0.000 000 014 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 014 9(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
0.000 000 014 9(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: 0.000 000 014 9.

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) 0.000 000 014 9 × 2 = 0 + 0.000 000 029 8;
  • 2) 0.000 000 029 8 × 2 = 0 + 0.000 000 059 6;
  • 3) 0.000 000 059 6 × 2 = 0 + 0.000 000 119 2;
  • 4) 0.000 000 119 2 × 2 = 0 + 0.000 000 238 4;
  • 5) 0.000 000 238 4 × 2 = 0 + 0.000 000 476 8;
  • 6) 0.000 000 476 8 × 2 = 0 + 0.000 000 953 6;
  • 7) 0.000 000 953 6 × 2 = 0 + 0.000 001 907 2;
  • 8) 0.000 001 907 2 × 2 = 0 + 0.000 003 814 4;
  • 9) 0.000 003 814 4 × 2 = 0 + 0.000 007 628 8;
  • 10) 0.000 007 628 8 × 2 = 0 + 0.000 015 257 6;
  • 11) 0.000 015 257 6 × 2 = 0 + 0.000 030 515 2;
  • 12) 0.000 030 515 2 × 2 = 0 + 0.000 061 030 4;
  • 13) 0.000 061 030 4 × 2 = 0 + 0.000 122 060 8;
  • 14) 0.000 122 060 8 × 2 = 0 + 0.000 244 121 6;
  • 15) 0.000 244 121 6 × 2 = 0 + 0.000 488 243 2;
  • 16) 0.000 488 243 2 × 2 = 0 + 0.000 976 486 4;
  • 17) 0.000 976 486 4 × 2 = 0 + 0.001 952 972 8;
  • 18) 0.001 952 972 8 × 2 = 0 + 0.003 905 945 6;
  • 19) 0.003 905 945 6 × 2 = 0 + 0.007 811 891 2;
  • 20) 0.007 811 891 2 × 2 = 0 + 0.015 623 782 4;
  • 21) 0.015 623 782 4 × 2 = 0 + 0.031 247 564 8;
  • 22) 0.031 247 564 8 × 2 = 0 + 0.062 495 129 6;
  • 23) 0.062 495 129 6 × 2 = 0 + 0.124 990 259 2;
  • 24) 0.124 990 259 2 × 2 = 0 + 0.249 980 518 4;
  • 25) 0.249 980 518 4 × 2 = 0 + 0.499 961 036 8;
  • 26) 0.499 961 036 8 × 2 = 0 + 0.999 922 073 6;
  • 27) 0.999 922 073 6 × 2 = 1 + 0.999 844 147 2;
  • 28) 0.999 844 147 2 × 2 = 1 + 0.999 688 294 4;
  • 29) 0.999 688 294 4 × 2 = 1 + 0.999 376 588 8;
  • 30) 0.999 376 588 8 × 2 = 1 + 0.998 753 177 6;
  • 31) 0.998 753 177 6 × 2 = 1 + 0.997 506 355 2;
  • 32) 0.997 506 355 2 × 2 = 1 + 0.995 012 710 4;
  • 33) 0.995 012 710 4 × 2 = 1 + 0.990 025 420 8;
  • 34) 0.990 025 420 8 × 2 = 1 + 0.980 050 841 6;
  • 35) 0.980 050 841 6 × 2 = 1 + 0.960 101 683 2;
  • 36) 0.960 101 683 2 × 2 = 1 + 0.920 203 366 4;
  • 37) 0.920 203 366 4 × 2 = 1 + 0.840 406 732 8;
  • 38) 0.840 406 732 8 × 2 = 1 + 0.680 813 465 6;
  • 39) 0.680 813 465 6 × 2 = 1 + 0.361 626 931 2;
  • 40) 0.361 626 931 2 × 2 = 0 + 0.723 253 862 4;
  • 41) 0.723 253 862 4 × 2 = 1 + 0.446 507 724 8;
  • 42) 0.446 507 724 8 × 2 = 0 + 0.893 015 449 6;
  • 43) 0.893 015 449 6 × 2 = 1 + 0.786 030 899 2;
  • 44) 0.786 030 899 2 × 2 = 1 + 0.572 061 798 4;
  • 45) 0.572 061 798 4 × 2 = 1 + 0.144 123 596 8;
  • 46) 0.144 123 596 8 × 2 = 0 + 0.288 247 193 6;
  • 47) 0.288 247 193 6 × 2 = 0 + 0.576 494 387 2;
  • 48) 0.576 494 387 2 × 2 = 1 + 0.152 988 774 4;
  • 49) 0.152 988 774 4 × 2 = 0 + 0.305 977 548 8;
  • 50) 0.305 977 548 8 × 2 = 0 + 0.611 955 097 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:


0.000 000 014 9(10) =

0.0000 0000 0000 0000 0000 0000 0011 1111 1111 1110 1011 1001 00(2)

5. Positive number before normalization:

0.000 000 014 9(10) =

0.0000 0000 0000 0000 0000 0000 0011 1111 1111 1110 1011 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:


0.000 000 014 9(10) =

0.0000 0000 0000 0000 0000 0000 0011 1111 1111 1110 1011 1001 00(2) =

0.0000 0000 0000 0000 0000 0000 0011 1111 1111 1110 1011 1001 00(2) × 20 =

1.1111 1111 1111 0101 1100 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):
1.1111 1111 1111 0101 1100 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. 111 1111 1111 1010 1110 0100 =

111 1111 1111 1010 1110 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) =
111 1111 1111 1010 1110 0100


Decimal number 0.000 000 014 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 111 1111 1111 1010 1110 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