Base ten decimal number -1 001 001 010.110 011 converted to 32 bit single precision IEEE 754 binary floating point standard

How to convert the decimal number -1 001 001 010.110 011(10)
to
32 bit single precision IEEE 754 binary floating point
(1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. We start with the positive version of the number:

|-1 001 001 010.110 011| = 1 001 001 010.110 011

2. First, convert to binary (base 2) the integer part: 1 001 001 010. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 1 001 001 010 ÷ 2 = 500 500 505 + 0;
  • 500 500 505 ÷ 2 = 250 250 252 + 1;
  • 250 250 252 ÷ 2 = 125 125 126 + 0;
  • 125 125 126 ÷ 2 = 62 562 563 + 0;
  • 62 562 563 ÷ 2 = 31 281 281 + 1;
  • 31 281 281 ÷ 2 = 15 640 640 + 1;
  • 15 640 640 ÷ 2 = 7 820 320 + 0;
  • 7 820 320 ÷ 2 = 3 910 160 + 0;
  • 3 910 160 ÷ 2 = 1 955 080 + 0;
  • 1 955 080 ÷ 2 = 977 540 + 0;
  • 977 540 ÷ 2 = 488 770 + 0;
  • 488 770 ÷ 2 = 244 385 + 0;
  • 244 385 ÷ 2 = 122 192 + 1;
  • 122 192 ÷ 2 = 61 096 + 0;
  • 61 096 ÷ 2 = 30 548 + 0;
  • 30 548 ÷ 2 = 15 274 + 0;
  • 15 274 ÷ 2 = 7 637 + 0;
  • 7 637 ÷ 2 = 3 818 + 1;
  • 3 818 ÷ 2 = 1 909 + 0;
  • 1 909 ÷ 2 = 954 + 1;
  • 954 ÷ 2 = 477 + 0;
  • 477 ÷ 2 = 238 + 1;
  • 238 ÷ 2 = 119 + 0;
  • 119 ÷ 2 = 59 + 1;
  • 59 ÷ 2 = 29 + 1;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

1 001 001 010(10) =


11 1011 1010 1010 0001 0000 0011 0010(2)

4. Convert to binary (base 2) the fractional part: 0.110 011. Multiply it 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) 0.110 011 × 2 = 0 + 0.220 022;
  • 2) 0.220 022 × 2 = 0 + 0.440 044;
  • 3) 0.440 044 × 2 = 0 + 0.880 088;
  • 4) 0.880 088 × 2 = 1 + 0.760 176;
  • 5) 0.760 176 × 2 = 1 + 0.520 352;
  • 6) 0.520 352 × 2 = 1 + 0.040 704;
  • 7) 0.040 704 × 2 = 0 + 0.081 408;
  • 8) 0.081 408 × 2 = 0 + 0.162 816;
  • 9) 0.162 816 × 2 = 0 + 0.325 632;
  • 10) 0.325 632 × 2 = 0 + 0.651 264;
  • 11) 0.651 264 × 2 = 1 + 0.302 528;
  • 12) 0.302 528 × 2 = 0 + 0.605 056;
  • 13) 0.605 056 × 2 = 1 + 0.210 112;
  • 14) 0.210 112 × 2 = 0 + 0.420 224;
  • 15) 0.420 224 × 2 = 0 + 0.840 448;
  • 16) 0.840 448 × 2 = 1 + 0.680 896;
  • 17) 0.680 896 × 2 = 1 + 0.361 792;
  • 18) 0.361 792 × 2 = 0 + 0.723 584;
  • 19) 0.723 584 × 2 = 1 + 0.447 168;
  • 20) 0.447 168 × 2 = 0 + 0.894 336;
  • 21) 0.894 336 × 2 = 1 + 0.788 672;
  • 22) 0.788 672 × 2 = 1 + 0.577 344;
  • 23) 0.577 344 × 2 = 1 + 0.154 688;
  • 24) 0.154 688 × 2 = 0 + 0.309 376;

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, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.110 011(10) =


0.0001 1100 0010 1001 1010 1110(2)

Positive number before normalization:

1 001 001 010.110 011(10) =


11 1011 1010 1010 0001 0000 0011 0010.0001 1100 0010 1001 1010 1110(2)

6. Normalize the binary representation of the number, shifting the decimal mark 29 positions to the left so that only one non zero digit remains to the left of it:

1 001 001 010.110 011(10) =


11 1011 1010 1010 0001 0000 0011 0010.0001 1100 0010 1001 1010 1110(2) =


11 1011 1010 1010 0001 0000 0011 0010.0001 1100 0010 1001 1010 1110(2) × 20 =


1.1101 1101 0101 0000 1000 0001 1001 0000 1110 0001 0100 1101 0111 0(2) × 229

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): 29


Mantissa (not normalized): 1.1101 1101 0101 0000 1000 0001 1001 0000 1110 0001 0100 1101 0111 0

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:

Exponent (adjusted) =


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


29 + 2(8-1) - 1 =


(29 + 127)(10) =


156(10)


  • division = quotient + remainder;
  • 156 ÷ 2 = 78 + 0;
  • 78 ÷ 2 = 39 + 0;
  • 39 ÷ 2 = 19 + 1;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


156(10) =


1001 1100(2)

8. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 23 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...):

Mantissa (normalized) =


1. 110 1110 1010 1000 0100 0000 11 0010 0001 1100 0010 1001 1010 1110 =


110 1110 1010 1000 0100 0000

Conclusion:

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) =
1001 1100


Mantissa (23 bits) =
110 1110 1010 1000 0100 0000

Number -1 001 001 010.110 011, a decimal, converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:


1 - 1001 1100 - 110 1110 1010 1000 0100 0000

(32 bits IEEE 754)
  • Sign (1 bit):

    • 1

      31
  • Exponent (8 bits):

    • 1

      30
    • 0

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 0

      24
    • 0

      23
  • Mantissa (23 bits):

    • 1

      22
    • 1

      21
    • 0

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 0

      14
    • 1

      13
    • 0

      12
    • 1

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 0

      0

Convert decimal numbers from base ten 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), 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