-0.000 000 005 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 005 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
-0.000 000 005 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:

|-0.000 000 005 1| = 0.000 000 005 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: 0.000 000 005 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) 0.000 000 005 1 × 2 = 0 + 0.000 000 010 2;
  • 2) 0.000 000 010 2 × 2 = 0 + 0.000 000 020 4;
  • 3) 0.000 000 020 4 × 2 = 0 + 0.000 000 040 8;
  • 4) 0.000 000 040 8 × 2 = 0 + 0.000 000 081 6;
  • 5) 0.000 000 081 6 × 2 = 0 + 0.000 000 163 2;
  • 6) 0.000 000 163 2 × 2 = 0 + 0.000 000 326 4;
  • 7) 0.000 000 326 4 × 2 = 0 + 0.000 000 652 8;
  • 8) 0.000 000 652 8 × 2 = 0 + 0.000 001 305 6;
  • 9) 0.000 001 305 6 × 2 = 0 + 0.000 002 611 2;
  • 10) 0.000 002 611 2 × 2 = 0 + 0.000 005 222 4;
  • 11) 0.000 005 222 4 × 2 = 0 + 0.000 010 444 8;
  • 12) 0.000 010 444 8 × 2 = 0 + 0.000 020 889 6;
  • 13) 0.000 020 889 6 × 2 = 0 + 0.000 041 779 2;
  • 14) 0.000 041 779 2 × 2 = 0 + 0.000 083 558 4;
  • 15) 0.000 083 558 4 × 2 = 0 + 0.000 167 116 8;
  • 16) 0.000 167 116 8 × 2 = 0 + 0.000 334 233 6;
  • 17) 0.000 334 233 6 × 2 = 0 + 0.000 668 467 2;
  • 18) 0.000 668 467 2 × 2 = 0 + 0.001 336 934 4;
  • 19) 0.001 336 934 4 × 2 = 0 + 0.002 673 868 8;
  • 20) 0.002 673 868 8 × 2 = 0 + 0.005 347 737 6;
  • 21) 0.005 347 737 6 × 2 = 0 + 0.010 695 475 2;
  • 22) 0.010 695 475 2 × 2 = 0 + 0.021 390 950 4;
  • 23) 0.021 390 950 4 × 2 = 0 + 0.042 781 900 8;
  • 24) 0.042 781 900 8 × 2 = 0 + 0.085 563 801 6;
  • 25) 0.085 563 801 6 × 2 = 0 + 0.171 127 603 2;
  • 26) 0.171 127 603 2 × 2 = 0 + 0.342 255 206 4;
  • 27) 0.342 255 206 4 × 2 = 0 + 0.684 510 412 8;
  • 28) 0.684 510 412 8 × 2 = 1 + 0.369 020 825 6;
  • 29) 0.369 020 825 6 × 2 = 0 + 0.738 041 651 2;
  • 30) 0.738 041 651 2 × 2 = 1 + 0.476 083 302 4;
  • 31) 0.476 083 302 4 × 2 = 0 + 0.952 166 604 8;
  • 32) 0.952 166 604 8 × 2 = 1 + 0.904 333 209 6;
  • 33) 0.904 333 209 6 × 2 = 1 + 0.808 666 419 2;
  • 34) 0.808 666 419 2 × 2 = 1 + 0.617 332 838 4;
  • 35) 0.617 332 838 4 × 2 = 1 + 0.234 665 676 8;
  • 36) 0.234 665 676 8 × 2 = 0 + 0.469 331 353 6;
  • 37) 0.469 331 353 6 × 2 = 0 + 0.938 662 707 2;
  • 38) 0.938 662 707 2 × 2 = 1 + 0.877 325 414 4;
  • 39) 0.877 325 414 4 × 2 = 1 + 0.754 650 828 8;
  • 40) 0.754 650 828 8 × 2 = 1 + 0.509 301 657 6;
  • 41) 0.509 301 657 6 × 2 = 1 + 0.018 603 315 2;
  • 42) 0.018 603 315 2 × 2 = 0 + 0.037 206 630 4;
  • 43) 0.037 206 630 4 × 2 = 0 + 0.074 413 260 8;
  • 44) 0.074 413 260 8 × 2 = 0 + 0.148 826 521 6;
  • 45) 0.148 826 521 6 × 2 = 0 + 0.297 653 043 2;
  • 46) 0.297 653 043 2 × 2 = 0 + 0.595 306 086 4;
  • 47) 0.595 306 086 4 × 2 = 1 + 0.190 612 172 8;
  • 48) 0.190 612 172 8 × 2 = 0 + 0.381 224 345 6;
  • 49) 0.381 224 345 6 × 2 = 0 + 0.762 448 691 2;
  • 50) 0.762 448 691 2 × 2 = 1 + 0.524 897 382 4;
  • 51) 0.524 897 382 4 × 2 = 1 + 0.049 794 764 8;

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:


0.000 000 005 1(10) =

0.0000 0000 0000 0000 0000 0000 0001 0101 1110 0111 1000 0010 011(2)

6. Positive number before normalization:

0.000 000 005 1(10) =

0.0000 0000 0000 0000 0000 0000 0001 0101 1110 0111 1000 0010 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 005 1(10) =

0.0000 0000 0000 0000 0000 0000 0001 0101 1110 0111 1000 0010 011(2) =

0.0000 0000 0000 0000 0000 0000 0001 0101 1110 0111 1000 0010 011(2) × 20 =

1.0101 1110 0111 1000 0010 011(2) × 2-28


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): -28

Mantissa (not normalized):
1.0101 1110 0111 1000 0010 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-28 + 2(8-1) - 1 =

(-28 + 127)(10) =

99(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;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

99(10) =

0110 0011(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. 010 1111 0011 1100 0001 0011 =

010 1111 0011 1100 0001 0011


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 0011

Mantissa (23 bits) =
010 1111 0011 1100 0001 0011


Decimal number -0.000 000 005 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0011 - 010 1111 0011 1100 0001 0011

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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