-0.000 000 000 742 147 676 646 709 41 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 709 41(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 41(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 41| = 0.000 000 000 742 147 676 646 709 41


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 41.

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 41 × 2 = 0 + 0.000 000 001 484 295 353 293 418 82;
  • 2) 0.000 000 001 484 295 353 293 418 82 × 2 = 0 + 0.000 000 002 968 590 706 586 837 64;
  • 3) 0.000 000 002 968 590 706 586 837 64 × 2 = 0 + 0.000 000 005 937 181 413 173 675 28;
  • 4) 0.000 000 005 937 181 413 173 675 28 × 2 = 0 + 0.000 000 011 874 362 826 347 350 56;
  • 5) 0.000 000 011 874 362 826 347 350 56 × 2 = 0 + 0.000 000 023 748 725 652 694 701 12;
  • 6) 0.000 000 023 748 725 652 694 701 12 × 2 = 0 + 0.000 000 047 497 451 305 389 402 24;
  • 7) 0.000 000 047 497 451 305 389 402 24 × 2 = 0 + 0.000 000 094 994 902 610 778 804 48;
  • 8) 0.000 000 094 994 902 610 778 804 48 × 2 = 0 + 0.000 000 189 989 805 221 557 608 96;
  • 9) 0.000 000 189 989 805 221 557 608 96 × 2 = 0 + 0.000 000 379 979 610 443 115 217 92;
  • 10) 0.000 000 379 979 610 443 115 217 92 × 2 = 0 + 0.000 000 759 959 220 886 230 435 84;
  • 11) 0.000 000 759 959 220 886 230 435 84 × 2 = 0 + 0.000 001 519 918 441 772 460 871 68;
  • 12) 0.000 001 519 918 441 772 460 871 68 × 2 = 0 + 0.000 003 039 836 883 544 921 743 36;
  • 13) 0.000 003 039 836 883 544 921 743 36 × 2 = 0 + 0.000 006 079 673 767 089 843 486 72;
  • 14) 0.000 006 079 673 767 089 843 486 72 × 2 = 0 + 0.000 012 159 347 534 179 686 973 44;
  • 15) 0.000 012 159 347 534 179 686 973 44 × 2 = 0 + 0.000 024 318 695 068 359 373 946 88;
  • 16) 0.000 024 318 695 068 359 373 946 88 × 2 = 0 + 0.000 048 637 390 136 718 747 893 76;
  • 17) 0.000 048 637 390 136 718 747 893 76 × 2 = 0 + 0.000 097 274 780 273 437 495 787 52;
  • 18) 0.000 097 274 780 273 437 495 787 52 × 2 = 0 + 0.000 194 549 560 546 874 991 575 04;
  • 19) 0.000 194 549 560 546 874 991 575 04 × 2 = 0 + 0.000 389 099 121 093 749 983 150 08;
  • 20) 0.000 389 099 121 093 749 983 150 08 × 2 = 0 + 0.000 778 198 242 187 499 966 300 16;
  • 21) 0.000 778 198 242 187 499 966 300 16 × 2 = 0 + 0.001 556 396 484 374 999 932 600 32;
  • 22) 0.001 556 396 484 374 999 932 600 32 × 2 = 0 + 0.003 112 792 968 749 999 865 200 64;
  • 23) 0.003 112 792 968 749 999 865 200 64 × 2 = 0 + 0.006 225 585 937 499 999 730 401 28;
  • 24) 0.006 225 585 937 499 999 730 401 28 × 2 = 0 + 0.012 451 171 874 999 999 460 802 56;
  • 25) 0.012 451 171 874 999 999 460 802 56 × 2 = 0 + 0.024 902 343 749 999 998 921 605 12;
  • 26) 0.024 902 343 749 999 998 921 605 12 × 2 = 0 + 0.049 804 687 499 999 997 843 210 24;
  • 27) 0.049 804 687 499 999 997 843 210 24 × 2 = 0 + 0.099 609 374 999 999 995 686 420 48;
  • 28) 0.099 609 374 999 999 995 686 420 48 × 2 = 0 + 0.199 218 749 999 999 991 372 840 96;
  • 29) 0.199 218 749 999 999 991 372 840 96 × 2 = 0 + 0.398 437 499 999 999 982 745 681 92;
  • 30) 0.398 437 499 999 999 982 745 681 92 × 2 = 0 + 0.796 874 999 999 999 965 491 363 84;
  • 31) 0.796 874 999 999 999 965 491 363 84 × 2 = 1 + 0.593 749 999 999 999 930 982 727 68;
  • 32) 0.593 749 999 999 999 930 982 727 68 × 2 = 1 + 0.187 499 999 999 999 861 965 455 36;
  • 33) 0.187 499 999 999 999 861 965 455 36 × 2 = 0 + 0.374 999 999 999 999 723 930 910 72;
  • 34) 0.374 999 999 999 999 723 930 910 72 × 2 = 0 + 0.749 999 999 999 999 447 861 821 44;
  • 35) 0.749 999 999 999 999 447 861 821 44 × 2 = 1 + 0.499 999 999 999 998 895 723 642 88;
  • 36) 0.499 999 999 999 998 895 723 642 88 × 2 = 0 + 0.999 999 999 999 997 791 447 285 76;
  • 37) 0.999 999 999 999 997 791 447 285 76 × 2 = 1 + 0.999 999 999 999 995 582 894 571 52;
  • 38) 0.999 999 999 999 995 582 894 571 52 × 2 = 1 + 0.999 999 999 999 991 165 789 143 04;
  • 39) 0.999 999 999 999 991 165 789 143 04 × 2 = 1 + 0.999 999 999 999 982 331 578 286 08;
  • 40) 0.999 999 999 999 982 331 578 286 08 × 2 = 1 + 0.999 999 999 999 964 663 156 572 16;
  • 41) 0.999 999 999 999 964 663 156 572 16 × 2 = 1 + 0.999 999 999 999 929 326 313 144 32;
  • 42) 0.999 999 999 999 929 326 313 144 32 × 2 = 1 + 0.999 999 999 999 858 652 626 288 64;
  • 43) 0.999 999 999 999 858 652 626 288 64 × 2 = 1 + 0.999 999 999 999 717 305 252 577 28;
  • 44) 0.999 999 999 999 717 305 252 577 28 × 2 = 1 + 0.999 999 999 999 434 610 505 154 56;
  • 45) 0.999 999 999 999 434 610 505 154 56 × 2 = 1 + 0.999 999 999 998 869 221 010 309 12;
  • 46) 0.999 999 999 998 869 221 010 309 12 × 2 = 1 + 0.999 999 999 997 738 442 020 618 24;
  • 47) 0.999 999 999 997 738 442 020 618 24 × 2 = 1 + 0.999 999 999 995 476 884 041 236 48;
  • 48) 0.999 999 999 995 476 884 041 236 48 × 2 = 1 + 0.999 999 999 990 953 768 082 472 96;
  • 49) 0.999 999 999 990 953 768 082 472 96 × 2 = 1 + 0.999 999 999 981 907 536 164 945 92;
  • 50) 0.999 999 999 981 907 536 164 945 92 × 2 = 1 + 0.999 999 999 963 815 072 329 891 84;
  • 51) 0.999 999 999 963 815 072 329 891 84 × 2 = 1 + 0.999 999 999 927 630 144 659 783 68;
  • 52) 0.999 999 999 927 630 144 659 783 68 × 2 = 1 + 0.999 999 999 855 260 289 319 567 36;
  • 53) 0.999 999 999 855 260 289 319 567 36 × 2 = 1 + 0.999 999 999 710 520 578 639 134 72;
  • 54) 0.999 999 999 710 520 578 639 134 72 × 2 = 1 + 0.999 999 999 421 041 157 278 269 44;

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 41(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 41(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 41(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 41 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