-11.851 100 000 100 111 101 093 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -11.851 100 000 100 111 101 093 1(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
-11.851 100 000 100 111 101 093 1(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:

|-11.851 100 000 100 111 101 093 1| = 11.851 100 000 100 111 101 093 1


2. First, convert to binary (in base 2) the integer part: 11.
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;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

11(10) =

1011(2)


4. Convert to binary (base 2) the fractional part: 0.851 100 000 100 111 101 093 1.

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) 0.851 100 000 100 111 101 093 1 × 2 = 1 + 0.702 200 000 200 222 202 186 2;
  • 2) 0.702 200 000 200 222 202 186 2 × 2 = 1 + 0.404 400 000 400 444 404 372 4;
  • 3) 0.404 400 000 400 444 404 372 4 × 2 = 0 + 0.808 800 000 800 888 808 744 8;
  • 4) 0.808 800 000 800 888 808 744 8 × 2 = 1 + 0.617 600 001 601 777 617 489 6;
  • 5) 0.617 600 001 601 777 617 489 6 × 2 = 1 + 0.235 200 003 203 555 234 979 2;
  • 6) 0.235 200 003 203 555 234 979 2 × 2 = 0 + 0.470 400 006 407 110 469 958 4;
  • 7) 0.470 400 006 407 110 469 958 4 × 2 = 0 + 0.940 800 012 814 220 939 916 8;
  • 8) 0.940 800 012 814 220 939 916 8 × 2 = 1 + 0.881 600 025 628 441 879 833 6;
  • 9) 0.881 600 025 628 441 879 833 6 × 2 = 1 + 0.763 200 051 256 883 759 667 2;
  • 10) 0.763 200 051 256 883 759 667 2 × 2 = 1 + 0.526 400 102 513 767 519 334 4;
  • 11) 0.526 400 102 513 767 519 334 4 × 2 = 1 + 0.052 800 205 027 535 038 668 8;
  • 12) 0.052 800 205 027 535 038 668 8 × 2 = 0 + 0.105 600 410 055 070 077 337 6;
  • 13) 0.105 600 410 055 070 077 337 6 × 2 = 0 + 0.211 200 820 110 140 154 675 2;
  • 14) 0.211 200 820 110 140 154 675 2 × 2 = 0 + 0.422 401 640 220 280 309 350 4;
  • 15) 0.422 401 640 220 280 309 350 4 × 2 = 0 + 0.844 803 280 440 560 618 700 8;
  • 16) 0.844 803 280 440 560 618 700 8 × 2 = 1 + 0.689 606 560 881 121 237 401 6;
  • 17) 0.689 606 560 881 121 237 401 6 × 2 = 1 + 0.379 213 121 762 242 474 803 2;
  • 18) 0.379 213 121 762 242 474 803 2 × 2 = 0 + 0.758 426 243 524 484 949 606 4;
  • 19) 0.758 426 243 524 484 949 606 4 × 2 = 1 + 0.516 852 487 048 969 899 212 8;
  • 20) 0.516 852 487 048 969 899 212 8 × 2 = 1 + 0.033 704 974 097 939 798 425 6;
  • 21) 0.033 704 974 097 939 798 425 6 × 2 = 0 + 0.067 409 948 195 879 596 851 2;
  • 22) 0.067 409 948 195 879 596 851 2 × 2 = 0 + 0.134 819 896 391 759 193 702 4;
  • 23) 0.134 819 896 391 759 193 702 4 × 2 = 0 + 0.269 639 792 783 518 387 404 8;
  • 24) 0.269 639 792 783 518 387 404 8 × 2 = 0 + 0.539 279 585 567 036 774 809 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:


0.851 100 000 100 111 101 093 1(10) =

0.1101 1001 1110 0001 1011 0000(2)

6. Positive number before normalization:

11.851 100 000 100 111 101 093 1(10) =

1011.1101 1001 1110 0001 1011 0000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


11.851 100 000 100 111 101 093 1(10) =

1011.1101 1001 1110 0001 1011 0000(2) =

1011.1101 1001 1110 0001 1011 0000(2) × 20 =

1.0111 1011 0011 1100 0011 0110 000(2) × 23


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): 3

Mantissa (not normalized):
1.0111 1011 0011 1100 0011 0110 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

3 + 2(8-1) - 1 =

(3 + 127)(10) =

130(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;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

130(10) =

1000 0010(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, 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. 011 1101 1001 1110 0001 1011 0000 =

011 1101 1001 1110 0001 1011


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) =
1000 0010

Mantissa (23 bits) =
011 1101 1001 1110 0001 1011


Decimal number -11.851 100 000 100 111 101 093 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 1000 0010 - 011 1101 1001 1110 0001 1011

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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