0.000 000 110 006 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 006 9(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
0.000 000 110 006 9(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: 0.000 000 110 006 9.

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) 0.000 000 110 006 9 × 2 = 0 + 0.000 000 220 013 8;
  • 2) 0.000 000 220 013 8 × 2 = 0 + 0.000 000 440 027 6;
  • 3) 0.000 000 440 027 6 × 2 = 0 + 0.000 000 880 055 2;
  • 4) 0.000 000 880 055 2 × 2 = 0 + 0.000 001 760 110 4;
  • 5) 0.000 001 760 110 4 × 2 = 0 + 0.000 003 520 220 8;
  • 6) 0.000 003 520 220 8 × 2 = 0 + 0.000 007 040 441 6;
  • 7) 0.000 007 040 441 6 × 2 = 0 + 0.000 014 080 883 2;
  • 8) 0.000 014 080 883 2 × 2 = 0 + 0.000 028 161 766 4;
  • 9) 0.000 028 161 766 4 × 2 = 0 + 0.000 056 323 532 8;
  • 10) 0.000 056 323 532 8 × 2 = 0 + 0.000 112 647 065 6;
  • 11) 0.000 112 647 065 6 × 2 = 0 + 0.000 225 294 131 2;
  • 12) 0.000 225 294 131 2 × 2 = 0 + 0.000 450 588 262 4;
  • 13) 0.000 450 588 262 4 × 2 = 0 + 0.000 901 176 524 8;
  • 14) 0.000 901 176 524 8 × 2 = 0 + 0.001 802 353 049 6;
  • 15) 0.001 802 353 049 6 × 2 = 0 + 0.003 604 706 099 2;
  • 16) 0.003 604 706 099 2 × 2 = 0 + 0.007 209 412 198 4;
  • 17) 0.007 209 412 198 4 × 2 = 0 + 0.014 418 824 396 8;
  • 18) 0.014 418 824 396 8 × 2 = 0 + 0.028 837 648 793 6;
  • 19) 0.028 837 648 793 6 × 2 = 0 + 0.057 675 297 587 2;
  • 20) 0.057 675 297 587 2 × 2 = 0 + 0.115 350 595 174 4;
  • 21) 0.115 350 595 174 4 × 2 = 0 + 0.230 701 190 348 8;
  • 22) 0.230 701 190 348 8 × 2 = 0 + 0.461 402 380 697 6;
  • 23) 0.461 402 380 697 6 × 2 = 0 + 0.922 804 761 395 2;
  • 24) 0.922 804 761 395 2 × 2 = 1 + 0.845 609 522 790 4;
  • 25) 0.845 609 522 790 4 × 2 = 1 + 0.691 219 045 580 8;
  • 26) 0.691 219 045 580 8 × 2 = 1 + 0.382 438 091 161 6;
  • 27) 0.382 438 091 161 6 × 2 = 0 + 0.764 876 182 323 2;
  • 28) 0.764 876 182 323 2 × 2 = 1 + 0.529 752 364 646 4;
  • 29) 0.529 752 364 646 4 × 2 = 1 + 0.059 504 729 292 8;
  • 30) 0.059 504 729 292 8 × 2 = 0 + 0.119 009 458 585 6;
  • 31) 0.119 009 458 585 6 × 2 = 0 + 0.238 018 917 171 2;
  • 32) 0.238 018 917 171 2 × 2 = 0 + 0.476 037 834 342 4;
  • 33) 0.476 037 834 342 4 × 2 = 0 + 0.952 075 668 684 8;
  • 34) 0.952 075 668 684 8 × 2 = 1 + 0.904 151 337 369 6;
  • 35) 0.904 151 337 369 6 × 2 = 1 + 0.808 302 674 739 2;
  • 36) 0.808 302 674 739 2 × 2 = 1 + 0.616 605 349 478 4;
  • 37) 0.616 605 349 478 4 × 2 = 1 + 0.233 210 698 956 8;
  • 38) 0.233 210 698 956 8 × 2 = 0 + 0.466 421 397 913 6;
  • 39) 0.466 421 397 913 6 × 2 = 0 + 0.932 842 795 827 2;
  • 40) 0.932 842 795 827 2 × 2 = 1 + 0.865 685 591 654 4;
  • 41) 0.865 685 591 654 4 × 2 = 1 + 0.731 371 183 308 8;
  • 42) 0.731 371 183 308 8 × 2 = 1 + 0.462 742 366 617 6;
  • 43) 0.462 742 366 617 6 × 2 = 0 + 0.925 484 733 235 2;
  • 44) 0.925 484 733 235 2 × 2 = 1 + 0.850 969 466 470 4;
  • 45) 0.850 969 466 470 4 × 2 = 1 + 0.701 938 932 940 8;
  • 46) 0.701 938 932 940 8 × 2 = 1 + 0.403 877 865 881 6;
  • 47) 0.403 877 865 881 6 × 2 = 0 + 0.807 755 731 763 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:


0.000 000 110 006 9(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1001 1101 110(2)

5. Positive number before normalization:

0.000 000 110 006 9(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1001 1101 110(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:


0.000 000 110 006 9(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1001 1101 110(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1001 1101 110(2) × 20 =

1.1101 1000 0111 1001 1101 110(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 0111 1001 1101 110


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 0011 1100 1110 1110 =

110 1100 0011 1100 1110 1110


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 0011 1100 1110 1110


Decimal number 0.000 000 110 006 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1100 1110 1110

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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