Convert -0.000 637 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

How to convert the decimal number -0.000 637(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:

|-0.000 637| = 0.000 637

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: 0.000 637.

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) 0.000 637 × 2 = 0 + 0.001 274;
  • 2) 0.001 274 × 2 = 0 + 0.002 548;
  • 3) 0.002 548 × 2 = 0 + 0.005 096;
  • 4) 0.005 096 × 2 = 0 + 0.010 192;
  • 5) 0.010 192 × 2 = 0 + 0.020 384;
  • 6) 0.020 384 × 2 = 0 + 0.040 768;
  • 7) 0.040 768 × 2 = 0 + 0.081 536;
  • 8) 0.081 536 × 2 = 0 + 0.163 072;
  • 9) 0.163 072 × 2 = 0 + 0.326 144;
  • 10) 0.326 144 × 2 = 0 + 0.652 288;
  • 11) 0.652 288 × 2 = 1 + 0.304 576;
  • 12) 0.304 576 × 2 = 0 + 0.609 152;
  • 13) 0.609 152 × 2 = 1 + 0.218 304;
  • 14) 0.218 304 × 2 = 0 + 0.436 608;
  • 15) 0.436 608 × 2 = 0 + 0.873 216;
  • 16) 0.873 216 × 2 = 1 + 0.746 432;
  • 17) 0.746 432 × 2 = 1 + 0.492 864;
  • 18) 0.492 864 × 2 = 0 + 0.985 728;
  • 19) 0.985 728 × 2 = 1 + 0.971 456;
  • 20) 0.971 456 × 2 = 1 + 0.942 912;
  • 21) 0.942 912 × 2 = 1 + 0.885 824;
  • 22) 0.885 824 × 2 = 1 + 0.771 648;
  • 23) 0.771 648 × 2 = 1 + 0.543 296;
  • 24) 0.543 296 × 2 = 1 + 0.086 592;
  • 25) 0.086 592 × 2 = 0 + 0.173 184;
  • 26) 0.173 184 × 2 = 0 + 0.346 368;
  • 27) 0.346 368 × 2 = 0 + 0.692 736;
  • 28) 0.692 736 × 2 = 1 + 0.385 472;
  • 29) 0.385 472 × 2 = 0 + 0.770 944;
  • 30) 0.770 944 × 2 = 1 + 0.541 888;
  • 31) 0.541 888 × 2 = 1 + 0.083 776;
  • 32) 0.083 776 × 2 = 0 + 0.167 552;
  • 33) 0.167 552 × 2 = 0 + 0.335 104;
  • 34) 0.335 104 × 2 = 0 + 0.670 208;

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:

0.000 637(10) =


0.0000 0000 0010 1001 1011 1111 0001 0110 00(2)


6. Positive number before normalization:

0.000 637(10) =


0.0000 0000 0010 1001 1011 1111 0001 0110 00(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right so that only one non zero digit remains to the left of it:

0.000 637(10) =


0.0000 0000 0010 1001 1011 1111 0001 0110 00(2) =


0.0000 0000 0010 1001 1011 1111 0001 0110 00(2) × 20 =


1.0100 1101 1111 1000 1011 000(2) × 2-11


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): -11


Mantissa (not normalized):
1.0100 1101 1111 1000 1011 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


-11 + 2(8-1) - 1 =


(-11 + 127)(10) =


116(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;
  • 116 ÷ 2 = 58 + 0;
  • 58 ÷ 2 = 29 + 0;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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) =


116(10) =


0111 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. 010 0110 1111 1100 0101 1000 =


010 0110 1111 1100 0101 1000


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) =
0111 0100


Mantissa (23 bits) =
010 0110 1111 1100 0101 1000


Conclusion:

Number -0.000 637 converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:
1 - 0111 0100 - 010 0110 1111 1100 0101 1000

(32 bits IEEE 754)
  • Sign (1 bit):

    • 1

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 0

      26
    • 1

      25
    • 0

      24
    • 0

      23
  • Mantissa (23 bits):

    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 1

      3
    • 0

      2
    • 0

      1
    • 0

      0

More operations of this kind:

-0.000 638 = ? ... -0.000 636 = ?


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)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

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