-0.000 000 000 742 138 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 138 9(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
-0.000 000 000 742 138 9(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:

|-0.000 000 000 742 138 9| = 0.000 000 000 742 138 9


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: 0.000 000 000 742 138 9.

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) 0.000 000 000 742 138 9 × 2 = 0 + 0.000 000 001 484 277 8;
  • 2) 0.000 000 001 484 277 8 × 2 = 0 + 0.000 000 002 968 555 6;
  • 3) 0.000 000 002 968 555 6 × 2 = 0 + 0.000 000 005 937 111 2;
  • 4) 0.000 000 005 937 111 2 × 2 = 0 + 0.000 000 011 874 222 4;
  • 5) 0.000 000 011 874 222 4 × 2 = 0 + 0.000 000 023 748 444 8;
  • 6) 0.000 000 023 748 444 8 × 2 = 0 + 0.000 000 047 496 889 6;
  • 7) 0.000 000 047 496 889 6 × 2 = 0 + 0.000 000 094 993 779 2;
  • 8) 0.000 000 094 993 779 2 × 2 = 0 + 0.000 000 189 987 558 4;
  • 9) 0.000 000 189 987 558 4 × 2 = 0 + 0.000 000 379 975 116 8;
  • 10) 0.000 000 379 975 116 8 × 2 = 0 + 0.000 000 759 950 233 6;
  • 11) 0.000 000 759 950 233 6 × 2 = 0 + 0.000 001 519 900 467 2;
  • 12) 0.000 001 519 900 467 2 × 2 = 0 + 0.000 003 039 800 934 4;
  • 13) 0.000 003 039 800 934 4 × 2 = 0 + 0.000 006 079 601 868 8;
  • 14) 0.000 006 079 601 868 8 × 2 = 0 + 0.000 012 159 203 737 6;
  • 15) 0.000 012 159 203 737 6 × 2 = 0 + 0.000 024 318 407 475 2;
  • 16) 0.000 024 318 407 475 2 × 2 = 0 + 0.000 048 636 814 950 4;
  • 17) 0.000 048 636 814 950 4 × 2 = 0 + 0.000 097 273 629 900 8;
  • 18) 0.000 097 273 629 900 8 × 2 = 0 + 0.000 194 547 259 801 6;
  • 19) 0.000 194 547 259 801 6 × 2 = 0 + 0.000 389 094 519 603 2;
  • 20) 0.000 389 094 519 603 2 × 2 = 0 + 0.000 778 189 039 206 4;
  • 21) 0.000 778 189 039 206 4 × 2 = 0 + 0.001 556 378 078 412 8;
  • 22) 0.001 556 378 078 412 8 × 2 = 0 + 0.003 112 756 156 825 6;
  • 23) 0.003 112 756 156 825 6 × 2 = 0 + 0.006 225 512 313 651 2;
  • 24) 0.006 225 512 313 651 2 × 2 = 0 + 0.012 451 024 627 302 4;
  • 25) 0.012 451 024 627 302 4 × 2 = 0 + 0.024 902 049 254 604 8;
  • 26) 0.024 902 049 254 604 8 × 2 = 0 + 0.049 804 098 509 209 6;
  • 27) 0.049 804 098 509 209 6 × 2 = 0 + 0.099 608 197 018 419 2;
  • 28) 0.099 608 197 018 419 2 × 2 = 0 + 0.199 216 394 036 838 4;
  • 29) 0.199 216 394 036 838 4 × 2 = 0 + 0.398 432 788 073 676 8;
  • 30) 0.398 432 788 073 676 8 × 2 = 0 + 0.796 865 576 147 353 6;
  • 31) 0.796 865 576 147 353 6 × 2 = 1 + 0.593 731 152 294 707 2;
  • 32) 0.593 731 152 294 707 2 × 2 = 1 + 0.187 462 304 589 414 4;
  • 33) 0.187 462 304 589 414 4 × 2 = 0 + 0.374 924 609 178 828 8;
  • 34) 0.374 924 609 178 828 8 × 2 = 0 + 0.749 849 218 357 657 6;
  • 35) 0.749 849 218 357 657 6 × 2 = 1 + 0.499 698 436 715 315 2;
  • 36) 0.499 698 436 715 315 2 × 2 = 0 + 0.999 396 873 430 630 4;
  • 37) 0.999 396 873 430 630 4 × 2 = 1 + 0.998 793 746 861 260 8;
  • 38) 0.998 793 746 861 260 8 × 2 = 1 + 0.997 587 493 722 521 6;
  • 39) 0.997 587 493 722 521 6 × 2 = 1 + 0.995 174 987 445 043 2;
  • 40) 0.995 174 987 445 043 2 × 2 = 1 + 0.990 349 974 890 086 4;
  • 41) 0.990 349 974 890 086 4 × 2 = 1 + 0.980 699 949 780 172 8;
  • 42) 0.980 699 949 780 172 8 × 2 = 1 + 0.961 399 899 560 345 6;
  • 43) 0.961 399 899 560 345 6 × 2 = 1 + 0.922 799 799 120 691 2;
  • 44) 0.922 799 799 120 691 2 × 2 = 1 + 0.845 599 598 241 382 4;
  • 45) 0.845 599 598 241 382 4 × 2 = 1 + 0.691 199 196 482 764 8;
  • 46) 0.691 199 196 482 764 8 × 2 = 1 + 0.382 398 392 965 529 6;
  • 47) 0.382 398 392 965 529 6 × 2 = 0 + 0.764 796 785 931 059 2;
  • 48) 0.764 796 785 931 059 2 × 2 = 1 + 0.529 593 571 862 118 4;
  • 49) 0.529 593 571 862 118 4 × 2 = 1 + 0.059 187 143 724 236 8;
  • 50) 0.059 187 143 724 236 8 × 2 = 0 + 0.118 374 287 448 473 6;
  • 51) 0.118 374 287 448 473 6 × 2 = 0 + 0.236 748 574 896 947 2;
  • 52) 0.236 748 574 896 947 2 × 2 = 0 + 0.473 497 149 793 894 4;
  • 53) 0.473 497 149 793 894 4 × 2 = 0 + 0.946 994 299 587 788 8;
  • 54) 0.946 994 299 587 788 8 × 2 = 1 + 0.893 988 599 175 577 6;

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:


0.000 000 000 742 138 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1000 01(2)

6. Positive number before normalization:

0.000 000 000 742 138 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1000 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:


0.000 000 000 742 138 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 1000 01(2) × 20 =

1.1001 0111 1111 1110 1100 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 1111 1110 1100 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 1111 1111 0110 0001 =

100 1011 1111 1111 0110 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 1111 1111 0110 0001


Decimal number -0.000 000 000 742 138 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 0110 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