-0.000 010 411 948 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 010 411 948 8(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
-0.000 010 411 948 8(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:

|-0.000 010 411 948 8| = 0.000 010 411 948 8


2. 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;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 010 411 948 8.

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) 0.000 010 411 948 8 × 2 = 0 + 0.000 020 823 897 6;
  • 2) 0.000 020 823 897 6 × 2 = 0 + 0.000 041 647 795 2;
  • 3) 0.000 041 647 795 2 × 2 = 0 + 0.000 083 295 590 4;
  • 4) 0.000 083 295 590 4 × 2 = 0 + 0.000 166 591 180 8;
  • 5) 0.000 166 591 180 8 × 2 = 0 + 0.000 333 182 361 6;
  • 6) 0.000 333 182 361 6 × 2 = 0 + 0.000 666 364 723 2;
  • 7) 0.000 666 364 723 2 × 2 = 0 + 0.001 332 729 446 4;
  • 8) 0.001 332 729 446 4 × 2 = 0 + 0.002 665 458 892 8;
  • 9) 0.002 665 458 892 8 × 2 = 0 + 0.005 330 917 785 6;
  • 10) 0.005 330 917 785 6 × 2 = 0 + 0.010 661 835 571 2;
  • 11) 0.010 661 835 571 2 × 2 = 0 + 0.021 323 671 142 4;
  • 12) 0.021 323 671 142 4 × 2 = 0 + 0.042 647 342 284 8;
  • 13) 0.042 647 342 284 8 × 2 = 0 + 0.085 294 684 569 6;
  • 14) 0.085 294 684 569 6 × 2 = 0 + 0.170 589 369 139 2;
  • 15) 0.170 589 369 139 2 × 2 = 0 + 0.341 178 738 278 4;
  • 16) 0.341 178 738 278 4 × 2 = 0 + 0.682 357 476 556 8;
  • 17) 0.682 357 476 556 8 × 2 = 1 + 0.364 714 953 113 6;
  • 18) 0.364 714 953 113 6 × 2 = 0 + 0.729 429 906 227 2;
  • 19) 0.729 429 906 227 2 × 2 = 1 + 0.458 859 812 454 4;
  • 20) 0.458 859 812 454 4 × 2 = 0 + 0.917 719 624 908 8;
  • 21) 0.917 719 624 908 8 × 2 = 1 + 0.835 439 249 817 6;
  • 22) 0.835 439 249 817 6 × 2 = 1 + 0.670 878 499 635 2;
  • 23) 0.670 878 499 635 2 × 2 = 1 + 0.341 756 999 270 4;
  • 24) 0.341 756 999 270 4 × 2 = 0 + 0.683 513 998 540 8;
  • 25) 0.683 513 998 540 8 × 2 = 1 + 0.367 027 997 081 6;
  • 26) 0.367 027 997 081 6 × 2 = 0 + 0.734 055 994 163 2;
  • 27) 0.734 055 994 163 2 × 2 = 1 + 0.468 111 988 326 4;
  • 28) 0.468 111 988 326 4 × 2 = 0 + 0.936 223 976 652 8;
  • 29) 0.936 223 976 652 8 × 2 = 1 + 0.872 447 953 305 6;
  • 30) 0.872 447 953 305 6 × 2 = 1 + 0.744 895 906 611 2;
  • 31) 0.744 895 906 611 2 × 2 = 1 + 0.489 791 813 222 4;
  • 32) 0.489 791 813 222 4 × 2 = 0 + 0.979 583 626 444 8;
  • 33) 0.979 583 626 444 8 × 2 = 1 + 0.959 167 252 889 6;
  • 34) 0.959 167 252 889 6 × 2 = 1 + 0.918 334 505 779 2;
  • 35) 0.918 334 505 779 2 × 2 = 1 + 0.836 669 011 558 4;
  • 36) 0.836 669 011 558 4 × 2 = 1 + 0.673 338 023 116 8;
  • 37) 0.673 338 023 116 8 × 2 = 1 + 0.346 676 046 233 6;
  • 38) 0.346 676 046 233 6 × 2 = 0 + 0.693 352 092 467 2;
  • 39) 0.693 352 092 467 2 × 2 = 1 + 0.386 704 184 934 4;
  • 40) 0.386 704 184 934 4 × 2 = 0 + 0.773 408 369 868 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).


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:


0.000 010 411 948 8(10) =

0.0000 0000 0000 0000 1010 1110 1010 1110 1111 1010(2)

6. Positive number before normalization:

0.000 010 411 948 8(10) =

0.0000 0000 0000 0000 1010 1110 1010 1110 1111 1010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 010 411 948 8(10) =

0.0000 0000 0000 0000 1010 1110 1010 1110 1111 1010(2) =

0.0000 0000 0000 0000 1010 1110 1010 1110 1111 1010(2) × 20 =

1.0101 1101 0101 1101 1111 010(2) × 2-17


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): -17

Mantissa (not normalized):
1.0101 1101 0101 1101 1111 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(8-1) - 1 =

(-17 + 127)(10) =

110(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;
  • 110 ÷ 2 = 55 + 0;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

110(10) =

0110 1110(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 010 1110 1010 1110 1111 1010 =

010 1110 1010 1110 1111 1010


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) =
0110 1110

Mantissa (23 bits) =
010 1110 1010 1110 1111 1010


Decimal number -0.000 010 411 948 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1110 - 010 1110 1010 1110 1111 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