0.000 000 019 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 019 3(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
0.000 000 019 3(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: 0.000 000 019 3.

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) 0.000 000 019 3 × 2 = 0 + 0.000 000 038 6;
  • 2) 0.000 000 038 6 × 2 = 0 + 0.000 000 077 2;
  • 3) 0.000 000 077 2 × 2 = 0 + 0.000 000 154 4;
  • 4) 0.000 000 154 4 × 2 = 0 + 0.000 000 308 8;
  • 5) 0.000 000 308 8 × 2 = 0 + 0.000 000 617 6;
  • 6) 0.000 000 617 6 × 2 = 0 + 0.000 001 235 2;
  • 7) 0.000 001 235 2 × 2 = 0 + 0.000 002 470 4;
  • 8) 0.000 002 470 4 × 2 = 0 + 0.000 004 940 8;
  • 9) 0.000 004 940 8 × 2 = 0 + 0.000 009 881 6;
  • 10) 0.000 009 881 6 × 2 = 0 + 0.000 019 763 2;
  • 11) 0.000 019 763 2 × 2 = 0 + 0.000 039 526 4;
  • 12) 0.000 039 526 4 × 2 = 0 + 0.000 079 052 8;
  • 13) 0.000 079 052 8 × 2 = 0 + 0.000 158 105 6;
  • 14) 0.000 158 105 6 × 2 = 0 + 0.000 316 211 2;
  • 15) 0.000 316 211 2 × 2 = 0 + 0.000 632 422 4;
  • 16) 0.000 632 422 4 × 2 = 0 + 0.001 264 844 8;
  • 17) 0.001 264 844 8 × 2 = 0 + 0.002 529 689 6;
  • 18) 0.002 529 689 6 × 2 = 0 + 0.005 059 379 2;
  • 19) 0.005 059 379 2 × 2 = 0 + 0.010 118 758 4;
  • 20) 0.010 118 758 4 × 2 = 0 + 0.020 237 516 8;
  • 21) 0.020 237 516 8 × 2 = 0 + 0.040 475 033 6;
  • 22) 0.040 475 033 6 × 2 = 0 + 0.080 950 067 2;
  • 23) 0.080 950 067 2 × 2 = 0 + 0.161 900 134 4;
  • 24) 0.161 900 134 4 × 2 = 0 + 0.323 800 268 8;
  • 25) 0.323 800 268 8 × 2 = 0 + 0.647 600 537 6;
  • 26) 0.647 600 537 6 × 2 = 1 + 0.295 201 075 2;
  • 27) 0.295 201 075 2 × 2 = 0 + 0.590 402 150 4;
  • 28) 0.590 402 150 4 × 2 = 1 + 0.180 804 300 8;
  • 29) 0.180 804 300 8 × 2 = 0 + 0.361 608 601 6;
  • 30) 0.361 608 601 6 × 2 = 0 + 0.723 217 203 2;
  • 31) 0.723 217 203 2 × 2 = 1 + 0.446 434 406 4;
  • 32) 0.446 434 406 4 × 2 = 0 + 0.892 868 812 8;
  • 33) 0.892 868 812 8 × 2 = 1 + 0.785 737 625 6;
  • 34) 0.785 737 625 6 × 2 = 1 + 0.571 475 251 2;
  • 35) 0.571 475 251 2 × 2 = 1 + 0.142 950 502 4;
  • 36) 0.142 950 502 4 × 2 = 0 + 0.285 901 004 8;
  • 37) 0.285 901 004 8 × 2 = 0 + 0.571 802 009 6;
  • 38) 0.571 802 009 6 × 2 = 1 + 0.143 604 019 2;
  • 39) 0.143 604 019 2 × 2 = 0 + 0.287 208 038 4;
  • 40) 0.287 208 038 4 × 2 = 0 + 0.574 416 076 8;
  • 41) 0.574 416 076 8 × 2 = 1 + 0.148 832 153 6;
  • 42) 0.148 832 153 6 × 2 = 0 + 0.297 664 307 2;
  • 43) 0.297 664 307 2 × 2 = 0 + 0.595 328 614 4;
  • 44) 0.595 328 614 4 × 2 = 1 + 0.190 657 228 8;
  • 45) 0.190 657 228 8 × 2 = 0 + 0.381 314 457 6;
  • 46) 0.381 314 457 6 × 2 = 0 + 0.762 628 915 2;
  • 47) 0.762 628 915 2 × 2 = 1 + 0.525 257 830 4;
  • 48) 0.525 257 830 4 × 2 = 1 + 0.050 515 660 8;
  • 49) 0.050 515 660 8 × 2 = 0 + 0.101 031 321 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:


0.000 000 019 3(10) =

0.0000 0000 0000 0000 0000 0000 0101 0010 1110 0100 1001 0011 0(2)

5. Positive number before normalization:

0.000 000 019 3(10) =

0.0000 0000 0000 0000 0000 0000 0101 0010 1110 0100 1001 0011 0(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:


0.000 000 019 3(10) =

0.0000 0000 0000 0000 0000 0000 0101 0010 1110 0100 1001 0011 0(2) =

0.0000 0000 0000 0000 0000 0000 0101 0010 1110 0100 1001 0011 0(2) × 20 =

1.0100 1011 1001 0010 0100 110(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):
1.0100 1011 1001 0010 0100 110


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. 010 0101 1100 1001 0010 0110 =

010 0101 1100 1001 0010 0110


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) =
010 0101 1100 1001 0010 0110


Decimal number 0.000 000 019 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 010 0101 1100 1001 0010 0110

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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