-0.000 000 000 742 147 676 630 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 630 6(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
-0.000 000 000 742 147 676 630 6(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:

|-0.000 000 000 742 147 676 630 6| = 0.000 000 000 742 147 676 630 6


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: 0.000 000 000 742 147 676 630 6.

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) 0.000 000 000 742 147 676 630 6 × 2 = 0 + 0.000 000 001 484 295 353 261 2;
  • 2) 0.000 000 001 484 295 353 261 2 × 2 = 0 + 0.000 000 002 968 590 706 522 4;
  • 3) 0.000 000 002 968 590 706 522 4 × 2 = 0 + 0.000 000 005 937 181 413 044 8;
  • 4) 0.000 000 005 937 181 413 044 8 × 2 = 0 + 0.000 000 011 874 362 826 089 6;
  • 5) 0.000 000 011 874 362 826 089 6 × 2 = 0 + 0.000 000 023 748 725 652 179 2;
  • 6) 0.000 000 023 748 725 652 179 2 × 2 = 0 + 0.000 000 047 497 451 304 358 4;
  • 7) 0.000 000 047 497 451 304 358 4 × 2 = 0 + 0.000 000 094 994 902 608 716 8;
  • 8) 0.000 000 094 994 902 608 716 8 × 2 = 0 + 0.000 000 189 989 805 217 433 6;
  • 9) 0.000 000 189 989 805 217 433 6 × 2 = 0 + 0.000 000 379 979 610 434 867 2;
  • 10) 0.000 000 379 979 610 434 867 2 × 2 = 0 + 0.000 000 759 959 220 869 734 4;
  • 11) 0.000 000 759 959 220 869 734 4 × 2 = 0 + 0.000 001 519 918 441 739 468 8;
  • 12) 0.000 001 519 918 441 739 468 8 × 2 = 0 + 0.000 003 039 836 883 478 937 6;
  • 13) 0.000 003 039 836 883 478 937 6 × 2 = 0 + 0.000 006 079 673 766 957 875 2;
  • 14) 0.000 006 079 673 766 957 875 2 × 2 = 0 + 0.000 012 159 347 533 915 750 4;
  • 15) 0.000 012 159 347 533 915 750 4 × 2 = 0 + 0.000 024 318 695 067 831 500 8;
  • 16) 0.000 024 318 695 067 831 500 8 × 2 = 0 + 0.000 048 637 390 135 663 001 6;
  • 17) 0.000 048 637 390 135 663 001 6 × 2 = 0 + 0.000 097 274 780 271 326 003 2;
  • 18) 0.000 097 274 780 271 326 003 2 × 2 = 0 + 0.000 194 549 560 542 652 006 4;
  • 19) 0.000 194 549 560 542 652 006 4 × 2 = 0 + 0.000 389 099 121 085 304 012 8;
  • 20) 0.000 389 099 121 085 304 012 8 × 2 = 0 + 0.000 778 198 242 170 608 025 6;
  • 21) 0.000 778 198 242 170 608 025 6 × 2 = 0 + 0.001 556 396 484 341 216 051 2;
  • 22) 0.001 556 396 484 341 216 051 2 × 2 = 0 + 0.003 112 792 968 682 432 102 4;
  • 23) 0.003 112 792 968 682 432 102 4 × 2 = 0 + 0.006 225 585 937 364 864 204 8;
  • 24) 0.006 225 585 937 364 864 204 8 × 2 = 0 + 0.012 451 171 874 729 728 409 6;
  • 25) 0.012 451 171 874 729 728 409 6 × 2 = 0 + 0.024 902 343 749 459 456 819 2;
  • 26) 0.024 902 343 749 459 456 819 2 × 2 = 0 + 0.049 804 687 498 918 913 638 4;
  • 27) 0.049 804 687 498 918 913 638 4 × 2 = 0 + 0.099 609 374 997 837 827 276 8;
  • 28) 0.099 609 374 997 837 827 276 8 × 2 = 0 + 0.199 218 749 995 675 654 553 6;
  • 29) 0.199 218 749 995 675 654 553 6 × 2 = 0 + 0.398 437 499 991 351 309 107 2;
  • 30) 0.398 437 499 991 351 309 107 2 × 2 = 0 + 0.796 874 999 982 702 618 214 4;
  • 31) 0.796 874 999 982 702 618 214 4 × 2 = 1 + 0.593 749 999 965 405 236 428 8;
  • 32) 0.593 749 999 965 405 236 428 8 × 2 = 1 + 0.187 499 999 930 810 472 857 6;
  • 33) 0.187 499 999 930 810 472 857 6 × 2 = 0 + 0.374 999 999 861 620 945 715 2;
  • 34) 0.374 999 999 861 620 945 715 2 × 2 = 0 + 0.749 999 999 723 241 891 430 4;
  • 35) 0.749 999 999 723 241 891 430 4 × 2 = 1 + 0.499 999 999 446 483 782 860 8;
  • 36) 0.499 999 999 446 483 782 860 8 × 2 = 0 + 0.999 999 998 892 967 565 721 6;
  • 37) 0.999 999 998 892 967 565 721 6 × 2 = 1 + 0.999 999 997 785 935 131 443 2;
  • 38) 0.999 999 997 785 935 131 443 2 × 2 = 1 + 0.999 999 995 571 870 262 886 4;
  • 39) 0.999 999 995 571 870 262 886 4 × 2 = 1 + 0.999 999 991 143 740 525 772 8;
  • 40) 0.999 999 991 143 740 525 772 8 × 2 = 1 + 0.999 999 982 287 481 051 545 6;
  • 41) 0.999 999 982 287 481 051 545 6 × 2 = 1 + 0.999 999 964 574 962 103 091 2;
  • 42) 0.999 999 964 574 962 103 091 2 × 2 = 1 + 0.999 999 929 149 924 206 182 4;
  • 43) 0.999 999 929 149 924 206 182 4 × 2 = 1 + 0.999 999 858 299 848 412 364 8;
  • 44) 0.999 999 858 299 848 412 364 8 × 2 = 1 + 0.999 999 716 599 696 824 729 6;
  • 45) 0.999 999 716 599 696 824 729 6 × 2 = 1 + 0.999 999 433 199 393 649 459 2;
  • 46) 0.999 999 433 199 393 649 459 2 × 2 = 1 + 0.999 998 866 398 787 298 918 4;
  • 47) 0.999 998 866 398 787 298 918 4 × 2 = 1 + 0.999 997 732 797 574 597 836 8;
  • 48) 0.999 997 732 797 574 597 836 8 × 2 = 1 + 0.999 995 465 595 149 195 673 6;
  • 49) 0.999 995 465 595 149 195 673 6 × 2 = 1 + 0.999 990 931 190 298 391 347 2;
  • 50) 0.999 990 931 190 298 391 347 2 × 2 = 1 + 0.999 981 862 380 596 782 694 4;
  • 51) 0.999 981 862 380 596 782 694 4 × 2 = 1 + 0.999 963 724 761 193 565 388 8;
  • 52) 0.999 963 724 761 193 565 388 8 × 2 = 1 + 0.999 927 449 522 387 130 777 6;
  • 53) 0.999 927 449 522 387 130 777 6 × 2 = 1 + 0.999 854 899 044 774 261 555 2;
  • 54) 0.999 854 899 044 774 261 555 2 × 2 = 1 + 0.999 709 798 089 548 523 110 4;

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:


0.000 000 000 742 147 676 630 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 630 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 630 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 111


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 1111 =

100 1011 1111 1111 1111 1111


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 1111


Decimal number -0.000 000 000 742 147 676 630 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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