-0.001 139 523 422 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.001 139 523 422(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
-0.001 139 523 422(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:

|-0.001 139 523 422| = 0.001 139 523 422


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: 0.001 139 523 422.

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) 0.001 139 523 422 × 2 = 0 + 0.002 279 046 844;
  • 2) 0.002 279 046 844 × 2 = 0 + 0.004 558 093 688;
  • 3) 0.004 558 093 688 × 2 = 0 + 0.009 116 187 376;
  • 4) 0.009 116 187 376 × 2 = 0 + 0.018 232 374 752;
  • 5) 0.018 232 374 752 × 2 = 0 + 0.036 464 749 504;
  • 6) 0.036 464 749 504 × 2 = 0 + 0.072 929 499 008;
  • 7) 0.072 929 499 008 × 2 = 0 + 0.145 858 998 016;
  • 8) 0.145 858 998 016 × 2 = 0 + 0.291 717 996 032;
  • 9) 0.291 717 996 032 × 2 = 0 + 0.583 435 992 064;
  • 10) 0.583 435 992 064 × 2 = 1 + 0.166 871 984 128;
  • 11) 0.166 871 984 128 × 2 = 0 + 0.333 743 968 256;
  • 12) 0.333 743 968 256 × 2 = 0 + 0.667 487 936 512;
  • 13) 0.667 487 936 512 × 2 = 1 + 0.334 975 873 024;
  • 14) 0.334 975 873 024 × 2 = 0 + 0.669 951 746 048;
  • 15) 0.669 951 746 048 × 2 = 1 + 0.339 903 492 096;
  • 16) 0.339 903 492 096 × 2 = 0 + 0.679 806 984 192;
  • 17) 0.679 806 984 192 × 2 = 1 + 0.359 613 968 384;
  • 18) 0.359 613 968 384 × 2 = 0 + 0.719 227 936 768;
  • 19) 0.719 227 936 768 × 2 = 1 + 0.438 455 873 536;
  • 20) 0.438 455 873 536 × 2 = 0 + 0.876 911 747 072;
  • 21) 0.876 911 747 072 × 2 = 1 + 0.753 823 494 144;
  • 22) 0.753 823 494 144 × 2 = 1 + 0.507 646 988 288;
  • 23) 0.507 646 988 288 × 2 = 1 + 0.015 293 976 576;
  • 24) 0.015 293 976 576 × 2 = 0 + 0.030 587 953 152;
  • 25) 0.030 587 953 152 × 2 = 0 + 0.061 175 906 304;
  • 26) 0.061 175 906 304 × 2 = 0 + 0.122 351 812 608;
  • 27) 0.122 351 812 608 × 2 = 0 + 0.244 703 625 216;
  • 28) 0.244 703 625 216 × 2 = 0 + 0.489 407 250 432;
  • 29) 0.489 407 250 432 × 2 = 0 + 0.978 814 500 864;
  • 30) 0.978 814 500 864 × 2 = 1 + 0.957 629 001 728;
  • 31) 0.957 629 001 728 × 2 = 1 + 0.915 258 003 456;
  • 32) 0.915 258 003 456 × 2 = 1 + 0.830 516 006 912;
  • 33) 0.830 516 006 912 × 2 = 1 + 0.661 032 013 824;

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:


0.001 139 523 422(10) =

0.0000 0000 0100 1010 1010 1110 0000 0111 1(2)

6. Positive number before normalization:

0.001 139 523 422(10) =

0.0000 0000 0100 1010 1010 1110 0000 0111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.001 139 523 422(10) =

0.0000 0000 0100 1010 1010 1110 0000 0111 1(2) =

0.0000 0000 0100 1010 1010 1110 0000 0111 1(2) × 20 =

1.0010 1010 1011 1000 0001 111(2) × 2-10


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): -10

Mantissa (not normalized):
1.0010 1010 1011 1000 0001 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-10 + 2(8-1) - 1 =

(-10 + 127)(10) =

117(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;
  • 117 ÷ 2 = 58 + 1;
  • 58 ÷ 2 = 29 + 0;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

117(10) =

0111 0101(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. 001 0101 0101 1100 0000 1111 =

001 0101 0101 1100 0000 1111


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) =
0111 0101

Mantissa (23 bits) =
001 0101 0101 1100 0000 1111


Decimal number -0.001 139 523 422 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0111 0101 - 001 0101 0101 1100 0000 1111

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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