-0.000 000 000 882 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 882(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
-0.000 000 000 882(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:

|-0.000 000 000 882| = 0.000 000 000 882


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: 0.000 000 000 882.

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) 0.000 000 000 882 × 2 = 0 + 0.000 000 001 764;
  • 2) 0.000 000 001 764 × 2 = 0 + 0.000 000 003 528;
  • 3) 0.000 000 003 528 × 2 = 0 + 0.000 000 007 056;
  • 4) 0.000 000 007 056 × 2 = 0 + 0.000 000 014 112;
  • 5) 0.000 000 014 112 × 2 = 0 + 0.000 000 028 224;
  • 6) 0.000 000 028 224 × 2 = 0 + 0.000 000 056 448;
  • 7) 0.000 000 056 448 × 2 = 0 + 0.000 000 112 896;
  • 8) 0.000 000 112 896 × 2 = 0 + 0.000 000 225 792;
  • 9) 0.000 000 225 792 × 2 = 0 + 0.000 000 451 584;
  • 10) 0.000 000 451 584 × 2 = 0 + 0.000 000 903 168;
  • 11) 0.000 000 903 168 × 2 = 0 + 0.000 001 806 336;
  • 12) 0.000 001 806 336 × 2 = 0 + 0.000 003 612 672;
  • 13) 0.000 003 612 672 × 2 = 0 + 0.000 007 225 344;
  • 14) 0.000 007 225 344 × 2 = 0 + 0.000 014 450 688;
  • 15) 0.000 014 450 688 × 2 = 0 + 0.000 028 901 376;
  • 16) 0.000 028 901 376 × 2 = 0 + 0.000 057 802 752;
  • 17) 0.000 057 802 752 × 2 = 0 + 0.000 115 605 504;
  • 18) 0.000 115 605 504 × 2 = 0 + 0.000 231 211 008;
  • 19) 0.000 231 211 008 × 2 = 0 + 0.000 462 422 016;
  • 20) 0.000 462 422 016 × 2 = 0 + 0.000 924 844 032;
  • 21) 0.000 924 844 032 × 2 = 0 + 0.001 849 688 064;
  • 22) 0.001 849 688 064 × 2 = 0 + 0.003 699 376 128;
  • 23) 0.003 699 376 128 × 2 = 0 + 0.007 398 752 256;
  • 24) 0.007 398 752 256 × 2 = 0 + 0.014 797 504 512;
  • 25) 0.014 797 504 512 × 2 = 0 + 0.029 595 009 024;
  • 26) 0.029 595 009 024 × 2 = 0 + 0.059 190 018 048;
  • 27) 0.059 190 018 048 × 2 = 0 + 0.118 380 036 096;
  • 28) 0.118 380 036 096 × 2 = 0 + 0.236 760 072 192;
  • 29) 0.236 760 072 192 × 2 = 0 + 0.473 520 144 384;
  • 30) 0.473 520 144 384 × 2 = 0 + 0.947 040 288 768;
  • 31) 0.947 040 288 768 × 2 = 1 + 0.894 080 577 536;
  • 32) 0.894 080 577 536 × 2 = 1 + 0.788 161 155 072;
  • 33) 0.788 161 155 072 × 2 = 1 + 0.576 322 310 144;
  • 34) 0.576 322 310 144 × 2 = 1 + 0.152 644 620 288;
  • 35) 0.152 644 620 288 × 2 = 0 + 0.305 289 240 576;
  • 36) 0.305 289 240 576 × 2 = 0 + 0.610 578 481 152;
  • 37) 0.610 578 481 152 × 2 = 1 + 0.221 156 962 304;
  • 38) 0.221 156 962 304 × 2 = 0 + 0.442 313 924 608;
  • 39) 0.442 313 924 608 × 2 = 0 + 0.884 627 849 216;
  • 40) 0.884 627 849 216 × 2 = 1 + 0.769 255 698 432;
  • 41) 0.769 255 698 432 × 2 = 1 + 0.538 511 396 864;
  • 42) 0.538 511 396 864 × 2 = 1 + 0.077 022 793 728;
  • 43) 0.077 022 793 728 × 2 = 0 + 0.154 045 587 456;
  • 44) 0.154 045 587 456 × 2 = 0 + 0.308 091 174 912;
  • 45) 0.308 091 174 912 × 2 = 0 + 0.616 182 349 824;
  • 46) 0.616 182 349 824 × 2 = 1 + 0.232 364 699 648;
  • 47) 0.232 364 699 648 × 2 = 0 + 0.464 729 399 296;
  • 48) 0.464 729 399 296 × 2 = 0 + 0.929 458 798 592;
  • 49) 0.929 458 798 592 × 2 = 1 + 0.858 917 597 184;
  • 50) 0.858 917 597 184 × 2 = 1 + 0.717 835 194 368;
  • 51) 0.717 835 194 368 × 2 = 1 + 0.435 670 388 736;
  • 52) 0.435 670 388 736 × 2 = 0 + 0.871 340 777 472;
  • 53) 0.871 340 777 472 × 2 = 1 + 0.742 681 554 944;
  • 54) 0.742 681 554 944 × 2 = 1 + 0.485 363 109 888;

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:


0.000 000 000 882(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 1001 1100 0100 1110 11(2)

6. Positive number before normalization:

0.000 000 000 882(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 1001 1100 0100 1110 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 882(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 1001 1100 0100 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 1001 1100 0100 1110 11(2) × 20 =

1.1110 0100 1110 0010 0111 011(2) × 2-31


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): -31

Mantissa (not normalized):
1.1110 0100 1110 0010 0111 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 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) =

96(10) =

0110 0000(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. 111 0010 0111 0001 0011 1011 =

111 0010 0111 0001 0011 1011


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 0000

Mantissa (23 bits) =
111 0010 0111 0001 0011 1011


Decimal number -0.000 000 000 882 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 111 0010 0111 0001 0011 1011

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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