-0.000 000 030 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 030 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
-0.000 000 030 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. Start with the positive version of the number:

|-0.000 000 030 1| = 0.000 000 030 1


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: 0.000 000 030 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) 0.000 000 030 1 × 2 = 0 + 0.000 000 060 2;
  • 2) 0.000 000 060 2 × 2 = 0 + 0.000 000 120 4;
  • 3) 0.000 000 120 4 × 2 = 0 + 0.000 000 240 8;
  • 4) 0.000 000 240 8 × 2 = 0 + 0.000 000 481 6;
  • 5) 0.000 000 481 6 × 2 = 0 + 0.000 000 963 2;
  • 6) 0.000 000 963 2 × 2 = 0 + 0.000 001 926 4;
  • 7) 0.000 001 926 4 × 2 = 0 + 0.000 003 852 8;
  • 8) 0.000 003 852 8 × 2 = 0 + 0.000 007 705 6;
  • 9) 0.000 007 705 6 × 2 = 0 + 0.000 015 411 2;
  • 10) 0.000 015 411 2 × 2 = 0 + 0.000 030 822 4;
  • 11) 0.000 030 822 4 × 2 = 0 + 0.000 061 644 8;
  • 12) 0.000 061 644 8 × 2 = 0 + 0.000 123 289 6;
  • 13) 0.000 123 289 6 × 2 = 0 + 0.000 246 579 2;
  • 14) 0.000 246 579 2 × 2 = 0 + 0.000 493 158 4;
  • 15) 0.000 493 158 4 × 2 = 0 + 0.000 986 316 8;
  • 16) 0.000 986 316 8 × 2 = 0 + 0.001 972 633 6;
  • 17) 0.001 972 633 6 × 2 = 0 + 0.003 945 267 2;
  • 18) 0.003 945 267 2 × 2 = 0 + 0.007 890 534 4;
  • 19) 0.007 890 534 4 × 2 = 0 + 0.015 781 068 8;
  • 20) 0.015 781 068 8 × 2 = 0 + 0.031 562 137 6;
  • 21) 0.031 562 137 6 × 2 = 0 + 0.063 124 275 2;
  • 22) 0.063 124 275 2 × 2 = 0 + 0.126 248 550 4;
  • 23) 0.126 248 550 4 × 2 = 0 + 0.252 497 100 8;
  • 24) 0.252 497 100 8 × 2 = 0 + 0.504 994 201 6;
  • 25) 0.504 994 201 6 × 2 = 1 + 0.009 988 403 2;
  • 26) 0.009 988 403 2 × 2 = 0 + 0.019 976 806 4;
  • 27) 0.019 976 806 4 × 2 = 0 + 0.039 953 612 8;
  • 28) 0.039 953 612 8 × 2 = 0 + 0.079 907 225 6;
  • 29) 0.079 907 225 6 × 2 = 0 + 0.159 814 451 2;
  • 30) 0.159 814 451 2 × 2 = 0 + 0.319 628 902 4;
  • 31) 0.319 628 902 4 × 2 = 0 + 0.639 257 804 8;
  • 32) 0.639 257 804 8 × 2 = 1 + 0.278 515 609 6;
  • 33) 0.278 515 609 6 × 2 = 0 + 0.557 031 219 2;
  • 34) 0.557 031 219 2 × 2 = 1 + 0.114 062 438 4;
  • 35) 0.114 062 438 4 × 2 = 0 + 0.228 124 876 8;
  • 36) 0.228 124 876 8 × 2 = 0 + 0.456 249 753 6;
  • 37) 0.456 249 753 6 × 2 = 0 + 0.912 499 507 2;
  • 38) 0.912 499 507 2 × 2 = 1 + 0.824 999 014 4;
  • 39) 0.824 999 014 4 × 2 = 1 + 0.649 998 028 8;
  • 40) 0.649 998 028 8 × 2 = 1 + 0.299 996 057 6;
  • 41) 0.299 996 057 6 × 2 = 0 + 0.599 992 115 2;
  • 42) 0.599 992 115 2 × 2 = 1 + 0.199 984 230 4;
  • 43) 0.199 984 230 4 × 2 = 0 + 0.399 968 460 8;
  • 44) 0.399 968 460 8 × 2 = 0 + 0.799 936 921 6;
  • 45) 0.799 936 921 6 × 2 = 1 + 0.599 873 843 2;
  • 46) 0.599 873 843 2 × 2 = 1 + 0.199 747 686 4;
  • 47) 0.199 747 686 4 × 2 = 0 + 0.399 495 372 8;
  • 48) 0.399 495 372 8 × 2 = 0 + 0.798 990 745 6;

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:


0.000 000 030 1(10) =

0.0000 0000 0000 0000 0000 0000 1000 0001 0100 0111 0100 1100(2)

6. Positive number before normalization:

0.000 000 030 1(10) =

0.0000 0000 0000 0000 0000 0000 1000 0001 0100 0111 0100 1100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 030 1(10) =

0.0000 0000 0000 0000 0000 0000 1000 0001 0100 0111 0100 1100(2) =

0.0000 0000 0000 0000 0000 0000 1000 0001 0100 0111 0100 1100(2) × 20 =

1.0000 0010 1000 1110 1001 100(2) × 2-25


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): -25

Mantissa (not normalized):
1.0000 0010 1000 1110 1001 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-25 + 2(8-1) - 1 =

(-25 + 127)(10) =

102(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;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

102(10) =

0110 0110(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. 000 0001 0100 0111 0100 1100 =

000 0001 0100 0111 0100 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) =
0110 0110

Mantissa (23 bits) =
000 0001 0100 0111 0100 1100


Decimal number -0.000 000 030 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0110 - 000 0001 0100 0111 0100 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