-0.000 000 026 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 026 3(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
-0.000 000 026 3(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:

|-0.000 000 026 3| = 0.000 000 026 3


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: 0.000 000 026 3.

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) 0.000 000 026 3 × 2 = 0 + 0.000 000 052 6;
  • 2) 0.000 000 052 6 × 2 = 0 + 0.000 000 105 2;
  • 3) 0.000 000 105 2 × 2 = 0 + 0.000 000 210 4;
  • 4) 0.000 000 210 4 × 2 = 0 + 0.000 000 420 8;
  • 5) 0.000 000 420 8 × 2 = 0 + 0.000 000 841 6;
  • 6) 0.000 000 841 6 × 2 = 0 + 0.000 001 683 2;
  • 7) 0.000 001 683 2 × 2 = 0 + 0.000 003 366 4;
  • 8) 0.000 003 366 4 × 2 = 0 + 0.000 006 732 8;
  • 9) 0.000 006 732 8 × 2 = 0 + 0.000 013 465 6;
  • 10) 0.000 013 465 6 × 2 = 0 + 0.000 026 931 2;
  • 11) 0.000 026 931 2 × 2 = 0 + 0.000 053 862 4;
  • 12) 0.000 053 862 4 × 2 = 0 + 0.000 107 724 8;
  • 13) 0.000 107 724 8 × 2 = 0 + 0.000 215 449 6;
  • 14) 0.000 215 449 6 × 2 = 0 + 0.000 430 899 2;
  • 15) 0.000 430 899 2 × 2 = 0 + 0.000 861 798 4;
  • 16) 0.000 861 798 4 × 2 = 0 + 0.001 723 596 8;
  • 17) 0.001 723 596 8 × 2 = 0 + 0.003 447 193 6;
  • 18) 0.003 447 193 6 × 2 = 0 + 0.006 894 387 2;
  • 19) 0.006 894 387 2 × 2 = 0 + 0.013 788 774 4;
  • 20) 0.013 788 774 4 × 2 = 0 + 0.027 577 548 8;
  • 21) 0.027 577 548 8 × 2 = 0 + 0.055 155 097 6;
  • 22) 0.055 155 097 6 × 2 = 0 + 0.110 310 195 2;
  • 23) 0.110 310 195 2 × 2 = 0 + 0.220 620 390 4;
  • 24) 0.220 620 390 4 × 2 = 0 + 0.441 240 780 8;
  • 25) 0.441 240 780 8 × 2 = 0 + 0.882 481 561 6;
  • 26) 0.882 481 561 6 × 2 = 1 + 0.764 963 123 2;
  • 27) 0.764 963 123 2 × 2 = 1 + 0.529 926 246 4;
  • 28) 0.529 926 246 4 × 2 = 1 + 0.059 852 492 8;
  • 29) 0.059 852 492 8 × 2 = 0 + 0.119 704 985 6;
  • 30) 0.119 704 985 6 × 2 = 0 + 0.239 409 971 2;
  • 31) 0.239 409 971 2 × 2 = 0 + 0.478 819 942 4;
  • 32) 0.478 819 942 4 × 2 = 0 + 0.957 639 884 8;
  • 33) 0.957 639 884 8 × 2 = 1 + 0.915 279 769 6;
  • 34) 0.915 279 769 6 × 2 = 1 + 0.830 559 539 2;
  • 35) 0.830 559 539 2 × 2 = 1 + 0.661 119 078 4;
  • 36) 0.661 119 078 4 × 2 = 1 + 0.322 238 156 8;
  • 37) 0.322 238 156 8 × 2 = 0 + 0.644 476 313 6;
  • 38) 0.644 476 313 6 × 2 = 1 + 0.288 952 627 2;
  • 39) 0.288 952 627 2 × 2 = 0 + 0.577 905 254 4;
  • 40) 0.577 905 254 4 × 2 = 1 + 0.155 810 508 8;
  • 41) 0.155 810 508 8 × 2 = 0 + 0.311 621 017 6;
  • 42) 0.311 621 017 6 × 2 = 0 + 0.623 242 035 2;
  • 43) 0.623 242 035 2 × 2 = 1 + 0.246 484 070 4;
  • 44) 0.246 484 070 4 × 2 = 0 + 0.492 968 140 8;
  • 45) 0.492 968 140 8 × 2 = 0 + 0.985 936 281 6;
  • 46) 0.985 936 281 6 × 2 = 1 + 0.971 872 563 2;
  • 47) 0.971 872 563 2 × 2 = 1 + 0.943 745 126 4;
  • 48) 0.943 745 126 4 × 2 = 1 + 0.887 490 252 8;
  • 49) 0.887 490 252 8 × 2 = 1 + 0.774 980 505 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:


0.000 000 026 3(10) =

0.0000 0000 0000 0000 0000 0000 0111 0000 1111 0101 0010 0111 1(2)

6. Positive number before normalization:

0.000 000 026 3(10) =

0.0000 0000 0000 0000 0000 0000 0111 0000 1111 0101 0010 0111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 026 3(10) =

0.0000 0000 0000 0000 0000 0000 0111 0000 1111 0101 0010 0111 1(2) =

0.0000 0000 0000 0000 0000 0000 0111 0000 1111 0101 0010 0111 1(2) × 20 =

1.1100 0011 1101 0100 1001 111(2) × 2-26


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): -26

Mantissa (not normalized):
1.1100 0011 1101 0100 1001 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(8-1) - 1 =

(-26 + 127)(10) =

101(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;
  • 101 ÷ 2 = 50 + 1;
  • 50 ÷ 2 = 25 + 0;
  • 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) =

101(10) =

0110 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. 110 0001 1110 1010 0100 1111 =

110 0001 1110 1010 0100 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) =
0110 0101

Mantissa (23 bits) =
110 0001 1110 1010 0100 1111


Decimal number -0.000 000 026 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0101 - 110 0001 1110 1010 0100 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