0.000 000 000 58 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 58(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
0.000 000 000 58(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: 0.000 000 000 58.

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) 0.000 000 000 58 × 2 = 0 + 0.000 000 001 16;
  • 2) 0.000 000 001 16 × 2 = 0 + 0.000 000 002 32;
  • 3) 0.000 000 002 32 × 2 = 0 + 0.000 000 004 64;
  • 4) 0.000 000 004 64 × 2 = 0 + 0.000 000 009 28;
  • 5) 0.000 000 009 28 × 2 = 0 + 0.000 000 018 56;
  • 6) 0.000 000 018 56 × 2 = 0 + 0.000 000 037 12;
  • 7) 0.000 000 037 12 × 2 = 0 + 0.000 000 074 24;
  • 8) 0.000 000 074 24 × 2 = 0 + 0.000 000 148 48;
  • 9) 0.000 000 148 48 × 2 = 0 + 0.000 000 296 96;
  • 10) 0.000 000 296 96 × 2 = 0 + 0.000 000 593 92;
  • 11) 0.000 000 593 92 × 2 = 0 + 0.000 001 187 84;
  • 12) 0.000 001 187 84 × 2 = 0 + 0.000 002 375 68;
  • 13) 0.000 002 375 68 × 2 = 0 + 0.000 004 751 36;
  • 14) 0.000 004 751 36 × 2 = 0 + 0.000 009 502 72;
  • 15) 0.000 009 502 72 × 2 = 0 + 0.000 019 005 44;
  • 16) 0.000 019 005 44 × 2 = 0 + 0.000 038 010 88;
  • 17) 0.000 038 010 88 × 2 = 0 + 0.000 076 021 76;
  • 18) 0.000 076 021 76 × 2 = 0 + 0.000 152 043 52;
  • 19) 0.000 152 043 52 × 2 = 0 + 0.000 304 087 04;
  • 20) 0.000 304 087 04 × 2 = 0 + 0.000 608 174 08;
  • 21) 0.000 608 174 08 × 2 = 0 + 0.001 216 348 16;
  • 22) 0.001 216 348 16 × 2 = 0 + 0.002 432 696 32;
  • 23) 0.002 432 696 32 × 2 = 0 + 0.004 865 392 64;
  • 24) 0.004 865 392 64 × 2 = 0 + 0.009 730 785 28;
  • 25) 0.009 730 785 28 × 2 = 0 + 0.019 461 570 56;
  • 26) 0.019 461 570 56 × 2 = 0 + 0.038 923 141 12;
  • 27) 0.038 923 141 12 × 2 = 0 + 0.077 846 282 24;
  • 28) 0.077 846 282 24 × 2 = 0 + 0.155 692 564 48;
  • 29) 0.155 692 564 48 × 2 = 0 + 0.311 385 128 96;
  • 30) 0.311 385 128 96 × 2 = 0 + 0.622 770 257 92;
  • 31) 0.622 770 257 92 × 2 = 1 + 0.245 540 515 84;
  • 32) 0.245 540 515 84 × 2 = 0 + 0.491 081 031 68;
  • 33) 0.491 081 031 68 × 2 = 0 + 0.982 162 063 36;
  • 34) 0.982 162 063 36 × 2 = 1 + 0.964 324 126 72;
  • 35) 0.964 324 126 72 × 2 = 1 + 0.928 648 253 44;
  • 36) 0.928 648 253 44 × 2 = 1 + 0.857 296 506 88;
  • 37) 0.857 296 506 88 × 2 = 1 + 0.714 593 013 76;
  • 38) 0.714 593 013 76 × 2 = 1 + 0.429 186 027 52;
  • 39) 0.429 186 027 52 × 2 = 0 + 0.858 372 055 04;
  • 40) 0.858 372 055 04 × 2 = 1 + 0.716 744 110 08;
  • 41) 0.716 744 110 08 × 2 = 1 + 0.433 488 220 16;
  • 42) 0.433 488 220 16 × 2 = 0 + 0.866 976 440 32;
  • 43) 0.866 976 440 32 × 2 = 1 + 0.733 952 880 64;
  • 44) 0.733 952 880 64 × 2 = 1 + 0.467 905 761 28;
  • 45) 0.467 905 761 28 × 2 = 0 + 0.935 811 522 56;
  • 46) 0.935 811 522 56 × 2 = 1 + 0.871 623 045 12;
  • 47) 0.871 623 045 12 × 2 = 1 + 0.743 246 090 24;
  • 48) 0.743 246 090 24 × 2 = 1 + 0.486 492 180 48;
  • 49) 0.486 492 180 48 × 2 = 0 + 0.972 984 360 96;
  • 50) 0.972 984 360 96 × 2 = 1 + 0.945 968 721 92;
  • 51) 0.945 968 721 92 × 2 = 1 + 0.891 937 443 84;
  • 52) 0.891 937 443 84 × 2 = 1 + 0.783 874 887 68;
  • 53) 0.783 874 887 68 × 2 = 1 + 0.567 749 775 36;
  • 54) 0.567 749 775 36 × 2 = 1 + 0.135 499 550 72;

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:


0.000 000 000 58(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0111 1101 1011 0111 0111 11(2)

5. Positive number before normalization:

0.000 000 000 58(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0111 1101 1011 0111 0111 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 58(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0111 1101 1011 0111 0111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0111 1101 1011 0111 0111 11(2) × 20 =

1.0011 1110 1101 1011 1011 111(2) × 2-31


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): -31

Mantissa (not normalized):
1.0011 1110 1101 1011 1011 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 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) =

96(10) =

0110 0000(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. 001 1111 0110 1101 1101 1111 =

001 1111 0110 1101 1101 1111


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 0000

Mantissa (23 bits) =
001 1111 0110 1101 1101 1111


Decimal number 0.000 000 000 58 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0000 - 001 1111 0110 1101 1101 1111

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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