0.000 000 010 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 010 5(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
0.000 000 010 5(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: 0.000 000 010 5.

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) 0.000 000 010 5 × 2 = 0 + 0.000 000 021;
  • 2) 0.000 000 021 × 2 = 0 + 0.000 000 042;
  • 3) 0.000 000 042 × 2 = 0 + 0.000 000 084;
  • 4) 0.000 000 084 × 2 = 0 + 0.000 000 168;
  • 5) 0.000 000 168 × 2 = 0 + 0.000 000 336;
  • 6) 0.000 000 336 × 2 = 0 + 0.000 000 672;
  • 7) 0.000 000 672 × 2 = 0 + 0.000 001 344;
  • 8) 0.000 001 344 × 2 = 0 + 0.000 002 688;
  • 9) 0.000 002 688 × 2 = 0 + 0.000 005 376;
  • 10) 0.000 005 376 × 2 = 0 + 0.000 010 752;
  • 11) 0.000 010 752 × 2 = 0 + 0.000 021 504;
  • 12) 0.000 021 504 × 2 = 0 + 0.000 043 008;
  • 13) 0.000 043 008 × 2 = 0 + 0.000 086 016;
  • 14) 0.000 086 016 × 2 = 0 + 0.000 172 032;
  • 15) 0.000 172 032 × 2 = 0 + 0.000 344 064;
  • 16) 0.000 344 064 × 2 = 0 + 0.000 688 128;
  • 17) 0.000 688 128 × 2 = 0 + 0.001 376 256;
  • 18) 0.001 376 256 × 2 = 0 + 0.002 752 512;
  • 19) 0.002 752 512 × 2 = 0 + 0.005 505 024;
  • 20) 0.005 505 024 × 2 = 0 + 0.011 010 048;
  • 21) 0.011 010 048 × 2 = 0 + 0.022 020 096;
  • 22) 0.022 020 096 × 2 = 0 + 0.044 040 192;
  • 23) 0.044 040 192 × 2 = 0 + 0.088 080 384;
  • 24) 0.088 080 384 × 2 = 0 + 0.176 160 768;
  • 25) 0.176 160 768 × 2 = 0 + 0.352 321 536;
  • 26) 0.352 321 536 × 2 = 0 + 0.704 643 072;
  • 27) 0.704 643 072 × 2 = 1 + 0.409 286 144;
  • 28) 0.409 286 144 × 2 = 0 + 0.818 572 288;
  • 29) 0.818 572 288 × 2 = 1 + 0.637 144 576;
  • 30) 0.637 144 576 × 2 = 1 + 0.274 289 152;
  • 31) 0.274 289 152 × 2 = 0 + 0.548 578 304;
  • 32) 0.548 578 304 × 2 = 1 + 0.097 156 608;
  • 33) 0.097 156 608 × 2 = 0 + 0.194 313 216;
  • 34) 0.194 313 216 × 2 = 0 + 0.388 626 432;
  • 35) 0.388 626 432 × 2 = 0 + 0.777 252 864;
  • 36) 0.777 252 864 × 2 = 1 + 0.554 505 728;
  • 37) 0.554 505 728 × 2 = 1 + 0.109 011 456;
  • 38) 0.109 011 456 × 2 = 0 + 0.218 022 912;
  • 39) 0.218 022 912 × 2 = 0 + 0.436 045 824;
  • 40) 0.436 045 824 × 2 = 0 + 0.872 091 648;
  • 41) 0.872 091 648 × 2 = 1 + 0.744 183 296;
  • 42) 0.744 183 296 × 2 = 1 + 0.488 366 592;
  • 43) 0.488 366 592 × 2 = 0 + 0.976 733 184;
  • 44) 0.976 733 184 × 2 = 1 + 0.953 466 368;
  • 45) 0.953 466 368 × 2 = 1 + 0.906 932 736;
  • 46) 0.906 932 736 × 2 = 1 + 0.813 865 472;
  • 47) 0.813 865 472 × 2 = 1 + 0.627 730 944;
  • 48) 0.627 730 944 × 2 = 1 + 0.255 461 888;
  • 49) 0.255 461 888 × 2 = 0 + 0.510 923 776;
  • 50) 0.510 923 776 × 2 = 1 + 0.021 847 552;

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:


0.000 000 010 5(10) =

0.0000 0000 0000 0000 0000 0000 0010 1101 0001 1000 1101 1111 01(2)

5. Positive number before normalization:

0.000 000 010 5(10) =

0.0000 0000 0000 0000 0000 0000 0010 1101 0001 1000 1101 1111 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 010 5(10) =

0.0000 0000 0000 0000 0000 0000 0010 1101 0001 1000 1101 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0010 1101 0001 1000 1101 1111 01(2) × 20 =

1.0110 1000 1100 0110 1111 101(2) × 2-27


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): -27

Mantissa (not normalized):
1.0110 1000 1100 0110 1111 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-27 + 2(8-1) - 1 =

(-27 + 127)(10) =

100(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;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 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) =

100(10) =

0110 0100(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. 011 0100 0110 0011 0111 1101 =

011 0100 0110 0011 0111 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 0100

Mantissa (23 bits) =
011 0100 0110 0011 0111 1101


Decimal number 0.000 000 010 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0100 - 011 0100 0110 0011 0111 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