-7.325 299 739 837 646 458 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -7.325 299 739 837 646 458 7(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
-7.325 299 739 837 646 458 7(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:

|-7.325 299 739 837 646 458 7| = 7.325 299 739 837 646 458 7


2. First, convert to binary (in base 2) the integer part: 7.
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;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

7(10) =

111(2)


4. Convert to binary (base 2) the fractional part: 0.325 299 739 837 646 458 7.

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) 0.325 299 739 837 646 458 7 × 2 = 0 + 0.650 599 479 675 292 917 4;
  • 2) 0.650 599 479 675 292 917 4 × 2 = 1 + 0.301 198 959 350 585 834 8;
  • 3) 0.301 198 959 350 585 834 8 × 2 = 0 + 0.602 397 918 701 171 669 6;
  • 4) 0.602 397 918 701 171 669 6 × 2 = 1 + 0.204 795 837 402 343 339 2;
  • 5) 0.204 795 837 402 343 339 2 × 2 = 0 + 0.409 591 674 804 686 678 4;
  • 6) 0.409 591 674 804 686 678 4 × 2 = 0 + 0.819 183 349 609 373 356 8;
  • 7) 0.819 183 349 609 373 356 8 × 2 = 1 + 0.638 366 699 218 746 713 6;
  • 8) 0.638 366 699 218 746 713 6 × 2 = 1 + 0.276 733 398 437 493 427 2;
  • 9) 0.276 733 398 437 493 427 2 × 2 = 0 + 0.553 466 796 874 986 854 4;
  • 10) 0.553 466 796 874 986 854 4 × 2 = 1 + 0.106 933 593 749 973 708 8;
  • 11) 0.106 933 593 749 973 708 8 × 2 = 0 + 0.213 867 187 499 947 417 6;
  • 12) 0.213 867 187 499 947 417 6 × 2 = 0 + 0.427 734 374 999 894 835 2;
  • 13) 0.427 734 374 999 894 835 2 × 2 = 0 + 0.855 468 749 999 789 670 4;
  • 14) 0.855 468 749 999 789 670 4 × 2 = 1 + 0.710 937 499 999 579 340 8;
  • 15) 0.710 937 499 999 579 340 8 × 2 = 1 + 0.421 874 999 999 158 681 6;
  • 16) 0.421 874 999 999 158 681 6 × 2 = 0 + 0.843 749 999 998 317 363 2;
  • 17) 0.843 749 999 998 317 363 2 × 2 = 1 + 0.687 499 999 996 634 726 4;
  • 18) 0.687 499 999 996 634 726 4 × 2 = 1 + 0.374 999 999 993 269 452 8;
  • 19) 0.374 999 999 993 269 452 8 × 2 = 0 + 0.749 999 999 986 538 905 6;
  • 20) 0.749 999 999 986 538 905 6 × 2 = 1 + 0.499 999 999 973 077 811 2;
  • 21) 0.499 999 999 973 077 811 2 × 2 = 0 + 0.999 999 999 946 155 622 4;
  • 22) 0.999 999 999 946 155 622 4 × 2 = 1 + 0.999 999 999 892 311 244 8;
  • 23) 0.999 999 999 892 311 244 8 × 2 = 1 + 0.999 999 999 784 622 489 6;
  • 24) 0.999 999 999 784 622 489 6 × 2 = 1 + 0.999 999 999 569 244 979 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).


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:


0.325 299 739 837 646 458 7(10) =

0.0101 0011 0100 0110 1101 0111(2)

6. Positive number before normalization:

7.325 299 739 837 646 458 7(10) =

111.0101 0011 0100 0110 1101 0111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


7.325 299 739 837 646 458 7(10) =

111.0101 0011 0100 0110 1101 0111(2) =

111.0101 0011 0100 0110 1101 0111(2) × 20 =

1.1101 0100 1101 0001 1011 0101 11(2) × 22


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): 2

Mantissa (not normalized):
1.1101 0100 1101 0001 1011 0101 11


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(8-1) - 1 =

(2 + 127)(10) =

129(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;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

129(10) =

1000 0001(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 110 1010 0110 1000 1101 1010 111 =

110 1010 0110 1000 1101 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) =
1000 0001

Mantissa (23 bits) =
110 1010 0110 1000 1101 1010


Decimal number -7.325 299 739 837 646 458 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 1000 0001 - 110 1010 0110 1000 1101 1010

» Convert decimal -7.325 299 739 837 646 461 7 to 32 bit single precision IEEE 754 binary floating point representation standard

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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