-0.000 000 000 742 147 676 646 709 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 709 2(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
-0.000 000 000 742 147 676 646 709 2(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:

|-0.000 000 000 742 147 676 646 709 2| = 0.000 000 000 742 147 676 646 709 2


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: 0.000 000 000 742 147 676 646 709 2.

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) 0.000 000 000 742 147 676 646 709 2 × 2 = 0 + 0.000 000 001 484 295 353 293 418 4;
  • 2) 0.000 000 001 484 295 353 293 418 4 × 2 = 0 + 0.000 000 002 968 590 706 586 836 8;
  • 3) 0.000 000 002 968 590 706 586 836 8 × 2 = 0 + 0.000 000 005 937 181 413 173 673 6;
  • 4) 0.000 000 005 937 181 413 173 673 6 × 2 = 0 + 0.000 000 011 874 362 826 347 347 2;
  • 5) 0.000 000 011 874 362 826 347 347 2 × 2 = 0 + 0.000 000 023 748 725 652 694 694 4;
  • 6) 0.000 000 023 748 725 652 694 694 4 × 2 = 0 + 0.000 000 047 497 451 305 389 388 8;
  • 7) 0.000 000 047 497 451 305 389 388 8 × 2 = 0 + 0.000 000 094 994 902 610 778 777 6;
  • 8) 0.000 000 094 994 902 610 778 777 6 × 2 = 0 + 0.000 000 189 989 805 221 557 555 2;
  • 9) 0.000 000 189 989 805 221 557 555 2 × 2 = 0 + 0.000 000 379 979 610 443 115 110 4;
  • 10) 0.000 000 379 979 610 443 115 110 4 × 2 = 0 + 0.000 000 759 959 220 886 230 220 8;
  • 11) 0.000 000 759 959 220 886 230 220 8 × 2 = 0 + 0.000 001 519 918 441 772 460 441 6;
  • 12) 0.000 001 519 918 441 772 460 441 6 × 2 = 0 + 0.000 003 039 836 883 544 920 883 2;
  • 13) 0.000 003 039 836 883 544 920 883 2 × 2 = 0 + 0.000 006 079 673 767 089 841 766 4;
  • 14) 0.000 006 079 673 767 089 841 766 4 × 2 = 0 + 0.000 012 159 347 534 179 683 532 8;
  • 15) 0.000 012 159 347 534 179 683 532 8 × 2 = 0 + 0.000 024 318 695 068 359 367 065 6;
  • 16) 0.000 024 318 695 068 359 367 065 6 × 2 = 0 + 0.000 048 637 390 136 718 734 131 2;
  • 17) 0.000 048 637 390 136 718 734 131 2 × 2 = 0 + 0.000 097 274 780 273 437 468 262 4;
  • 18) 0.000 097 274 780 273 437 468 262 4 × 2 = 0 + 0.000 194 549 560 546 874 936 524 8;
  • 19) 0.000 194 549 560 546 874 936 524 8 × 2 = 0 + 0.000 389 099 121 093 749 873 049 6;
  • 20) 0.000 389 099 121 093 749 873 049 6 × 2 = 0 + 0.000 778 198 242 187 499 746 099 2;
  • 21) 0.000 778 198 242 187 499 746 099 2 × 2 = 0 + 0.001 556 396 484 374 999 492 198 4;
  • 22) 0.001 556 396 484 374 999 492 198 4 × 2 = 0 + 0.003 112 792 968 749 998 984 396 8;
  • 23) 0.003 112 792 968 749 998 984 396 8 × 2 = 0 + 0.006 225 585 937 499 997 968 793 6;
  • 24) 0.006 225 585 937 499 997 968 793 6 × 2 = 0 + 0.012 451 171 874 999 995 937 587 2;
  • 25) 0.012 451 171 874 999 995 937 587 2 × 2 = 0 + 0.024 902 343 749 999 991 875 174 4;
  • 26) 0.024 902 343 749 999 991 875 174 4 × 2 = 0 + 0.049 804 687 499 999 983 750 348 8;
  • 27) 0.049 804 687 499 999 983 750 348 8 × 2 = 0 + 0.099 609 374 999 999 967 500 697 6;
  • 28) 0.099 609 374 999 999 967 500 697 6 × 2 = 0 + 0.199 218 749 999 999 935 001 395 2;
  • 29) 0.199 218 749 999 999 935 001 395 2 × 2 = 0 + 0.398 437 499 999 999 870 002 790 4;
  • 30) 0.398 437 499 999 999 870 002 790 4 × 2 = 0 + 0.796 874 999 999 999 740 005 580 8;
  • 31) 0.796 874 999 999 999 740 005 580 8 × 2 = 1 + 0.593 749 999 999 999 480 011 161 6;
  • 32) 0.593 749 999 999 999 480 011 161 6 × 2 = 1 + 0.187 499 999 999 998 960 022 323 2;
  • 33) 0.187 499 999 999 998 960 022 323 2 × 2 = 0 + 0.374 999 999 999 997 920 044 646 4;
  • 34) 0.374 999 999 999 997 920 044 646 4 × 2 = 0 + 0.749 999 999 999 995 840 089 292 8;
  • 35) 0.749 999 999 999 995 840 089 292 8 × 2 = 1 + 0.499 999 999 999 991 680 178 585 6;
  • 36) 0.499 999 999 999 991 680 178 585 6 × 2 = 0 + 0.999 999 999 999 983 360 357 171 2;
  • 37) 0.999 999 999 999 983 360 357 171 2 × 2 = 1 + 0.999 999 999 999 966 720 714 342 4;
  • 38) 0.999 999 999 999 966 720 714 342 4 × 2 = 1 + 0.999 999 999 999 933 441 428 684 8;
  • 39) 0.999 999 999 999 933 441 428 684 8 × 2 = 1 + 0.999 999 999 999 866 882 857 369 6;
  • 40) 0.999 999 999 999 866 882 857 369 6 × 2 = 1 + 0.999 999 999 999 733 765 714 739 2;
  • 41) 0.999 999 999 999 733 765 714 739 2 × 2 = 1 + 0.999 999 999 999 467 531 429 478 4;
  • 42) 0.999 999 999 999 467 531 429 478 4 × 2 = 1 + 0.999 999 999 998 935 062 858 956 8;
  • 43) 0.999 999 999 998 935 062 858 956 8 × 2 = 1 + 0.999 999 999 997 870 125 717 913 6;
  • 44) 0.999 999 999 997 870 125 717 913 6 × 2 = 1 + 0.999 999 999 995 740 251 435 827 2;
  • 45) 0.999 999 999 995 740 251 435 827 2 × 2 = 1 + 0.999 999 999 991 480 502 871 654 4;
  • 46) 0.999 999 999 991 480 502 871 654 4 × 2 = 1 + 0.999 999 999 982 961 005 743 308 8;
  • 47) 0.999 999 999 982 961 005 743 308 8 × 2 = 1 + 0.999 999 999 965 922 011 486 617 6;
  • 48) 0.999 999 999 965 922 011 486 617 6 × 2 = 1 + 0.999 999 999 931 844 022 973 235 2;
  • 49) 0.999 999 999 931 844 022 973 235 2 × 2 = 1 + 0.999 999 999 863 688 045 946 470 4;
  • 50) 0.999 999 999 863 688 045 946 470 4 × 2 = 1 + 0.999 999 999 727 376 091 892 940 8;
  • 51) 0.999 999 999 727 376 091 892 940 8 × 2 = 1 + 0.999 999 999 454 752 183 785 881 6;
  • 52) 0.999 999 999 454 752 183 785 881 6 × 2 = 1 + 0.999 999 998 909 504 367 571 763 2;
  • 53) 0.999 999 998 909 504 367 571 763 2 × 2 = 1 + 0.999 999 997 819 008 735 143 526 4;
  • 54) 0.999 999 997 819 008 735 143 526 4 × 2 = 1 + 0.999 999 995 638 017 470 287 052 8;

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:


0.000 000 000 742 147 676 646 709 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 709 2(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:


0.000 000 000 742 147 676 646 709 2(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 -0.000 000 000 742 147 676 646 709 2 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