-0.000 000 000 742 147 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 5(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
-0.000 000 000 742 147 5(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:

|-0.000 000 000 742 147 5| = 0.000 000 000 742 147 5


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: 0.000 000 000 742 147 5.

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) 0.000 000 000 742 147 5 × 2 = 0 + 0.000 000 001 484 295;
  • 2) 0.000 000 001 484 295 × 2 = 0 + 0.000 000 002 968 59;
  • 3) 0.000 000 002 968 59 × 2 = 0 + 0.000 000 005 937 18;
  • 4) 0.000 000 005 937 18 × 2 = 0 + 0.000 000 011 874 36;
  • 5) 0.000 000 011 874 36 × 2 = 0 + 0.000 000 023 748 72;
  • 6) 0.000 000 023 748 72 × 2 = 0 + 0.000 000 047 497 44;
  • 7) 0.000 000 047 497 44 × 2 = 0 + 0.000 000 094 994 88;
  • 8) 0.000 000 094 994 88 × 2 = 0 + 0.000 000 189 989 76;
  • 9) 0.000 000 189 989 76 × 2 = 0 + 0.000 000 379 979 52;
  • 10) 0.000 000 379 979 52 × 2 = 0 + 0.000 000 759 959 04;
  • 11) 0.000 000 759 959 04 × 2 = 0 + 0.000 001 519 918 08;
  • 12) 0.000 001 519 918 08 × 2 = 0 + 0.000 003 039 836 16;
  • 13) 0.000 003 039 836 16 × 2 = 0 + 0.000 006 079 672 32;
  • 14) 0.000 006 079 672 32 × 2 = 0 + 0.000 012 159 344 64;
  • 15) 0.000 012 159 344 64 × 2 = 0 + 0.000 024 318 689 28;
  • 16) 0.000 024 318 689 28 × 2 = 0 + 0.000 048 637 378 56;
  • 17) 0.000 048 637 378 56 × 2 = 0 + 0.000 097 274 757 12;
  • 18) 0.000 097 274 757 12 × 2 = 0 + 0.000 194 549 514 24;
  • 19) 0.000 194 549 514 24 × 2 = 0 + 0.000 389 099 028 48;
  • 20) 0.000 389 099 028 48 × 2 = 0 + 0.000 778 198 056 96;
  • 21) 0.000 778 198 056 96 × 2 = 0 + 0.001 556 396 113 92;
  • 22) 0.001 556 396 113 92 × 2 = 0 + 0.003 112 792 227 84;
  • 23) 0.003 112 792 227 84 × 2 = 0 + 0.006 225 584 455 68;
  • 24) 0.006 225 584 455 68 × 2 = 0 + 0.012 451 168 911 36;
  • 25) 0.012 451 168 911 36 × 2 = 0 + 0.024 902 337 822 72;
  • 26) 0.024 902 337 822 72 × 2 = 0 + 0.049 804 675 645 44;
  • 27) 0.049 804 675 645 44 × 2 = 0 + 0.099 609 351 290 88;
  • 28) 0.099 609 351 290 88 × 2 = 0 + 0.199 218 702 581 76;
  • 29) 0.199 218 702 581 76 × 2 = 0 + 0.398 437 405 163 52;
  • 30) 0.398 437 405 163 52 × 2 = 0 + 0.796 874 810 327 04;
  • 31) 0.796 874 810 327 04 × 2 = 1 + 0.593 749 620 654 08;
  • 32) 0.593 749 620 654 08 × 2 = 1 + 0.187 499 241 308 16;
  • 33) 0.187 499 241 308 16 × 2 = 0 + 0.374 998 482 616 32;
  • 34) 0.374 998 482 616 32 × 2 = 0 + 0.749 996 965 232 64;
  • 35) 0.749 996 965 232 64 × 2 = 1 + 0.499 993 930 465 28;
  • 36) 0.499 993 930 465 28 × 2 = 0 + 0.999 987 860 930 56;
  • 37) 0.999 987 860 930 56 × 2 = 1 + 0.999 975 721 861 12;
  • 38) 0.999 975 721 861 12 × 2 = 1 + 0.999 951 443 722 24;
  • 39) 0.999 951 443 722 24 × 2 = 1 + 0.999 902 887 444 48;
  • 40) 0.999 902 887 444 48 × 2 = 1 + 0.999 805 774 888 96;
  • 41) 0.999 805 774 888 96 × 2 = 1 + 0.999 611 549 777 92;
  • 42) 0.999 611 549 777 92 × 2 = 1 + 0.999 223 099 555 84;
  • 43) 0.999 223 099 555 84 × 2 = 1 + 0.998 446 199 111 68;
  • 44) 0.998 446 199 111 68 × 2 = 1 + 0.996 892 398 223 36;
  • 45) 0.996 892 398 223 36 × 2 = 1 + 0.993 784 796 446 72;
  • 46) 0.993 784 796 446 72 × 2 = 1 + 0.987 569 592 893 44;
  • 47) 0.987 569 592 893 44 × 2 = 1 + 0.975 139 185 786 88;
  • 48) 0.975 139 185 786 88 × 2 = 1 + 0.950 278 371 573 76;
  • 49) 0.950 278 371 573 76 × 2 = 1 + 0.900 556 743 147 52;
  • 50) 0.900 556 743 147 52 × 2 = 1 + 0.801 113 486 295 04;
  • 51) 0.801 113 486 295 04 × 2 = 1 + 0.602 226 972 590 08;
  • 52) 0.602 226 972 590 08 × 2 = 1 + 0.204 453 945 180 16;
  • 53) 0.204 453 945 180 16 × 2 = 0 + 0.408 907 890 360 32;
  • 54) 0.408 907 890 360 32 × 2 = 0 + 0.817 815 780 720 64;

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:


0.000 000 000 742 147 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(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:


0.000 000 000 742 147 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 100(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 1111 1111 1111 100


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 1111 1111 1111 1100 =

100 1011 1111 1111 1111 1100


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 1111 1111 1111 1100


Decimal number -0.000 000 000 742 147 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1100

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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