0.000 000 060 542 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 060 542 2(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
0.000 000 060 542 2(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 060 542 2.

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) 0.000 000 060 542 2 × 2 = 0 + 0.000 000 121 084 4;
  • 2) 0.000 000 121 084 4 × 2 = 0 + 0.000 000 242 168 8;
  • 3) 0.000 000 242 168 8 × 2 = 0 + 0.000 000 484 337 6;
  • 4) 0.000 000 484 337 6 × 2 = 0 + 0.000 000 968 675 2;
  • 5) 0.000 000 968 675 2 × 2 = 0 + 0.000 001 937 350 4;
  • 6) 0.000 001 937 350 4 × 2 = 0 + 0.000 003 874 700 8;
  • 7) 0.000 003 874 700 8 × 2 = 0 + 0.000 007 749 401 6;
  • 8) 0.000 007 749 401 6 × 2 = 0 + 0.000 015 498 803 2;
  • 9) 0.000 015 498 803 2 × 2 = 0 + 0.000 030 997 606 4;
  • 10) 0.000 030 997 606 4 × 2 = 0 + 0.000 061 995 212 8;
  • 11) 0.000 061 995 212 8 × 2 = 0 + 0.000 123 990 425 6;
  • 12) 0.000 123 990 425 6 × 2 = 0 + 0.000 247 980 851 2;
  • 13) 0.000 247 980 851 2 × 2 = 0 + 0.000 495 961 702 4;
  • 14) 0.000 495 961 702 4 × 2 = 0 + 0.000 991 923 404 8;
  • 15) 0.000 991 923 404 8 × 2 = 0 + 0.001 983 846 809 6;
  • 16) 0.001 983 846 809 6 × 2 = 0 + 0.003 967 693 619 2;
  • 17) 0.003 967 693 619 2 × 2 = 0 + 0.007 935 387 238 4;
  • 18) 0.007 935 387 238 4 × 2 = 0 + 0.015 870 774 476 8;
  • 19) 0.015 870 774 476 8 × 2 = 0 + 0.031 741 548 953 6;
  • 20) 0.031 741 548 953 6 × 2 = 0 + 0.063 483 097 907 2;
  • 21) 0.063 483 097 907 2 × 2 = 0 + 0.126 966 195 814 4;
  • 22) 0.126 966 195 814 4 × 2 = 0 + 0.253 932 391 628 8;
  • 23) 0.253 932 391 628 8 × 2 = 0 + 0.507 864 783 257 6;
  • 24) 0.507 864 783 257 6 × 2 = 1 + 0.015 729 566 515 2;
  • 25) 0.015 729 566 515 2 × 2 = 0 + 0.031 459 133 030 4;
  • 26) 0.031 459 133 030 4 × 2 = 0 + 0.062 918 266 060 8;
  • 27) 0.062 918 266 060 8 × 2 = 0 + 0.125 836 532 121 6;
  • 28) 0.125 836 532 121 6 × 2 = 0 + 0.251 673 064 243 2;
  • 29) 0.251 673 064 243 2 × 2 = 0 + 0.503 346 128 486 4;
  • 30) 0.503 346 128 486 4 × 2 = 1 + 0.006 692 256 972 8;
  • 31) 0.006 692 256 972 8 × 2 = 0 + 0.013 384 513 945 6;
  • 32) 0.013 384 513 945 6 × 2 = 0 + 0.026 769 027 891 2;
  • 33) 0.026 769 027 891 2 × 2 = 0 + 0.053 538 055 782 4;
  • 34) 0.053 538 055 782 4 × 2 = 0 + 0.107 076 111 564 8;
  • 35) 0.107 076 111 564 8 × 2 = 0 + 0.214 152 223 129 6;
  • 36) 0.214 152 223 129 6 × 2 = 0 + 0.428 304 446 259 2;
  • 37) 0.428 304 446 259 2 × 2 = 0 + 0.856 608 892 518 4;
  • 38) 0.856 608 892 518 4 × 2 = 1 + 0.713 217 785 036 8;
  • 39) 0.713 217 785 036 8 × 2 = 1 + 0.426 435 570 073 6;
  • 40) 0.426 435 570 073 6 × 2 = 0 + 0.852 871 140 147 2;
  • 41) 0.852 871 140 147 2 × 2 = 1 + 0.705 742 280 294 4;
  • 42) 0.705 742 280 294 4 × 2 = 1 + 0.411 484 560 588 8;
  • 43) 0.411 484 560 588 8 × 2 = 0 + 0.822 969 121 177 6;
  • 44) 0.822 969 121 177 6 × 2 = 1 + 0.645 938 242 355 2;
  • 45) 0.645 938 242 355 2 × 2 = 1 + 0.291 876 484 710 4;
  • 46) 0.291 876 484 710 4 × 2 = 0 + 0.583 752 969 420 8;
  • 47) 0.583 752 969 420 8 × 2 = 1 + 0.167 505 938 841 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).


4. 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:


0.000 000 060 542 2(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0110 1101 101(2)

5. Positive number before normalization:

0.000 000 060 542 2(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0110 1101 101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 060 542 2(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0110 1101 101(2) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0110 1101 101(2) × 20 =

1.0000 0100 0000 0110 1101 101(2) × 2-24


7. 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 0 (a positive number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.0000 0100 0000 0110 1101 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(10)


9. 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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

103(10) =

0110 0111(2)


11. 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 0010 0000 0011 0110 1101 =

000 0010 0000 0011 0110 1101


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 0111

Mantissa (23 bits) =
000 0010 0000 0011 0110 1101


Decimal number 0.000 000 060 542 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 000 0010 0000 0011 0110 1101

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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