-0.000 000 000 22 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 22(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 22(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. Start with the positive version of the number:

|-0.000 000 000 22| = 0.000 000 000 22


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 22.

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 22 × 2 = 0 + 0.000 000 000 44;
  • 2) 0.000 000 000 44 × 2 = 0 + 0.000 000 000 88;
  • 3) 0.000 000 000 88 × 2 = 0 + 0.000 000 001 76;
  • 4) 0.000 000 001 76 × 2 = 0 + 0.000 000 003 52;
  • 5) 0.000 000 003 52 × 2 = 0 + 0.000 000 007 04;
  • 6) 0.000 000 007 04 × 2 = 0 + 0.000 000 014 08;
  • 7) 0.000 000 014 08 × 2 = 0 + 0.000 000 028 16;
  • 8) 0.000 000 028 16 × 2 = 0 + 0.000 000 056 32;
  • 9) 0.000 000 056 32 × 2 = 0 + 0.000 000 112 64;
  • 10) 0.000 000 112 64 × 2 = 0 + 0.000 000 225 28;
  • 11) 0.000 000 225 28 × 2 = 0 + 0.000 000 450 56;
  • 12) 0.000 000 450 56 × 2 = 0 + 0.000 000 901 12;
  • 13) 0.000 000 901 12 × 2 = 0 + 0.000 001 802 24;
  • 14) 0.000 001 802 24 × 2 = 0 + 0.000 003 604 48;
  • 15) 0.000 003 604 48 × 2 = 0 + 0.000 007 208 96;
  • 16) 0.000 007 208 96 × 2 = 0 + 0.000 014 417 92;
  • 17) 0.000 014 417 92 × 2 = 0 + 0.000 028 835 84;
  • 18) 0.000 028 835 84 × 2 = 0 + 0.000 057 671 68;
  • 19) 0.000 057 671 68 × 2 = 0 + 0.000 115 343 36;
  • 20) 0.000 115 343 36 × 2 = 0 + 0.000 230 686 72;
  • 21) 0.000 230 686 72 × 2 = 0 + 0.000 461 373 44;
  • 22) 0.000 461 373 44 × 2 = 0 + 0.000 922 746 88;
  • 23) 0.000 922 746 88 × 2 = 0 + 0.001 845 493 76;
  • 24) 0.001 845 493 76 × 2 = 0 + 0.003 690 987 52;
  • 25) 0.003 690 987 52 × 2 = 0 + 0.007 381 975 04;
  • 26) 0.007 381 975 04 × 2 = 0 + 0.014 763 950 08;
  • 27) 0.014 763 950 08 × 2 = 0 + 0.029 527 900 16;
  • 28) 0.029 527 900 16 × 2 = 0 + 0.059 055 800 32;
  • 29) 0.059 055 800 32 × 2 = 0 + 0.118 111 600 64;
  • 30) 0.118 111 600 64 × 2 = 0 + 0.236 223 201 28;
  • 31) 0.236 223 201 28 × 2 = 0 + 0.472 446 402 56;
  • 32) 0.472 446 402 56 × 2 = 0 + 0.944 892 805 12;
  • 33) 0.944 892 805 12 × 2 = 1 + 0.889 785 610 24;
  • 34) 0.889 785 610 24 × 2 = 1 + 0.779 571 220 48;
  • 35) 0.779 571 220 48 × 2 = 1 + 0.559 142 440 96;
  • 36) 0.559 142 440 96 × 2 = 1 + 0.118 284 881 92;
  • 37) 0.118 284 881 92 × 2 = 0 + 0.236 569 763 84;
  • 38) 0.236 569 763 84 × 2 = 0 + 0.473 139 527 68;
  • 39) 0.473 139 527 68 × 2 = 0 + 0.946 279 055 36;
  • 40) 0.946 279 055 36 × 2 = 1 + 0.892 558 110 72;
  • 41) 0.892 558 110 72 × 2 = 1 + 0.785 116 221 44;
  • 42) 0.785 116 221 44 × 2 = 1 + 0.570 232 442 88;
  • 43) 0.570 232 442 88 × 2 = 1 + 0.140 464 885 76;
  • 44) 0.140 464 885 76 × 2 = 0 + 0.280 929 771 52;
  • 45) 0.280 929 771 52 × 2 = 0 + 0.561 859 543 04;
  • 46) 0.561 859 543 04 × 2 = 1 + 0.123 719 086 08;
  • 47) 0.123 719 086 08 × 2 = 0 + 0.247 438 172 16;
  • 48) 0.247 438 172 16 × 2 = 0 + 0.494 876 344 32;
  • 49) 0.494 876 344 32 × 2 = 0 + 0.989 752 688 64;
  • 50) 0.989 752 688 64 × 2 = 1 + 0.979 505 377 28;
  • 51) 0.979 505 377 28 × 2 = 1 + 0.959 010 754 56;
  • 52) 0.959 010 754 56 × 2 = 1 + 0.918 021 509 12;
  • 53) 0.918 021 509 12 × 2 = 1 + 0.836 043 018 24;
  • 54) 0.836 043 018 24 × 2 = 1 + 0.672 086 036 48;
  • 55) 0.672 086 036 48 × 2 = 1 + 0.344 172 072 96;
  • 56) 0.344 172 072 96 × 2 = 0 + 0.688 344 145 92;

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).


5. 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 22(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1111 0001 1110 0100 0111 1110(2)

6. Positive number before normalization:

0.000 000 000 22(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1111 0001 1110 0100 0111 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 33 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 22(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1111 0001 1110 0100 0111 1110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 1111 0001 1110 0100 0111 1110(2) × 20 =

1.1110 0011 1100 1000 1111 110(2) × 2-33


8. 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 1 (a negative number)

Exponent (unadjusted): -33

Mantissa (not normalized):
1.1110 0011 1100 1000 1111 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-33 + 2(8-1) - 1 =

(-33 + 127)(10) =

94(10)


10. 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;
  • 94 ÷ 2 = 47 + 0;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

94(10) =

0101 1110(2)


12. 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 0001 1110 0100 0111 1110 =

111 0001 1110 0100 0111 1110


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0101 1110

Mantissa (23 bits) =
111 0001 1110 0100 0111 1110


Decimal number -0.000 000 000 22 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1110 - 111 0001 1110 0100 0111 1110

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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