-0.000 000 000 813 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 813(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
-0.000 000 000 813(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:

|-0.000 000 000 813| = 0.000 000 000 813


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: 0.000 000 000 813.

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) 0.000 000 000 813 × 2 = 0 + 0.000 000 001 626;
  • 2) 0.000 000 001 626 × 2 = 0 + 0.000 000 003 252;
  • 3) 0.000 000 003 252 × 2 = 0 + 0.000 000 006 504;
  • 4) 0.000 000 006 504 × 2 = 0 + 0.000 000 013 008;
  • 5) 0.000 000 013 008 × 2 = 0 + 0.000 000 026 016;
  • 6) 0.000 000 026 016 × 2 = 0 + 0.000 000 052 032;
  • 7) 0.000 000 052 032 × 2 = 0 + 0.000 000 104 064;
  • 8) 0.000 000 104 064 × 2 = 0 + 0.000 000 208 128;
  • 9) 0.000 000 208 128 × 2 = 0 + 0.000 000 416 256;
  • 10) 0.000 000 416 256 × 2 = 0 + 0.000 000 832 512;
  • 11) 0.000 000 832 512 × 2 = 0 + 0.000 001 665 024;
  • 12) 0.000 001 665 024 × 2 = 0 + 0.000 003 330 048;
  • 13) 0.000 003 330 048 × 2 = 0 + 0.000 006 660 096;
  • 14) 0.000 006 660 096 × 2 = 0 + 0.000 013 320 192;
  • 15) 0.000 013 320 192 × 2 = 0 + 0.000 026 640 384;
  • 16) 0.000 026 640 384 × 2 = 0 + 0.000 053 280 768;
  • 17) 0.000 053 280 768 × 2 = 0 + 0.000 106 561 536;
  • 18) 0.000 106 561 536 × 2 = 0 + 0.000 213 123 072;
  • 19) 0.000 213 123 072 × 2 = 0 + 0.000 426 246 144;
  • 20) 0.000 426 246 144 × 2 = 0 + 0.000 852 492 288;
  • 21) 0.000 852 492 288 × 2 = 0 + 0.001 704 984 576;
  • 22) 0.001 704 984 576 × 2 = 0 + 0.003 409 969 152;
  • 23) 0.003 409 969 152 × 2 = 0 + 0.006 819 938 304;
  • 24) 0.006 819 938 304 × 2 = 0 + 0.013 639 876 608;
  • 25) 0.013 639 876 608 × 2 = 0 + 0.027 279 753 216;
  • 26) 0.027 279 753 216 × 2 = 0 + 0.054 559 506 432;
  • 27) 0.054 559 506 432 × 2 = 0 + 0.109 119 012 864;
  • 28) 0.109 119 012 864 × 2 = 0 + 0.218 238 025 728;
  • 29) 0.218 238 025 728 × 2 = 0 + 0.436 476 051 456;
  • 30) 0.436 476 051 456 × 2 = 0 + 0.872 952 102 912;
  • 31) 0.872 952 102 912 × 2 = 1 + 0.745 904 205 824;
  • 32) 0.745 904 205 824 × 2 = 1 + 0.491 808 411 648;
  • 33) 0.491 808 411 648 × 2 = 0 + 0.983 616 823 296;
  • 34) 0.983 616 823 296 × 2 = 1 + 0.967 233 646 592;
  • 35) 0.967 233 646 592 × 2 = 1 + 0.934 467 293 184;
  • 36) 0.934 467 293 184 × 2 = 1 + 0.868 934 586 368;
  • 37) 0.868 934 586 368 × 2 = 1 + 0.737 869 172 736;
  • 38) 0.737 869 172 736 × 2 = 1 + 0.475 738 345 472;
  • 39) 0.475 738 345 472 × 2 = 0 + 0.951 476 690 944;
  • 40) 0.951 476 690 944 × 2 = 1 + 0.902 953 381 888;
  • 41) 0.902 953 381 888 × 2 = 1 + 0.805 906 763 776;
  • 42) 0.805 906 763 776 × 2 = 1 + 0.611 813 527 552;
  • 43) 0.611 813 527 552 × 2 = 1 + 0.223 627 055 104;
  • 44) 0.223 627 055 104 × 2 = 0 + 0.447 254 110 208;
  • 45) 0.447 254 110 208 × 2 = 0 + 0.894 508 220 416;
  • 46) 0.894 508 220 416 × 2 = 1 + 0.789 016 440 832;
  • 47) 0.789 016 440 832 × 2 = 1 + 0.578 032 881 664;
  • 48) 0.578 032 881 664 × 2 = 1 + 0.156 065 763 328;
  • 49) 0.156 065 763 328 × 2 = 0 + 0.312 131 526 656;
  • 50) 0.312 131 526 656 × 2 = 0 + 0.624 263 053 312;
  • 51) 0.624 263 053 312 × 2 = 1 + 0.248 526 106 624;
  • 52) 0.248 526 106 624 × 2 = 0 + 0.497 052 213 248;
  • 53) 0.497 052 213 248 × 2 = 0 + 0.994 104 426 496;
  • 54) 0.994 104 426 496 × 2 = 1 + 0.988 208 852 992;

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:


0.000 000 000 813(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1101 1110 0111 0010 01(2)

6. Positive number before normalization:

0.000 000 000 813(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1101 1110 0111 0010 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:


0.000 000 000 813(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1101 1110 0111 0010 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0111 1101 1110 0111 0010 01(2) × 20 =

1.1011 1110 1111 0011 1001 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):
1.1011 1110 1111 0011 1001 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. 101 1111 0111 1001 1100 1001 =

101 1111 0111 1001 1100 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) =
101 1111 0111 1001 1100 1001


Decimal number -0.000 000 000 813 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 1111 0111 1001 1100 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