-6.031 499 862 670 898 437 698 639 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -6.031 499 862 670 898 437 698 639 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
-6.031 499 862 670 898 437 698 639 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. Start with the positive version of the number:

|-6.031 499 862 670 898 437 698 639 9| = 6.031 499 862 670 898 437 698 639 9


2. First, convert to binary (in base 2) the integer part: 6.
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;
  • 6 ÷ 2 = 3 + 0;
  • 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.

6(10) =

110(2)


4. Convert to binary (base 2) the fractional part: 0.031 499 862 670 898 437 698 639 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) 0.031 499 862 670 898 437 698 639 9 × 2 = 0 + 0.062 999 725 341 796 875 397 279 8;
  • 2) 0.062 999 725 341 796 875 397 279 8 × 2 = 0 + 0.125 999 450 683 593 750 794 559 6;
  • 3) 0.125 999 450 683 593 750 794 559 6 × 2 = 0 + 0.251 998 901 367 187 501 589 119 2;
  • 4) 0.251 998 901 367 187 501 589 119 2 × 2 = 0 + 0.503 997 802 734 375 003 178 238 4;
  • 5) 0.503 997 802 734 375 003 178 238 4 × 2 = 1 + 0.007 995 605 468 750 006 356 476 8;
  • 6) 0.007 995 605 468 750 006 356 476 8 × 2 = 0 + 0.015 991 210 937 500 012 712 953 6;
  • 7) 0.015 991 210 937 500 012 712 953 6 × 2 = 0 + 0.031 982 421 875 000 025 425 907 2;
  • 8) 0.031 982 421 875 000 025 425 907 2 × 2 = 0 + 0.063 964 843 750 000 050 851 814 4;
  • 9) 0.063 964 843 750 000 050 851 814 4 × 2 = 0 + 0.127 929 687 500 000 101 703 628 8;
  • 10) 0.127 929 687 500 000 101 703 628 8 × 2 = 0 + 0.255 859 375 000 000 203 407 257 6;
  • 11) 0.255 859 375 000 000 203 407 257 6 × 2 = 0 + 0.511 718 750 000 000 406 814 515 2;
  • 12) 0.511 718 750 000 000 406 814 515 2 × 2 = 1 + 0.023 437 500 000 000 813 629 030 4;
  • 13) 0.023 437 500 000 000 813 629 030 4 × 2 = 0 + 0.046 875 000 000 001 627 258 060 8;
  • 14) 0.046 875 000 000 001 627 258 060 8 × 2 = 0 + 0.093 750 000 000 003 254 516 121 6;
  • 15) 0.093 750 000 000 003 254 516 121 6 × 2 = 0 + 0.187 500 000 000 006 509 032 243 2;
  • 16) 0.187 500 000 000 006 509 032 243 2 × 2 = 0 + 0.375 000 000 000 013 018 064 486 4;
  • 17) 0.375 000 000 000 013 018 064 486 4 × 2 = 0 + 0.750 000 000 000 026 036 128 972 8;
  • 18) 0.750 000 000 000 026 036 128 972 8 × 2 = 1 + 0.500 000 000 000 052 072 257 945 6;
  • 19) 0.500 000 000 000 052 072 257 945 6 × 2 = 1 + 0.000 000 000 000 104 144 515 891 2;
  • 20) 0.000 000 000 000 104 144 515 891 2 × 2 = 0 + 0.000 000 000 000 208 289 031 782 4;
  • 21) 0.000 000 000 000 208 289 031 782 4 × 2 = 0 + 0.000 000 000 000 416 578 063 564 8;
  • 22) 0.000 000 000 000 416 578 063 564 8 × 2 = 0 + 0.000 000 000 000 833 156 127 129 6;
  • 23) 0.000 000 000 000 833 156 127 129 6 × 2 = 0 + 0.000 000 000 001 666 312 254 259 2;
  • 24) 0.000 000 000 001 666 312 254 259 2 × 2 = 0 + 0.000 000 000 003 332 624 508 518 4;

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:


0.031 499 862 670 898 437 698 639 9(10) =

0.0000 1000 0001 0000 0110 0000(2)

6. Positive number before normalization:

6.031 499 862 670 898 437 698 639 9(10) =

110.0000 1000 0001 0000 0110 0000(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:


6.031 499 862 670 898 437 698 639 9(10) =

110.0000 1000 0001 0000 0110 0000(2) =

110.0000 1000 0001 0000 0110 0000(2) × 20 =

1.1000 0010 0000 0100 0001 1000 00(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):
1.1000 0010 0000 0100 0001 1000 00


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. 100 0001 0000 0010 0000 1100 000 =

100 0001 0000 0010 0000 1100


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) =
100 0001 0000 0010 0000 1100


Decimal number -6.031 499 862 670 898 437 698 639 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 1000 0001 - 100 0001 0000 0010 0000 1100

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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