0.000 000 110 007 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 007 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
0.000 000 110 007 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. 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: 0.000 000 110 007 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) 0.000 000 110 007 1 × 2 = 0 + 0.000 000 220 014 2;
  • 2) 0.000 000 220 014 2 × 2 = 0 + 0.000 000 440 028 4;
  • 3) 0.000 000 440 028 4 × 2 = 0 + 0.000 000 880 056 8;
  • 4) 0.000 000 880 056 8 × 2 = 0 + 0.000 001 760 113 6;
  • 5) 0.000 001 760 113 6 × 2 = 0 + 0.000 003 520 227 2;
  • 6) 0.000 003 520 227 2 × 2 = 0 + 0.000 007 040 454 4;
  • 7) 0.000 007 040 454 4 × 2 = 0 + 0.000 014 080 908 8;
  • 8) 0.000 014 080 908 8 × 2 = 0 + 0.000 028 161 817 6;
  • 9) 0.000 028 161 817 6 × 2 = 0 + 0.000 056 323 635 2;
  • 10) 0.000 056 323 635 2 × 2 = 0 + 0.000 112 647 270 4;
  • 11) 0.000 112 647 270 4 × 2 = 0 + 0.000 225 294 540 8;
  • 12) 0.000 225 294 540 8 × 2 = 0 + 0.000 450 589 081 6;
  • 13) 0.000 450 589 081 6 × 2 = 0 + 0.000 901 178 163 2;
  • 14) 0.000 901 178 163 2 × 2 = 0 + 0.001 802 356 326 4;
  • 15) 0.001 802 356 326 4 × 2 = 0 + 0.003 604 712 652 8;
  • 16) 0.003 604 712 652 8 × 2 = 0 + 0.007 209 425 305 6;
  • 17) 0.007 209 425 305 6 × 2 = 0 + 0.014 418 850 611 2;
  • 18) 0.014 418 850 611 2 × 2 = 0 + 0.028 837 701 222 4;
  • 19) 0.028 837 701 222 4 × 2 = 0 + 0.057 675 402 444 8;
  • 20) 0.057 675 402 444 8 × 2 = 0 + 0.115 350 804 889 6;
  • 21) 0.115 350 804 889 6 × 2 = 0 + 0.230 701 609 779 2;
  • 22) 0.230 701 609 779 2 × 2 = 0 + 0.461 403 219 558 4;
  • 23) 0.461 403 219 558 4 × 2 = 0 + 0.922 806 439 116 8;
  • 24) 0.922 806 439 116 8 × 2 = 1 + 0.845 612 878 233 6;
  • 25) 0.845 612 878 233 6 × 2 = 1 + 0.691 225 756 467 2;
  • 26) 0.691 225 756 467 2 × 2 = 1 + 0.382 451 512 934 4;
  • 27) 0.382 451 512 934 4 × 2 = 0 + 0.764 903 025 868 8;
  • 28) 0.764 903 025 868 8 × 2 = 1 + 0.529 806 051 737 6;
  • 29) 0.529 806 051 737 6 × 2 = 1 + 0.059 612 103 475 2;
  • 30) 0.059 612 103 475 2 × 2 = 0 + 0.119 224 206 950 4;
  • 31) 0.119 224 206 950 4 × 2 = 0 + 0.238 448 413 900 8;
  • 32) 0.238 448 413 900 8 × 2 = 0 + 0.476 896 827 801 6;
  • 33) 0.476 896 827 801 6 × 2 = 0 + 0.953 793 655 603 2;
  • 34) 0.953 793 655 603 2 × 2 = 1 + 0.907 587 311 206 4;
  • 35) 0.907 587 311 206 4 × 2 = 1 + 0.815 174 622 412 8;
  • 36) 0.815 174 622 412 8 × 2 = 1 + 0.630 349 244 825 6;
  • 37) 0.630 349 244 825 6 × 2 = 1 + 0.260 698 489 651 2;
  • 38) 0.260 698 489 651 2 × 2 = 0 + 0.521 396 979 302 4;
  • 39) 0.521 396 979 302 4 × 2 = 1 + 0.042 793 958 604 8;
  • 40) 0.042 793 958 604 8 × 2 = 0 + 0.085 587 917 209 6;
  • 41) 0.085 587 917 209 6 × 2 = 0 + 0.171 175 834 419 2;
  • 42) 0.171 175 834 419 2 × 2 = 0 + 0.342 351 668 838 4;
  • 43) 0.342 351 668 838 4 × 2 = 0 + 0.684 703 337 676 8;
  • 44) 0.684 703 337 676 8 × 2 = 1 + 0.369 406 675 353 6;
  • 45) 0.369 406 675 353 6 × 2 = 0 + 0.738 813 350 707 2;
  • 46) 0.738 813 350 707 2 × 2 = 1 + 0.477 626 701 414 4;
  • 47) 0.477 626 701 414 4 × 2 = 0 + 0.955 253 402 828 8;

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:


0.000 000 110 007 1(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1010 0001 010(2)

5. Positive number before normalization:

0.000 000 110 007 1(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1010 0001 010(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:


0.000 000 110 007 1(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1010 0001 010(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1010 0001 010(2) × 20 =

1.1101 1000 0111 1010 0001 010(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 1010 0001 010


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 1101 0000 1010 =

110 1100 0011 1101 0000 1010


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 1101 0000 1010


Decimal number 0.000 000 110 007 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1101 0000 1010

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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