-0.000 000 004 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 004 4(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
-0.000 000 004 4(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:

|-0.000 000 004 4| = 0.000 000 004 4


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: 0.000 000 004 4.

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) 0.000 000 004 4 × 2 = 0 + 0.000 000 008 8;
  • 2) 0.000 000 008 8 × 2 = 0 + 0.000 000 017 6;
  • 3) 0.000 000 017 6 × 2 = 0 + 0.000 000 035 2;
  • 4) 0.000 000 035 2 × 2 = 0 + 0.000 000 070 4;
  • 5) 0.000 000 070 4 × 2 = 0 + 0.000 000 140 8;
  • 6) 0.000 000 140 8 × 2 = 0 + 0.000 000 281 6;
  • 7) 0.000 000 281 6 × 2 = 0 + 0.000 000 563 2;
  • 8) 0.000 000 563 2 × 2 = 0 + 0.000 001 126 4;
  • 9) 0.000 001 126 4 × 2 = 0 + 0.000 002 252 8;
  • 10) 0.000 002 252 8 × 2 = 0 + 0.000 004 505 6;
  • 11) 0.000 004 505 6 × 2 = 0 + 0.000 009 011 2;
  • 12) 0.000 009 011 2 × 2 = 0 + 0.000 018 022 4;
  • 13) 0.000 018 022 4 × 2 = 0 + 0.000 036 044 8;
  • 14) 0.000 036 044 8 × 2 = 0 + 0.000 072 089 6;
  • 15) 0.000 072 089 6 × 2 = 0 + 0.000 144 179 2;
  • 16) 0.000 144 179 2 × 2 = 0 + 0.000 288 358 4;
  • 17) 0.000 288 358 4 × 2 = 0 + 0.000 576 716 8;
  • 18) 0.000 576 716 8 × 2 = 0 + 0.001 153 433 6;
  • 19) 0.001 153 433 6 × 2 = 0 + 0.002 306 867 2;
  • 20) 0.002 306 867 2 × 2 = 0 + 0.004 613 734 4;
  • 21) 0.004 613 734 4 × 2 = 0 + 0.009 227 468 8;
  • 22) 0.009 227 468 8 × 2 = 0 + 0.018 454 937 6;
  • 23) 0.018 454 937 6 × 2 = 0 + 0.036 909 875 2;
  • 24) 0.036 909 875 2 × 2 = 0 + 0.073 819 750 4;
  • 25) 0.073 819 750 4 × 2 = 0 + 0.147 639 500 8;
  • 26) 0.147 639 500 8 × 2 = 0 + 0.295 279 001 6;
  • 27) 0.295 279 001 6 × 2 = 0 + 0.590 558 003 2;
  • 28) 0.590 558 003 2 × 2 = 1 + 0.181 116 006 4;
  • 29) 0.181 116 006 4 × 2 = 0 + 0.362 232 012 8;
  • 30) 0.362 232 012 8 × 2 = 0 + 0.724 464 025 6;
  • 31) 0.724 464 025 6 × 2 = 1 + 0.448 928 051 2;
  • 32) 0.448 928 051 2 × 2 = 0 + 0.897 856 102 4;
  • 33) 0.897 856 102 4 × 2 = 1 + 0.795 712 204 8;
  • 34) 0.795 712 204 8 × 2 = 1 + 0.591 424 409 6;
  • 35) 0.591 424 409 6 × 2 = 1 + 0.182 848 819 2;
  • 36) 0.182 848 819 2 × 2 = 0 + 0.365 697 638 4;
  • 37) 0.365 697 638 4 × 2 = 0 + 0.731 395 276 8;
  • 38) 0.731 395 276 8 × 2 = 1 + 0.462 790 553 6;
  • 39) 0.462 790 553 6 × 2 = 0 + 0.925 581 107 2;
  • 40) 0.925 581 107 2 × 2 = 1 + 0.851 162 214 4;
  • 41) 0.851 162 214 4 × 2 = 1 + 0.702 324 428 8;
  • 42) 0.702 324 428 8 × 2 = 1 + 0.404 648 857 6;
  • 43) 0.404 648 857 6 × 2 = 0 + 0.809 297 715 2;
  • 44) 0.809 297 715 2 × 2 = 1 + 0.618 595 430 4;
  • 45) 0.618 595 430 4 × 2 = 1 + 0.237 190 860 8;
  • 46) 0.237 190 860 8 × 2 = 0 + 0.474 381 721 6;
  • 47) 0.474 381 721 6 × 2 = 0 + 0.948 763 443 2;
  • 48) 0.948 763 443 2 × 2 = 1 + 0.897 526 886 4;
  • 49) 0.897 526 886 4 × 2 = 1 + 0.795 053 772 8;
  • 50) 0.795 053 772 8 × 2 = 1 + 0.590 107 545 6;
  • 51) 0.590 107 545 6 × 2 = 1 + 0.180 215 091 2;

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:


0.000 000 004 4(10) =

0.0000 0000 0000 0000 0000 0000 0001 0010 1110 0101 1101 1001 111(2)

6. Positive number before normalization:

0.000 000 004 4(10) =

0.0000 0000 0000 0000 0000 0000 0001 0010 1110 0101 1101 1001 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 004 4(10) =

0.0000 0000 0000 0000 0000 0000 0001 0010 1110 0101 1101 1001 111(2) =

0.0000 0000 0000 0000 0000 0000 0001 0010 1110 0101 1101 1001 111(2) × 20 =

1.0010 1110 0101 1101 1001 111(2) × 2-28


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): -28

Mantissa (not normalized):
1.0010 1110 0101 1101 1001 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(8-1) - 1 =

(-28 + 127)(10) =

99(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;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

99(10) =

0110 0011(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. 001 0111 0010 1110 1100 1111 =

001 0111 0010 1110 1100 1111


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) =
0110 0011

Mantissa (23 bits) =
001 0111 0010 1110 1100 1111


Decimal number -0.000 000 004 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0011 - 001 0111 0010 1110 1100 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