × 2 = 0 + 0.098 146 713 6;
  • 24) 0.098 146 713 6 × 2 = 0 + 0.196 293 427 2;
  • 25) 0.196 293 427 2 × 2 = 0 + 0.392 586 854 4;
  • 26) 0.392 586 854 4 × 2 = 0 + 0.785 173 708 8;
  • 27) 0.785 173 708 8 × 2 = 1 + 0.570 347 417 6;
  • 28) 0.570 347 417 6 × 2 = 1 + 0.140 694 835 2;
  • 29) 0.140 694 835 2 × 2 = 0 + 0.281 389 670 4;
  • 30) 0.281 389 670 4 × 2 = 0 + 0.562 779 340 8;
  • 31) 0.562 779 340 8 × 2 = 1 + 0.125 558 681 6;
  • 32) 0.125 558 681 6 × 2 = 0 + 0.251 117 363 2;
  • 33) 0.251 117 363 2 × 2 = 0 + 0.502 234 726 4;
  • 34) 0.502 234 726 4 × 2 = 1 + 0.004 469 452 8;
  • 35) 0.004 469 452 8 × 2 = 0 + 0.008 938 905 6;
  • 36) 0.008 938 905 6 × 2 = 0 + 0.017 877 811 2;
  • 37) 0.017 877 811 2 × 2 = 0 + 0.035 755 622 4;
  • 38) 0.035 755 622 4 × 2 = 0 + 0.071 511 244 8;
  • 39) 0.071 511 244 8 × 2 = 0 + 0.143 022 489 6;
  • 40) 0.143 022 489 6 × 2 = 0 + 0.286 044 979 2;
  • 41) 0.286 044 979 2 × 2 = 0 + 0.572 089 958 4;
  • 42) 0.572 089 958 4 × 2 = 1 + 0.144 179 916 8;
  • 43) 0.144 179 916 8 × 2 = 0 + 0.288 359 833 6;
  • 44) 0.288 359 833 6 × 2 = 0 + 0.576 719 667 2;
  • 45) 0.576 719 667 2 × 2 = 1 + 0.153 439 334 4;
  • 46) 0.153 439 334 4 × 2 = 0 + 0.306 878 668 8;
  • 47) 0.306 878 668 8 × 2 = 0 + 0.613 757 337 6;
  • 48) 0.613 757 337 6 × 2 = 1 + 0.227 514 675 2;
  • 49) 0.227 514 675 2 × 2 = 0 + 0.455 029 350 4;
  • 50) 0.455 029 350 4 × 2 = 0 + 0.910 058 700 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:


0.000 000 011 7(10) =

0.0000 0000 0000 0000 0000 0000 0011 0010 0100 0000 0100 1001 00(2)

5. Positive number before normalization:

0.000 000 011 7(10) =

0.0000 0000 0000 0000 0000 0000 0011 0010 0100 0000 0100 1001 00(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:


0.000 000 011 7(10) =

0.0000 0000 0000 0000 0000 0000 0011 0010 0100 0000 0100 1001 00(2) =

0.0000 0000 0000 0000 0000 0000 0011 0010 0100 0000 0100 1001 00(2) × 20 =

1.1001 0010 0000 0010 0100 100(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):
1.1001 0010 0000 0010 0100 100


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. 100 1001 0000 0001 0010 0100 =

100 1001 0000 0001 0010 0100


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) =
100 1001 0000 0001 0010 0100


Decimal number 0.000 000 011 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 100 1001 0000 0001 0010 0100

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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