0.000 030 229 79 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 030 229 79(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
0.000 030 229 79(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: 0.000 030 229 79.

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) 0.000 030 229 79 × 2 = 0 + 0.000 060 459 58;
  • 2) 0.000 060 459 58 × 2 = 0 + 0.000 120 919 16;
  • 3) 0.000 120 919 16 × 2 = 0 + 0.000 241 838 32;
  • 4) 0.000 241 838 32 × 2 = 0 + 0.000 483 676 64;
  • 5) 0.000 483 676 64 × 2 = 0 + 0.000 967 353 28;
  • 6) 0.000 967 353 28 × 2 = 0 + 0.001 934 706 56;
  • 7) 0.001 934 706 56 × 2 = 0 + 0.003 869 413 12;
  • 8) 0.003 869 413 12 × 2 = 0 + 0.007 738 826 24;
  • 9) 0.007 738 826 24 × 2 = 0 + 0.015 477 652 48;
  • 10) 0.015 477 652 48 × 2 = 0 + 0.030 955 304 96;
  • 11) 0.030 955 304 96 × 2 = 0 + 0.061 910 609 92;
  • 12) 0.061 910 609 92 × 2 = 0 + 0.123 821 219 84;
  • 13) 0.123 821 219 84 × 2 = 0 + 0.247 642 439 68;
  • 14) 0.247 642 439 68 × 2 = 0 + 0.495 284 879 36;
  • 15) 0.495 284 879 36 × 2 = 0 + 0.990 569 758 72;
  • 16) 0.990 569 758 72 × 2 = 1 + 0.981 139 517 44;
  • 17) 0.981 139 517 44 × 2 = 1 + 0.962 279 034 88;
  • 18) 0.962 279 034 88 × 2 = 1 + 0.924 558 069 76;
  • 19) 0.924 558 069 76 × 2 = 1 + 0.849 116 139 52;
  • 20) 0.849 116 139 52 × 2 = 1 + 0.698 232 279 04;
  • 21) 0.698 232 279 04 × 2 = 1 + 0.396 464 558 08;
  • 22) 0.396 464 558 08 × 2 = 0 + 0.792 929 116 16;
  • 23) 0.792 929 116 16 × 2 = 1 + 0.585 858 232 32;
  • 24) 0.585 858 232 32 × 2 = 1 + 0.171 716 464 64;
  • 25) 0.171 716 464 64 × 2 = 0 + 0.343 432 929 28;
  • 26) 0.343 432 929 28 × 2 = 0 + 0.686 865 858 56;
  • 27) 0.686 865 858 56 × 2 = 1 + 0.373 731 717 12;
  • 28) 0.373 731 717 12 × 2 = 0 + 0.747 463 434 24;
  • 29) 0.747 463 434 24 × 2 = 1 + 0.494 926 868 48;
  • 30) 0.494 926 868 48 × 2 = 0 + 0.989 853 736 96;
  • 31) 0.989 853 736 96 × 2 = 1 + 0.979 707 473 92;
  • 32) 0.979 707 473 92 × 2 = 1 + 0.959 414 947 84;
  • 33) 0.959 414 947 84 × 2 = 1 + 0.918 829 895 68;
  • 34) 0.918 829 895 68 × 2 = 1 + 0.837 659 791 36;
  • 35) 0.837 659 791 36 × 2 = 1 + 0.675 319 582 72;
  • 36) 0.675 319 582 72 × 2 = 1 + 0.350 639 165 44;
  • 37) 0.350 639 165 44 × 2 = 0 + 0.701 278 330 88;
  • 38) 0.701 278 330 88 × 2 = 1 + 0.402 556 661 76;
  • 39) 0.402 556 661 76 × 2 = 0 + 0.805 113 323 52;

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:


0.000 030 229 79(10) =

0.0000 0000 0000 0001 1111 1011 0010 1011 1111 010(2)

5. Positive number before normalization:

0.000 030 229 79(10) =

0.0000 0000 0000 0001 1111 1011 0010 1011 1111 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the right, so that only one non zero digit remains to the left of it:


0.000 030 229 79(10) =

0.0000 0000 0000 0001 1111 1011 0010 1011 1111 010(2) =

0.0000 0000 0000 0001 1111 1011 0010 1011 1111 010(2) × 20 =

1.1111 1011 0010 1011 1111 010(2) × 2-16


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): -16

Mantissa (not normalized):
1.1111 1011 0010 1011 1111 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-16 + 2(8-1) - 1 =

(-16 + 127)(10) =

111(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;
  • 111 ÷ 2 = 55 + 1;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

111(10) =

0110 1111(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 1101 1001 0101 1111 1010 =

111 1101 1001 0101 1111 1010


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 1111

Mantissa (23 bits) =
111 1101 1001 0101 1111 1010


Decimal number 0.000 030 229 79 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1111 - 111 1101 1001 0101 1111 1010

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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