0.000 000 001 03 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 001 03(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
0.000 000 001 03(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: 0.000 000 001 03.

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) 0.000 000 001 03 × 2 = 0 + 0.000 000 002 06;
  • 2) 0.000 000 002 06 × 2 = 0 + 0.000 000 004 12;
  • 3) 0.000 000 004 12 × 2 = 0 + 0.000 000 008 24;
  • 4) 0.000 000 008 24 × 2 = 0 + 0.000 000 016 48;
  • 5) 0.000 000 016 48 × 2 = 0 + 0.000 000 032 96;
  • 6) 0.000 000 032 96 × 2 = 0 + 0.000 000 065 92;
  • 7) 0.000 000 065 92 × 2 = 0 + 0.000 000 131 84;
  • 8) 0.000 000 131 84 × 2 = 0 + 0.000 000 263 68;
  • 9) 0.000 000 263 68 × 2 = 0 + 0.000 000 527 36;
  • 10) 0.000 000 527 36 × 2 = 0 + 0.000 001 054 72;
  • 11) 0.000 001 054 72 × 2 = 0 + 0.000 002 109 44;
  • 12) 0.000 002 109 44 × 2 = 0 + 0.000 004 218 88;
  • 13) 0.000 004 218 88 × 2 = 0 + 0.000 008 437 76;
  • 14) 0.000 008 437 76 × 2 = 0 + 0.000 016 875 52;
  • 15) 0.000 016 875 52 × 2 = 0 + 0.000 033 751 04;
  • 16) 0.000 033 751 04 × 2 = 0 + 0.000 067 502 08;
  • 17) 0.000 067 502 08 × 2 = 0 + 0.000 135 004 16;
  • 18) 0.000 135 004 16 × 2 = 0 + 0.000 270 008 32;
  • 19) 0.000 270 008 32 × 2 = 0 + 0.000 540 016 64;
  • 20) 0.000 540 016 64 × 2 = 0 + 0.001 080 033 28;
  • 21) 0.001 080 033 28 × 2 = 0 + 0.002 160 066 56;
  • 22) 0.002 160 066 56 × 2 = 0 + 0.004 320 133 12;
  • 23) 0.004 320 133 12 × 2 = 0 + 0.008 640 266 24;
  • 24) 0.008 640 266 24 × 2 = 0 + 0.017 280 532 48;
  • 25) 0.017 280 532 48 × 2 = 0 + 0.034 561 064 96;
  • 26) 0.034 561 064 96 × 2 = 0 + 0.069 122 129 92;
  • 27) 0.069 122 129 92 × 2 = 0 + 0.138 244 259 84;
  • 28) 0.138 244 259 84 × 2 = 0 + 0.276 488 519 68;
  • 29) 0.276 488 519 68 × 2 = 0 + 0.552 977 039 36;
  • 30) 0.552 977 039 36 × 2 = 1 + 0.105 954 078 72;
  • 31) 0.105 954 078 72 × 2 = 0 + 0.211 908 157 44;
  • 32) 0.211 908 157 44 × 2 = 0 + 0.423 816 314 88;
  • 33) 0.423 816 314 88 × 2 = 0 + 0.847 632 629 76;
  • 34) 0.847 632 629 76 × 2 = 1 + 0.695 265 259 52;
  • 35) 0.695 265 259 52 × 2 = 1 + 0.390 530 519 04;
  • 36) 0.390 530 519 04 × 2 = 0 + 0.781 061 038 08;
  • 37) 0.781 061 038 08 × 2 = 1 + 0.562 122 076 16;
  • 38) 0.562 122 076 16 × 2 = 1 + 0.124 244 152 32;
  • 39) 0.124 244 152 32 × 2 = 0 + 0.248 488 304 64;
  • 40) 0.248 488 304 64 × 2 = 0 + 0.496 976 609 28;
  • 41) 0.496 976 609 28 × 2 = 0 + 0.993 953 218 56;
  • 42) 0.993 953 218 56 × 2 = 1 + 0.987 906 437 12;
  • 43) 0.987 906 437 12 × 2 = 1 + 0.975 812 874 24;
  • 44) 0.975 812 874 24 × 2 = 1 + 0.951 625 748 48;
  • 45) 0.951 625 748 48 × 2 = 1 + 0.903 251 496 96;
  • 46) 0.903 251 496 96 × 2 = 1 + 0.806 502 993 92;
  • 47) 0.806 502 993 92 × 2 = 1 + 0.613 005 987 84;
  • 48) 0.613 005 987 84 × 2 = 1 + 0.226 011 975 68;
  • 49) 0.226 011 975 68 × 2 = 0 + 0.452 023 951 36;
  • 50) 0.452 023 951 36 × 2 = 0 + 0.904 047 902 72;
  • 51) 0.904 047 902 72 × 2 = 1 + 0.808 095 805 44;
  • 52) 0.808 095 805 44 × 2 = 1 + 0.616 191 610 88;
  • 53) 0.616 191 610 88 × 2 = 1 + 0.232 383 221 76;

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:


0.000 000 001 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0110 1100 0111 1111 0011 1(2)

5. Positive number before normalization:

0.000 000 001 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0110 1100 0111 1111 0011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 30 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 001 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0110 1100 0111 1111 0011 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0110 1100 0111 1111 0011 1(2) × 20 =

1.0001 1011 0001 1111 1100 111(2) × 2-30


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): -30

Mantissa (not normalized):
1.0001 1011 0001 1111 1100 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-30 + 2(8-1) - 1 =

(-30 + 127)(10) =

97(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;
  • 97 ÷ 2 = 48 + 1;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

97(10) =

0110 0001(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 1101 1000 1111 1110 0111 =

000 1101 1000 1111 1110 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 0001

Mantissa (23 bits) =
000 1101 1000 1111 1110 0111


Decimal number 0.000 000 001 03 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0001 - 000 1101 1000 1111 1110 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