-0.000 000 000 818 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 818(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
-0.000 000 000 818(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:

|-0.000 000 000 818| = 0.000 000 000 818


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: 0.000 000 000 818.

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) 0.000 000 000 818 × 2 = 0 + 0.000 000 001 636;
  • 2) 0.000 000 001 636 × 2 = 0 + 0.000 000 003 272;
  • 3) 0.000 000 003 272 × 2 = 0 + 0.000 000 006 544;
  • 4) 0.000 000 006 544 × 2 = 0 + 0.000 000 013 088;
  • 5) 0.000 000 013 088 × 2 = 0 + 0.000 000 026 176;
  • 6) 0.000 000 026 176 × 2 = 0 + 0.000 000 052 352;
  • 7) 0.000 000 052 352 × 2 = 0 + 0.000 000 104 704;
  • 8) 0.000 000 104 704 × 2 = 0 + 0.000 000 209 408;
  • 9) 0.000 000 209 408 × 2 = 0 + 0.000 000 418 816;
  • 10) 0.000 000 418 816 × 2 = 0 + 0.000 000 837 632;
  • 11) 0.000 000 837 632 × 2 = 0 + 0.000 001 675 264;
  • 12) 0.000 001 675 264 × 2 = 0 + 0.000 003 350 528;
  • 13) 0.000 003 350 528 × 2 = 0 + 0.000 006 701 056;
  • 14) 0.000 006 701 056 × 2 = 0 + 0.000 013 402 112;
  • 15) 0.000 013 402 112 × 2 = 0 + 0.000 026 804 224;
  • 16) 0.000 026 804 224 × 2 = 0 + 0.000 053 608 448;
  • 17) 0.000 053 608 448 × 2 = 0 + 0.000 107 216 896;
  • 18) 0.000 107 216 896 × 2 = 0 + 0.000 214 433 792;
  • 19) 0.000 214 433 792 × 2 = 0 + 0.000 428 867 584;
  • 20) 0.000 428 867 584 × 2 = 0 + 0.000 857 735 168;
  • 21) 0.000 857 735 168 × 2 = 0 + 0.001 715 470 336;
  • 22) 0.001 715 470 336 × 2 = 0 + 0.003 430 940 672;
  • 23) 0.003 430 940 672 × 2 = 0 + 0.006 861 881 344;
  • 24) 0.006 861 881 344 × 2 = 0 + 0.013 723 762 688;
  • 25) 0.013 723 762 688 × 2 = 0 + 0.027 447 525 376;
  • 26) 0.027 447 525 376 × 2 = 0 + 0.054 895 050 752;
  • 27) 0.054 895 050 752 × 2 = 0 + 0.109 790 101 504;
  • 28) 0.109 790 101 504 × 2 = 0 + 0.219 580 203 008;
  • 29) 0.219 580 203 008 × 2 = 0 + 0.439 160 406 016;
  • 30) 0.439 160 406 016 × 2 = 0 + 0.878 320 812 032;
  • 31) 0.878 320 812 032 × 2 = 1 + 0.756 641 624 064;
  • 32) 0.756 641 624 064 × 2 = 1 + 0.513 283 248 128;
  • 33) 0.513 283 248 128 × 2 = 1 + 0.026 566 496 256;
  • 34) 0.026 566 496 256 × 2 = 0 + 0.053 132 992 512;
  • 35) 0.053 132 992 512 × 2 = 0 + 0.106 265 985 024;
  • 36) 0.106 265 985 024 × 2 = 0 + 0.212 531 970 048;
  • 37) 0.212 531 970 048 × 2 = 0 + 0.425 063 940 096;
  • 38) 0.425 063 940 096 × 2 = 0 + 0.850 127 880 192;
  • 39) 0.850 127 880 192 × 2 = 1 + 0.700 255 760 384;
  • 40) 0.700 255 760 384 × 2 = 1 + 0.400 511 520 768;
  • 41) 0.400 511 520 768 × 2 = 0 + 0.801 023 041 536;
  • 42) 0.801 023 041 536 × 2 = 1 + 0.602 046 083 072;
  • 43) 0.602 046 083 072 × 2 = 1 + 0.204 092 166 144;
  • 44) 0.204 092 166 144 × 2 = 0 + 0.408 184 332 288;
  • 45) 0.408 184 332 288 × 2 = 0 + 0.816 368 664 576;
  • 46) 0.816 368 664 576 × 2 = 1 + 0.632 737 329 152;
  • 47) 0.632 737 329 152 × 2 = 1 + 0.265 474 658 304;
  • 48) 0.265 474 658 304 × 2 = 0 + 0.530 949 316 608;
  • 49) 0.530 949 316 608 × 2 = 1 + 0.061 898 633 216;
  • 50) 0.061 898 633 216 × 2 = 0 + 0.123 797 266 432;
  • 51) 0.123 797 266 432 × 2 = 0 + 0.247 594 532 864;
  • 52) 0.247 594 532 864 × 2 = 0 + 0.495 189 065 728;
  • 53) 0.495 189 065 728 × 2 = 0 + 0.990 378 131 456;
  • 54) 0.990 378 131 456 × 2 = 1 + 0.980 756 262 912;

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:


0.000 000 000 818(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 0011 0110 0110 1000 01(2)

6. Positive number before normalization:

0.000 000 000 818(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 0011 0110 0110 1000 01(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:


0.000 000 000 818(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 0011 0110 0110 1000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 0011 0110 0110 1000 01(2) × 20 =

1.1100 0001 1011 0011 0100 001(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):
1.1100 0001 1011 0011 0100 001


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. 110 0000 1101 1001 1010 0001 =

110 0000 1101 1001 1010 0001


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) =
110 0000 1101 1001 1010 0001


Decimal number -0.000 000 000 818 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 0000 1101 1001 1010 0001

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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