-0.000 000 007 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 007 9(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
-0.000 000 007 9(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:

|-0.000 000 007 9| = 0.000 000 007 9


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: 0.000 000 007 9.

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) 0.000 000 007 9 × 2 = 0 + 0.000 000 015 8;
  • 2) 0.000 000 015 8 × 2 = 0 + 0.000 000 031 6;
  • 3) 0.000 000 031 6 × 2 = 0 + 0.000 000 063 2;
  • 4) 0.000 000 063 2 × 2 = 0 + 0.000 000 126 4;
  • 5) 0.000 000 126 4 × 2 = 0 + 0.000 000 252 8;
  • 6) 0.000 000 252 8 × 2 = 0 + 0.000 000 505 6;
  • 7) 0.000 000 505 6 × 2 = 0 + 0.000 001 011 2;
  • 8) 0.000 001 011 2 × 2 = 0 + 0.000 002 022 4;
  • 9) 0.000 002 022 4 × 2 = 0 + 0.000 004 044 8;
  • 10) 0.000 004 044 8 × 2 = 0 + 0.000 008 089 6;
  • 11) 0.000 008 089 6 × 2 = 0 + 0.000 016 179 2;
  • 12) 0.000 016 179 2 × 2 = 0 + 0.000 032 358 4;
  • 13) 0.000 032 358 4 × 2 = 0 + 0.000 064 716 8;
  • 14) 0.000 064 716 8 × 2 = 0 + 0.000 129 433 6;
  • 15) 0.000 129 433 6 × 2 = 0 + 0.000 258 867 2;
  • 16) 0.000 258 867 2 × 2 = 0 + 0.000 517 734 4;
  • 17) 0.000 517 734 4 × 2 = 0 + 0.001 035 468 8;
  • 18) 0.001 035 468 8 × 2 = 0 + 0.002 070 937 6;
  • 19) 0.002 070 937 6 × 2 = 0 + 0.004 141 875 2;
  • 20) 0.004 141 875 2 × 2 = 0 + 0.008 283 750 4;
  • 21) 0.008 283 750 4 × 2 = 0 + 0.016 567 500 8;
  • 22) 0.016 567 500 8 × 2 = 0 + 0.033 135 001 6;
  • 23) 0.033 135 001 6 × 2 = 0 + 0.066 270 003 2;
  • 24) 0.066 270 003 2 × 2 = 0 + 0.132 540 006 4;
  • 25) 0.132 540 006 4 × 2 = 0 + 0.265 080 012 8;
  • 26) 0.265 080 012 8 × 2 = 0 + 0.530 160 025 6;
  • 27) 0.530 160 025 6 × 2 = 1 + 0.060 320 051 2;
  • 28) 0.060 320 051 2 × 2 = 0 + 0.120 640 102 4;
  • 29) 0.120 640 102 4 × 2 = 0 + 0.241 280 204 8;
  • 30) 0.241 280 204 8 × 2 = 0 + 0.482 560 409 6;
  • 31) 0.482 560 409 6 × 2 = 0 + 0.965 120 819 2;
  • 32) 0.965 120 819 2 × 2 = 1 + 0.930 241 638 4;
  • 33) 0.930 241 638 4 × 2 = 1 + 0.860 483 276 8;
  • 34) 0.860 483 276 8 × 2 = 1 + 0.720 966 553 6;
  • 35) 0.720 966 553 6 × 2 = 1 + 0.441 933 107 2;
  • 36) 0.441 933 107 2 × 2 = 0 + 0.883 866 214 4;
  • 37) 0.883 866 214 4 × 2 = 1 + 0.767 732 428 8;
  • 38) 0.767 732 428 8 × 2 = 1 + 0.535 464 857 6;
  • 39) 0.535 464 857 6 × 2 = 1 + 0.070 929 715 2;
  • 40) 0.070 929 715 2 × 2 = 0 + 0.141 859 430 4;
  • 41) 0.141 859 430 4 × 2 = 0 + 0.283 718 860 8;
  • 42) 0.283 718 860 8 × 2 = 0 + 0.567 437 721 6;
  • 43) 0.567 437 721 6 × 2 = 1 + 0.134 875 443 2;
  • 44) 0.134 875 443 2 × 2 = 0 + 0.269 750 886 4;
  • 45) 0.269 750 886 4 × 2 = 0 + 0.539 501 772 8;
  • 46) 0.539 501 772 8 × 2 = 1 + 0.079 003 545 6;
  • 47) 0.079 003 545 6 × 2 = 0 + 0.158 007 091 2;
  • 48) 0.158 007 091 2 × 2 = 0 + 0.316 014 182 4;
  • 49) 0.316 014 182 4 × 2 = 0 + 0.632 028 364 8;
  • 50) 0.632 028 364 8 × 2 = 1 + 0.264 056 729 6;

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:


0.000 000 007 9(10) =

0.0000 0000 0000 0000 0000 0000 0010 0001 1110 1110 0010 0100 01(2)

6. Positive number before normalization:

0.000 000 007 9(10) =

0.0000 0000 0000 0000 0000 0000 0010 0001 1110 1110 0010 0100 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 007 9(10) =

0.0000 0000 0000 0000 0000 0000 0010 0001 1110 1110 0010 0100 01(2) =

0.0000 0000 0000 0000 0000 0000 0010 0001 1110 1110 0010 0100 01(2) × 20 =

1.0000 1111 0111 0001 0010 001(2) × 2-27


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): -27

Mantissa (not normalized):
1.0000 1111 0111 0001 0010 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-27 + 2(8-1) - 1 =

(-27 + 127)(10) =

100(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;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

100(10) =

0110 0100(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. 000 0111 1011 1000 1001 0001 =

000 0111 1011 1000 1001 0001


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 0100

Mantissa (23 bits) =
000 0111 1011 1000 1001 0001


Decimal number -0.000 000 007 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0100 - 000 0111 1011 1000 1001 0001

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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