-0.000 000 000 741 82 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 82(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
-0.000 000 000 741 82(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:

|-0.000 000 000 741 82| = 0.000 000 000 741 82


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: 0.000 000 000 741 82.

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) 0.000 000 000 741 82 × 2 = 0 + 0.000 000 001 483 64;
  • 2) 0.000 000 001 483 64 × 2 = 0 + 0.000 000 002 967 28;
  • 3) 0.000 000 002 967 28 × 2 = 0 + 0.000 000 005 934 56;
  • 4) 0.000 000 005 934 56 × 2 = 0 + 0.000 000 011 869 12;
  • 5) 0.000 000 011 869 12 × 2 = 0 + 0.000 000 023 738 24;
  • 6) 0.000 000 023 738 24 × 2 = 0 + 0.000 000 047 476 48;
  • 7) 0.000 000 047 476 48 × 2 = 0 + 0.000 000 094 952 96;
  • 8) 0.000 000 094 952 96 × 2 = 0 + 0.000 000 189 905 92;
  • 9) 0.000 000 189 905 92 × 2 = 0 + 0.000 000 379 811 84;
  • 10) 0.000 000 379 811 84 × 2 = 0 + 0.000 000 759 623 68;
  • 11) 0.000 000 759 623 68 × 2 = 0 + 0.000 001 519 247 36;
  • 12) 0.000 001 519 247 36 × 2 = 0 + 0.000 003 038 494 72;
  • 13) 0.000 003 038 494 72 × 2 = 0 + 0.000 006 076 989 44;
  • 14) 0.000 006 076 989 44 × 2 = 0 + 0.000 012 153 978 88;
  • 15) 0.000 012 153 978 88 × 2 = 0 + 0.000 024 307 957 76;
  • 16) 0.000 024 307 957 76 × 2 = 0 + 0.000 048 615 915 52;
  • 17) 0.000 048 615 915 52 × 2 = 0 + 0.000 097 231 831 04;
  • 18) 0.000 097 231 831 04 × 2 = 0 + 0.000 194 463 662 08;
  • 19) 0.000 194 463 662 08 × 2 = 0 + 0.000 388 927 324 16;
  • 20) 0.000 388 927 324 16 × 2 = 0 + 0.000 777 854 648 32;
  • 21) 0.000 777 854 648 32 × 2 = 0 + 0.001 555 709 296 64;
  • 22) 0.001 555 709 296 64 × 2 = 0 + 0.003 111 418 593 28;
  • 23) 0.003 111 418 593 28 × 2 = 0 + 0.006 222 837 186 56;
  • 24) 0.006 222 837 186 56 × 2 = 0 + 0.012 445 674 373 12;
  • 25) 0.012 445 674 373 12 × 2 = 0 + 0.024 891 348 746 24;
  • 26) 0.024 891 348 746 24 × 2 = 0 + 0.049 782 697 492 48;
  • 27) 0.049 782 697 492 48 × 2 = 0 + 0.099 565 394 984 96;
  • 28) 0.099 565 394 984 96 × 2 = 0 + 0.199 130 789 969 92;
  • 29) 0.199 130 789 969 92 × 2 = 0 + 0.398 261 579 939 84;
  • 30) 0.398 261 579 939 84 × 2 = 0 + 0.796 523 159 879 68;
  • 31) 0.796 523 159 879 68 × 2 = 1 + 0.593 046 319 759 36;
  • 32) 0.593 046 319 759 36 × 2 = 1 + 0.186 092 639 518 72;
  • 33) 0.186 092 639 518 72 × 2 = 0 + 0.372 185 279 037 44;
  • 34) 0.372 185 279 037 44 × 2 = 0 + 0.744 370 558 074 88;
  • 35) 0.744 370 558 074 88 × 2 = 1 + 0.488 741 116 149 76;
  • 36) 0.488 741 116 149 76 × 2 = 0 + 0.977 482 232 299 52;
  • 37) 0.977 482 232 299 52 × 2 = 1 + 0.954 964 464 599 04;
  • 38) 0.954 964 464 599 04 × 2 = 1 + 0.909 928 929 198 08;
  • 39) 0.909 928 929 198 08 × 2 = 1 + 0.819 857 858 396 16;
  • 40) 0.819 857 858 396 16 × 2 = 1 + 0.639 715 716 792 32;
  • 41) 0.639 715 716 792 32 × 2 = 1 + 0.279 431 433 584 64;
  • 42) 0.279 431 433 584 64 × 2 = 0 + 0.558 862 867 169 28;
  • 43) 0.558 862 867 169 28 × 2 = 1 + 0.117 725 734 338 56;
  • 44) 0.117 725 734 338 56 × 2 = 0 + 0.235 451 468 677 12;
  • 45) 0.235 451 468 677 12 × 2 = 0 + 0.470 902 937 354 24;
  • 46) 0.470 902 937 354 24 × 2 = 0 + 0.941 805 874 708 48;
  • 47) 0.941 805 874 708 48 × 2 = 1 + 0.883 611 749 416 96;
  • 48) 0.883 611 749 416 96 × 2 = 1 + 0.767 223 498 833 92;
  • 49) 0.767 223 498 833 92 × 2 = 1 + 0.534 446 997 667 84;
  • 50) 0.534 446 997 667 84 × 2 = 1 + 0.068 893 995 335 68;
  • 51) 0.068 893 995 335 68 × 2 = 0 + 0.137 787 990 671 36;
  • 52) 0.137 787 990 671 36 × 2 = 0 + 0.275 575 981 342 72;
  • 53) 0.275 575 981 342 72 × 2 = 0 + 0.551 151 962 685 44;
  • 54) 0.551 151 962 685 44 × 2 = 1 + 0.102 303 925 370 88;

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:


0.000 000 000 741 82(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1010 0011 1100 01(2)

6. Positive number before normalization:

0.000 000 000 741 82(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1010 0011 1100 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:


0.000 000 000 741 82(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1010 0011 1100 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1010 0011 1100 01(2) × 20 =

1.1001 0111 1101 0001 1110 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):
1.1001 0111 1101 0001 1110 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. 100 1011 1110 1000 1111 0001 =

100 1011 1110 1000 1111 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) =
100 1011 1110 1000 1111 0001


Decimal number -0.000 000 000 741 82 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1110 1000 1111 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