-0.000 000 000 823 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 823(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
-0.000 000 000 823(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:

|-0.000 000 000 823| = 0.000 000 000 823


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: 0.000 000 000 823.

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) 0.000 000 000 823 × 2 = 0 + 0.000 000 001 646;
  • 2) 0.000 000 001 646 × 2 = 0 + 0.000 000 003 292;
  • 3) 0.000 000 003 292 × 2 = 0 + 0.000 000 006 584;
  • 4) 0.000 000 006 584 × 2 = 0 + 0.000 000 013 168;
  • 5) 0.000 000 013 168 × 2 = 0 + 0.000 000 026 336;
  • 6) 0.000 000 026 336 × 2 = 0 + 0.000 000 052 672;
  • 7) 0.000 000 052 672 × 2 = 0 + 0.000 000 105 344;
  • 8) 0.000 000 105 344 × 2 = 0 + 0.000 000 210 688;
  • 9) 0.000 000 210 688 × 2 = 0 + 0.000 000 421 376;
  • 10) 0.000 000 421 376 × 2 = 0 + 0.000 000 842 752;
  • 11) 0.000 000 842 752 × 2 = 0 + 0.000 001 685 504;
  • 12) 0.000 001 685 504 × 2 = 0 + 0.000 003 371 008;
  • 13) 0.000 003 371 008 × 2 = 0 + 0.000 006 742 016;
  • 14) 0.000 006 742 016 × 2 = 0 + 0.000 013 484 032;
  • 15) 0.000 013 484 032 × 2 = 0 + 0.000 026 968 064;
  • 16) 0.000 026 968 064 × 2 = 0 + 0.000 053 936 128;
  • 17) 0.000 053 936 128 × 2 = 0 + 0.000 107 872 256;
  • 18) 0.000 107 872 256 × 2 = 0 + 0.000 215 744 512;
  • 19) 0.000 215 744 512 × 2 = 0 + 0.000 431 489 024;
  • 20) 0.000 431 489 024 × 2 = 0 + 0.000 862 978 048;
  • 21) 0.000 862 978 048 × 2 = 0 + 0.001 725 956 096;
  • 22) 0.001 725 956 096 × 2 = 0 + 0.003 451 912 192;
  • 23) 0.003 451 912 192 × 2 = 0 + 0.006 903 824 384;
  • 24) 0.006 903 824 384 × 2 = 0 + 0.013 807 648 768;
  • 25) 0.013 807 648 768 × 2 = 0 + 0.027 615 297 536;
  • 26) 0.027 615 297 536 × 2 = 0 + 0.055 230 595 072;
  • 27) 0.055 230 595 072 × 2 = 0 + 0.110 461 190 144;
  • 28) 0.110 461 190 144 × 2 = 0 + 0.220 922 380 288;
  • 29) 0.220 922 380 288 × 2 = 0 + 0.441 844 760 576;
  • 30) 0.441 844 760 576 × 2 = 0 + 0.883 689 521 152;
  • 31) 0.883 689 521 152 × 2 = 1 + 0.767 379 042 304;
  • 32) 0.767 379 042 304 × 2 = 1 + 0.534 758 084 608;
  • 33) 0.534 758 084 608 × 2 = 1 + 0.069 516 169 216;
  • 34) 0.069 516 169 216 × 2 = 0 + 0.139 032 338 432;
  • 35) 0.139 032 338 432 × 2 = 0 + 0.278 064 676 864;
  • 36) 0.278 064 676 864 × 2 = 0 + 0.556 129 353 728;
  • 37) 0.556 129 353 728 × 2 = 1 + 0.112 258 707 456;
  • 38) 0.112 258 707 456 × 2 = 0 + 0.224 517 414 912;
  • 39) 0.224 517 414 912 × 2 = 0 + 0.449 034 829 824;
  • 40) 0.449 034 829 824 × 2 = 0 + 0.898 069 659 648;
  • 41) 0.898 069 659 648 × 2 = 1 + 0.796 139 319 296;
  • 42) 0.796 139 319 296 × 2 = 1 + 0.592 278 638 592;
  • 43) 0.592 278 638 592 × 2 = 1 + 0.184 557 277 184;
  • 44) 0.184 557 277 184 × 2 = 0 + 0.369 114 554 368;
  • 45) 0.369 114 554 368 × 2 = 0 + 0.738 229 108 736;
  • 46) 0.738 229 108 736 × 2 = 1 + 0.476 458 217 472;
  • 47) 0.476 458 217 472 × 2 = 0 + 0.952 916 434 944;
  • 48) 0.952 916 434 944 × 2 = 1 + 0.905 832 869 888;
  • 49) 0.905 832 869 888 × 2 = 1 + 0.811 665 739 776;
  • 50) 0.811 665 739 776 × 2 = 1 + 0.623 331 479 552;
  • 51) 0.623 331 479 552 × 2 = 1 + 0.246 662 959 104;
  • 52) 0.246 662 959 104 × 2 = 0 + 0.493 325 918 208;
  • 53) 0.493 325 918 208 × 2 = 0 + 0.986 651 836 416;
  • 54) 0.986 651 836 416 × 2 = 1 + 0.973 303 672 832;

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:


0.000 000 000 823(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 1000 1110 0101 1110 01(2)

6. Positive number before normalization:

0.000 000 000 823(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 1000 1110 0101 1110 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:


0.000 000 000 823(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 1000 1110 0101 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 1000 1110 0101 1110 01(2) × 20 =

1.1100 0100 0111 0010 1111 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 0100 0111 0010 1111 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 0010 0011 1001 0111 1001 =

110 0010 0011 1001 0111 1001


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 0010 0011 1001 0111 1001


Decimal number -0.000 000 000 823 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 110 0010 0011 1001 0111 1001

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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