0.000 000 110 018 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 018 4(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
0.000 000 110 018 4(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 110 018 4.

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) 0.000 000 110 018 4 × 2 = 0 + 0.000 000 220 036 8;
  • 2) 0.000 000 220 036 8 × 2 = 0 + 0.000 000 440 073 6;
  • 3) 0.000 000 440 073 6 × 2 = 0 + 0.000 000 880 147 2;
  • 4) 0.000 000 880 147 2 × 2 = 0 + 0.000 001 760 294 4;
  • 5) 0.000 001 760 294 4 × 2 = 0 + 0.000 003 520 588 8;
  • 6) 0.000 003 520 588 8 × 2 = 0 + 0.000 007 041 177 6;
  • 7) 0.000 007 041 177 6 × 2 = 0 + 0.000 014 082 355 2;
  • 8) 0.000 014 082 355 2 × 2 = 0 + 0.000 028 164 710 4;
  • 9) 0.000 028 164 710 4 × 2 = 0 + 0.000 056 329 420 8;
  • 10) 0.000 056 329 420 8 × 2 = 0 + 0.000 112 658 841 6;
  • 11) 0.000 112 658 841 6 × 2 = 0 + 0.000 225 317 683 2;
  • 12) 0.000 225 317 683 2 × 2 = 0 + 0.000 450 635 366 4;
  • 13) 0.000 450 635 366 4 × 2 = 0 + 0.000 901 270 732 8;
  • 14) 0.000 901 270 732 8 × 2 = 0 + 0.001 802 541 465 6;
  • 15) 0.001 802 541 465 6 × 2 = 0 + 0.003 605 082 931 2;
  • 16) 0.003 605 082 931 2 × 2 = 0 + 0.007 210 165 862 4;
  • 17) 0.007 210 165 862 4 × 2 = 0 + 0.014 420 331 724 8;
  • 18) 0.014 420 331 724 8 × 2 = 0 + 0.028 840 663 449 6;
  • 19) 0.028 840 663 449 6 × 2 = 0 + 0.057 681 326 899 2;
  • 20) 0.057 681 326 899 2 × 2 = 0 + 0.115 362 653 798 4;
  • 21) 0.115 362 653 798 4 × 2 = 0 + 0.230 725 307 596 8;
  • 22) 0.230 725 307 596 8 × 2 = 0 + 0.461 450 615 193 6;
  • 23) 0.461 450 615 193 6 × 2 = 0 + 0.922 901 230 387 2;
  • 24) 0.922 901 230 387 2 × 2 = 1 + 0.845 802 460 774 4;
  • 25) 0.845 802 460 774 4 × 2 = 1 + 0.691 604 921 548 8;
  • 26) 0.691 604 921 548 8 × 2 = 1 + 0.383 209 843 097 6;
  • 27) 0.383 209 843 097 6 × 2 = 0 + 0.766 419 686 195 2;
  • 28) 0.766 419 686 195 2 × 2 = 1 + 0.532 839 372 390 4;
  • 29) 0.532 839 372 390 4 × 2 = 1 + 0.065 678 744 780 8;
  • 30) 0.065 678 744 780 8 × 2 = 0 + 0.131 357 489 561 6;
  • 31) 0.131 357 489 561 6 × 2 = 0 + 0.262 714 979 123 2;
  • 32) 0.262 714 979 123 2 × 2 = 0 + 0.525 429 958 246 4;
  • 33) 0.525 429 958 246 4 × 2 = 1 + 0.050 859 916 492 8;
  • 34) 0.050 859 916 492 8 × 2 = 0 + 0.101 719 832 985 6;
  • 35) 0.101 719 832 985 6 × 2 = 0 + 0.203 439 665 971 2;
  • 36) 0.203 439 665 971 2 × 2 = 0 + 0.406 879 331 942 4;
  • 37) 0.406 879 331 942 4 × 2 = 0 + 0.813 758 663 884 8;
  • 38) 0.813 758 663 884 8 × 2 = 1 + 0.627 517 327 769 6;
  • 39) 0.627 517 327 769 6 × 2 = 1 + 0.255 034 655 539 2;
  • 40) 0.255 034 655 539 2 × 2 = 0 + 0.510 069 311 078 4;
  • 41) 0.510 069 311 078 4 × 2 = 1 + 0.020 138 622 156 8;
  • 42) 0.020 138 622 156 8 × 2 = 0 + 0.040 277 244 313 6;
  • 43) 0.040 277 244 313 6 × 2 = 0 + 0.080 554 488 627 2;
  • 44) 0.080 554 488 627 2 × 2 = 0 + 0.161 108 977 254 4;
  • 45) 0.161 108 977 254 4 × 2 = 0 + 0.322 217 954 508 8;
  • 46) 0.322 217 954 508 8 × 2 = 0 + 0.644 435 909 017 6;
  • 47) 0.644 435 909 017 6 × 2 = 1 + 0.288 871 818 035 2;

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).


4. 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:


0.000 000 110 018 4(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0110 1000 001(2)

5. Positive number before normalization:

0.000 000 110 018 4(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0110 1000 001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 110 018 4(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0110 1000 001(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0110 1000 001(2) × 20 =

1.1101 1000 1000 0110 1000 001(2) × 2-24


7. 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 0 (a positive number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1101 1000 1000 0110 1000 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(10)


9. 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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

103(10) =

0110 0111(2)


11. 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 1100 0100 0011 0100 0001 =

110 1100 0100 0011 0100 0001


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 0111

Mantissa (23 bits) =
110 1100 0100 0011 0100 0001


Decimal number 0.000 000 110 018 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0100 0011 0100 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