0.000 000 001 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 001 7(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
0.000 000 001 7(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: 0.000 000 001 7.

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) 0.000 000 001 7 × 2 = 0 + 0.000 000 003 4;
  • 2) 0.000 000 003 4 × 2 = 0 + 0.000 000 006 8;
  • 3) 0.000 000 006 8 × 2 = 0 + 0.000 000 013 6;
  • 4) 0.000 000 013 6 × 2 = 0 + 0.000 000 027 2;
  • 5) 0.000 000 027 2 × 2 = 0 + 0.000 000 054 4;
  • 6) 0.000 000 054 4 × 2 = 0 + 0.000 000 108 8;
  • 7) 0.000 000 108 8 × 2 = 0 + 0.000 000 217 6;
  • 8) 0.000 000 217 6 × 2 = 0 + 0.000 000 435 2;
  • 9) 0.000 000 435 2 × 2 = 0 + 0.000 000 870 4;
  • 10) 0.000 000 870 4 × 2 = 0 + 0.000 001 740 8;
  • 11) 0.000 001 740 8 × 2 = 0 + 0.000 003 481 6;
  • 12) 0.000 003 481 6 × 2 = 0 + 0.000 006 963 2;
  • 13) 0.000 006 963 2 × 2 = 0 + 0.000 013 926 4;
  • 14) 0.000 013 926 4 × 2 = 0 + 0.000 027 852 8;
  • 15) 0.000 027 852 8 × 2 = 0 + 0.000 055 705 6;
  • 16) 0.000 055 705 6 × 2 = 0 + 0.000 111 411 2;
  • 17) 0.000 111 411 2 × 2 = 0 + 0.000 222 822 4;
  • 18) 0.000 222 822 4 × 2 = 0 + 0.000 445 644 8;
  • 19) 0.000 445 644 8 × 2 = 0 + 0.000 891 289 6;
  • 20) 0.000 891 289 6 × 2 = 0 + 0.001 782 579 2;
  • 21) 0.001 782 579 2 × 2 = 0 + 0.003 565 158 4;
  • 22) 0.003 565 158 4 × 2 = 0 + 0.007 130 316 8;
  • 23) 0.007 130 316 8 × 2 = 0 + 0.014 260 633 6;
  • 24) 0.014 260 633 6 × 2 = 0 + 0.028 521 267 2;
  • 25) 0.028 521 267 2 × 2 = 0 + 0.057 042 534 4;
  • 26) 0.057 042 534 4 × 2 = 0 + 0.114 085 068 8;
  • 27) 0.114 085 068 8 × 2 = 0 + 0.228 170 137 6;
  • 28) 0.228 170 137 6 × 2 = 0 + 0.456 340 275 2;
  • 29) 0.456 340 275 2 × 2 = 0 + 0.912 680 550 4;
  • 30) 0.912 680 550 4 × 2 = 1 + 0.825 361 100 8;
  • 31) 0.825 361 100 8 × 2 = 1 + 0.650 722 201 6;
  • 32) 0.650 722 201 6 × 2 = 1 + 0.301 444 403 2;
  • 33) 0.301 444 403 2 × 2 = 0 + 0.602 888 806 4;
  • 34) 0.602 888 806 4 × 2 = 1 + 0.205 777 612 8;
  • 35) 0.205 777 612 8 × 2 = 0 + 0.411 555 225 6;
  • 36) 0.411 555 225 6 × 2 = 0 + 0.823 110 451 2;
  • 37) 0.823 110 451 2 × 2 = 1 + 0.646 220 902 4;
  • 38) 0.646 220 902 4 × 2 = 1 + 0.292 441 804 8;
  • 39) 0.292 441 804 8 × 2 = 0 + 0.584 883 609 6;
  • 40) 0.584 883 609 6 × 2 = 1 + 0.169 767 219 2;
  • 41) 0.169 767 219 2 × 2 = 0 + 0.339 534 438 4;
  • 42) 0.339 534 438 4 × 2 = 0 + 0.679 068 876 8;
  • 43) 0.679 068 876 8 × 2 = 1 + 0.358 137 753 6;
  • 44) 0.358 137 753 6 × 2 = 0 + 0.716 275 507 2;
  • 45) 0.716 275 507 2 × 2 = 1 + 0.432 551 014 4;
  • 46) 0.432 551 014 4 × 2 = 0 + 0.865 102 028 8;
  • 47) 0.865 102 028 8 × 2 = 1 + 0.730 204 057 6;
  • 48) 0.730 204 057 6 × 2 = 1 + 0.460 408 115 2;
  • 49) 0.460 408 115 2 × 2 = 0 + 0.920 816 230 4;
  • 50) 0.920 816 230 4 × 2 = 1 + 0.841 632 460 8;
  • 51) 0.841 632 460 8 × 2 = 1 + 0.683 264 921 6;
  • 52) 0.683 264 921 6 × 2 = 1 + 0.366 529 843 2;
  • 53) 0.366 529 843 2 × 2 = 0 + 0.733 059 686 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).


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:


0.000 000 001 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0111 0100 1101 0010 1011 0111 0(2)

5. Positive number before normalization:

0.000 000 001 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0111 0100 1101 0010 1011 0111 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 30 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 001 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0111 0100 1101 0010 1011 0111 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0111 0100 1101 0010 1011 0111 0(2) × 20 =

1.1101 0011 0100 1010 1101 110(2) × 2-30


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): -30

Mantissa (not normalized):
1.1101 0011 0100 1010 1101 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-30 + 2(8-1) - 1 =

(-30 + 127)(10) =

97(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;
  • 97 ÷ 2 = 48 + 1;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

97(10) =

0110 0001(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. 110 1001 1010 0101 0110 1110 =

110 1001 1010 0101 0110 1110


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 0001

Mantissa (23 bits) =
110 1001 1010 0101 0110 1110


Decimal number 0.000 000 001 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0001 - 110 1001 1010 0101 0110 1110

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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