Convert -0.000 003 39 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

How to convert the decimal number -0.000 003 39(10)
to
32 bit single precision IEEE 754 binary floating point
(1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 003 39| = 0.000 003 39

2. First, convert to the binary (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 the binary (base 2) the fractional part: 0.000 003 39.

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) 0.000 003 39 × 2 = 0 + 0.000 006 78;
  • 2) 0.000 006 78 × 2 = 0 + 0.000 013 56;
  • 3) 0.000 013 56 × 2 = 0 + 0.000 027 12;
  • 4) 0.000 027 12 × 2 = 0 + 0.000 054 24;
  • 5) 0.000 054 24 × 2 = 0 + 0.000 108 48;
  • 6) 0.000 108 48 × 2 = 0 + 0.000 216 96;
  • 7) 0.000 216 96 × 2 = 0 + 0.000 433 92;
  • 8) 0.000 433 92 × 2 = 0 + 0.000 867 84;
  • 9) 0.000 867 84 × 2 = 0 + 0.001 735 68;
  • 10) 0.001 735 68 × 2 = 0 + 0.003 471 36;
  • 11) 0.003 471 36 × 2 = 0 + 0.006 942 72;
  • 12) 0.006 942 72 × 2 = 0 + 0.013 885 44;
  • 13) 0.013 885 44 × 2 = 0 + 0.027 770 88;
  • 14) 0.027 770 88 × 2 = 0 + 0.055 541 76;
  • 15) 0.055 541 76 × 2 = 0 + 0.111 083 52;
  • 16) 0.111 083 52 × 2 = 0 + 0.222 167 04;
  • 17) 0.222 167 04 × 2 = 0 + 0.444 334 08;
  • 18) 0.444 334 08 × 2 = 0 + 0.888 668 16;
  • 19) 0.888 668 16 × 2 = 1 + 0.777 336 32;
  • 20) 0.777 336 32 × 2 = 1 + 0.554 672 64;
  • 21) 0.554 672 64 × 2 = 1 + 0.109 345 28;
  • 22) 0.109 345 28 × 2 = 0 + 0.218 690 56;
  • 23) 0.218 690 56 × 2 = 0 + 0.437 381 12;
  • 24) 0.437 381 12 × 2 = 0 + 0.874 762 24;
  • 25) 0.874 762 24 × 2 = 1 + 0.749 524 48;
  • 26) 0.749 524 48 × 2 = 1 + 0.499 048 96;
  • 27) 0.499 048 96 × 2 = 0 + 0.998 097 92;
  • 28) 0.998 097 92 × 2 = 1 + 0.996 195 84;
  • 29) 0.996 195 84 × 2 = 1 + 0.992 391 68;
  • 30) 0.992 391 68 × 2 = 1 + 0.984 783 36;
  • 31) 0.984 783 36 × 2 = 1 + 0.969 566 72;
  • 32) 0.969 566 72 × 2 = 1 + 0.939 133 44;
  • 33) 0.939 133 44 × 2 = 1 + 0.878 266 88;
  • 34) 0.878 266 88 × 2 = 1 + 0.756 533 76;
  • 35) 0.756 533 76 × 2 = 1 + 0.513 067 52;
  • 36) 0.513 067 52 × 2 = 1 + 0.026 135 04;
  • 37) 0.026 135 04 × 2 = 0 + 0.052 270 08;
  • 38) 0.052 270 08 × 2 = 0 + 0.104 540 16;
  • 39) 0.104 540 16 × 2 = 0 + 0.209 080 32;
  • 40) 0.209 080 32 × 2 = 0 + 0.418 160 64;
  • 41) 0.418 160 64 × 2 = 0 + 0.836 321 28;
  • 42) 0.836 321 28 × 2 = 1 + 0.672 642 56;

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...)


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:

0.000 003 39(10) =


0.0000 0000 0000 0000 0011 1000 1101 1111 1111 0000 01(2)


6. Positive number before normalization:

0.000 003 39(10) =


0.0000 0000 0000 0000 0011 1000 1101 1111 1111 0000 01(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right so that only one non zero digit remains to the left of it:

0.000 003 39(10) =


0.0000 0000 0000 0000 0011 1000 1101 1111 1111 0000 01(2) =


0.0000 0000 0000 0000 0011 1000 1101 1111 1111 0000 01(2) × 20 =


1.1100 0110 1111 1111 1000 001(2) × 2-19


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): -19


Mantissa (not normalized):
1.1100 0110 1111 1111 1000 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


-19 + 2(8-1) - 1 =


(-19 + 127)(10) =


108(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;
  • 108 ÷ 2 = 54 + 0;
  • 54 ÷ 2 = 27 + 0;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 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) =


108(10) =


0110 1100(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. 110 0011 0111 1111 1100 0001 =


110 0011 0111 1111 1100 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 1100


Mantissa (23 bits) =
110 0011 0111 1111 1100 0001


Conclusion:

Number -0.000 003 39 converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:
1 - 0110 1100 - 110 0011 0111 1111 1100 0001

(32 bits IEEE 754)
  • Sign (1 bit):

    • 1

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 1

      25
    • 0

      24
    • 0

      23
  • Mantissa (23 bits):

    • 1

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 1

      17
    • 1

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 1

      7
    • 1

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 1

      0

More operations of this kind:

-0.000 003 4 = ? ... -0.000 003 38 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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