-0.000 000 000 742 17 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 17(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
-0.000 000 000 742 17(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:

|-0.000 000 000 742 17| = 0.000 000 000 742 17


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: 0.000 000 000 742 17.

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) 0.000 000 000 742 17 × 2 = 0 + 0.000 000 001 484 34;
  • 2) 0.000 000 001 484 34 × 2 = 0 + 0.000 000 002 968 68;
  • 3) 0.000 000 002 968 68 × 2 = 0 + 0.000 000 005 937 36;
  • 4) 0.000 000 005 937 36 × 2 = 0 + 0.000 000 011 874 72;
  • 5) 0.000 000 011 874 72 × 2 = 0 + 0.000 000 023 749 44;
  • 6) 0.000 000 023 749 44 × 2 = 0 + 0.000 000 047 498 88;
  • 7) 0.000 000 047 498 88 × 2 = 0 + 0.000 000 094 997 76;
  • 8) 0.000 000 094 997 76 × 2 = 0 + 0.000 000 189 995 52;
  • 9) 0.000 000 189 995 52 × 2 = 0 + 0.000 000 379 991 04;
  • 10) 0.000 000 379 991 04 × 2 = 0 + 0.000 000 759 982 08;
  • 11) 0.000 000 759 982 08 × 2 = 0 + 0.000 001 519 964 16;
  • 12) 0.000 001 519 964 16 × 2 = 0 + 0.000 003 039 928 32;
  • 13) 0.000 003 039 928 32 × 2 = 0 + 0.000 006 079 856 64;
  • 14) 0.000 006 079 856 64 × 2 = 0 + 0.000 012 159 713 28;
  • 15) 0.000 012 159 713 28 × 2 = 0 + 0.000 024 319 426 56;
  • 16) 0.000 024 319 426 56 × 2 = 0 + 0.000 048 638 853 12;
  • 17) 0.000 048 638 853 12 × 2 = 0 + 0.000 097 277 706 24;
  • 18) 0.000 097 277 706 24 × 2 = 0 + 0.000 194 555 412 48;
  • 19) 0.000 194 555 412 48 × 2 = 0 + 0.000 389 110 824 96;
  • 20) 0.000 389 110 824 96 × 2 = 0 + 0.000 778 221 649 92;
  • 21) 0.000 778 221 649 92 × 2 = 0 + 0.001 556 443 299 84;
  • 22) 0.001 556 443 299 84 × 2 = 0 + 0.003 112 886 599 68;
  • 23) 0.003 112 886 599 68 × 2 = 0 + 0.006 225 773 199 36;
  • 24) 0.006 225 773 199 36 × 2 = 0 + 0.012 451 546 398 72;
  • 25) 0.012 451 546 398 72 × 2 = 0 + 0.024 903 092 797 44;
  • 26) 0.024 903 092 797 44 × 2 = 0 + 0.049 806 185 594 88;
  • 27) 0.049 806 185 594 88 × 2 = 0 + 0.099 612 371 189 76;
  • 28) 0.099 612 371 189 76 × 2 = 0 + 0.199 224 742 379 52;
  • 29) 0.199 224 742 379 52 × 2 = 0 + 0.398 449 484 759 04;
  • 30) 0.398 449 484 759 04 × 2 = 0 + 0.796 898 969 518 08;
  • 31) 0.796 898 969 518 08 × 2 = 1 + 0.593 797 939 036 16;
  • 32) 0.593 797 939 036 16 × 2 = 1 + 0.187 595 878 072 32;
  • 33) 0.187 595 878 072 32 × 2 = 0 + 0.375 191 756 144 64;
  • 34) 0.375 191 756 144 64 × 2 = 0 + 0.750 383 512 289 28;
  • 35) 0.750 383 512 289 28 × 2 = 1 + 0.500 767 024 578 56;
  • 36) 0.500 767 024 578 56 × 2 = 1 + 0.001 534 049 157 12;
  • 37) 0.001 534 049 157 12 × 2 = 0 + 0.003 068 098 314 24;
  • 38) 0.003 068 098 314 24 × 2 = 0 + 0.006 136 196 628 48;
  • 39) 0.006 136 196 628 48 × 2 = 0 + 0.012 272 393 256 96;
  • 40) 0.012 272 393 256 96 × 2 = 0 + 0.024 544 786 513 92;
  • 41) 0.024 544 786 513 92 × 2 = 0 + 0.049 089 573 027 84;
  • 42) 0.049 089 573 027 84 × 2 = 0 + 0.098 179 146 055 68;
  • 43) 0.098 179 146 055 68 × 2 = 0 + 0.196 358 292 111 36;
  • 44) 0.196 358 292 111 36 × 2 = 0 + 0.392 716 584 222 72;
  • 45) 0.392 716 584 222 72 × 2 = 0 + 0.785 433 168 445 44;
  • 46) 0.785 433 168 445 44 × 2 = 1 + 0.570 866 336 890 88;
  • 47) 0.570 866 336 890 88 × 2 = 1 + 0.141 732 673 781 76;
  • 48) 0.141 732 673 781 76 × 2 = 0 + 0.283 465 347 563 52;
  • 49) 0.283 465 347 563 52 × 2 = 0 + 0.566 930 695 127 04;
  • 50) 0.566 930 695 127 04 × 2 = 1 + 0.133 861 390 254 08;
  • 51) 0.133 861 390 254 08 × 2 = 0 + 0.267 722 780 508 16;
  • 52) 0.267 722 780 508 16 × 2 = 0 + 0.535 445 561 016 32;
  • 53) 0.535 445 561 016 32 × 2 = 1 + 0.070 891 122 032 64;
  • 54) 0.070 891 122 032 64 × 2 = 0 + 0.141 782 244 065 28;

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:


0.000 000 000 742 17(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0110 0100 10(2)

6. Positive number before normalization:

0.000 000 000 742 17(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0110 0100 10(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:


0.000 000 000 742 17(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0110 0100 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0110 0100 10(2) × 20 =

1.1001 1000 0000 0011 0010 010(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 1000 0000 0011 0010 010


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 1100 0000 0001 1001 0010 =

100 1100 0000 0001 1001 0010


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 1100 0000 0001 1001 0010


Decimal number -0.000 000 000 742 17 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0001 1001 0010

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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