-0.000 000 000 742 147 676 639 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 639 1(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
-0.000 000 000 742 147 676 639 1(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:

|-0.000 000 000 742 147 676 639 1| = 0.000 000 000 742 147 676 639 1


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: 0.000 000 000 742 147 676 639 1.

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) 0.000 000 000 742 147 676 639 1 × 2 = 0 + 0.000 000 001 484 295 353 278 2;
  • 2) 0.000 000 001 484 295 353 278 2 × 2 = 0 + 0.000 000 002 968 590 706 556 4;
  • 3) 0.000 000 002 968 590 706 556 4 × 2 = 0 + 0.000 000 005 937 181 413 112 8;
  • 4) 0.000 000 005 937 181 413 112 8 × 2 = 0 + 0.000 000 011 874 362 826 225 6;
  • 5) 0.000 000 011 874 362 826 225 6 × 2 = 0 + 0.000 000 023 748 725 652 451 2;
  • 6) 0.000 000 023 748 725 652 451 2 × 2 = 0 + 0.000 000 047 497 451 304 902 4;
  • 7) 0.000 000 047 497 451 304 902 4 × 2 = 0 + 0.000 000 094 994 902 609 804 8;
  • 8) 0.000 000 094 994 902 609 804 8 × 2 = 0 + 0.000 000 189 989 805 219 609 6;
  • 9) 0.000 000 189 989 805 219 609 6 × 2 = 0 + 0.000 000 379 979 610 439 219 2;
  • 10) 0.000 000 379 979 610 439 219 2 × 2 = 0 + 0.000 000 759 959 220 878 438 4;
  • 11) 0.000 000 759 959 220 878 438 4 × 2 = 0 + 0.000 001 519 918 441 756 876 8;
  • 12) 0.000 001 519 918 441 756 876 8 × 2 = 0 + 0.000 003 039 836 883 513 753 6;
  • 13) 0.000 003 039 836 883 513 753 6 × 2 = 0 + 0.000 006 079 673 767 027 507 2;
  • 14) 0.000 006 079 673 767 027 507 2 × 2 = 0 + 0.000 012 159 347 534 055 014 4;
  • 15) 0.000 012 159 347 534 055 014 4 × 2 = 0 + 0.000 024 318 695 068 110 028 8;
  • 16) 0.000 024 318 695 068 110 028 8 × 2 = 0 + 0.000 048 637 390 136 220 057 6;
  • 17) 0.000 048 637 390 136 220 057 6 × 2 = 0 + 0.000 097 274 780 272 440 115 2;
  • 18) 0.000 097 274 780 272 440 115 2 × 2 = 0 + 0.000 194 549 560 544 880 230 4;
  • 19) 0.000 194 549 560 544 880 230 4 × 2 = 0 + 0.000 389 099 121 089 760 460 8;
  • 20) 0.000 389 099 121 089 760 460 8 × 2 = 0 + 0.000 778 198 242 179 520 921 6;
  • 21) 0.000 778 198 242 179 520 921 6 × 2 = 0 + 0.001 556 396 484 359 041 843 2;
  • 22) 0.001 556 396 484 359 041 843 2 × 2 = 0 + 0.003 112 792 968 718 083 686 4;
  • 23) 0.003 112 792 968 718 083 686 4 × 2 = 0 + 0.006 225 585 937 436 167 372 8;
  • 24) 0.006 225 585 937 436 167 372 8 × 2 = 0 + 0.012 451 171 874 872 334 745 6;
  • 25) 0.012 451 171 874 872 334 745 6 × 2 = 0 + 0.024 902 343 749 744 669 491 2;
  • 26) 0.024 902 343 749 744 669 491 2 × 2 = 0 + 0.049 804 687 499 489 338 982 4;
  • 27) 0.049 804 687 499 489 338 982 4 × 2 = 0 + 0.099 609 374 998 978 677 964 8;
  • 28) 0.099 609 374 998 978 677 964 8 × 2 = 0 + 0.199 218 749 997 957 355 929 6;
  • 29) 0.199 218 749 997 957 355 929 6 × 2 = 0 + 0.398 437 499 995 914 711 859 2;
  • 30) 0.398 437 499 995 914 711 859 2 × 2 = 0 + 0.796 874 999 991 829 423 718 4;
  • 31) 0.796 874 999 991 829 423 718 4 × 2 = 1 + 0.593 749 999 983 658 847 436 8;
  • 32) 0.593 749 999 983 658 847 436 8 × 2 = 1 + 0.187 499 999 967 317 694 873 6;
  • 33) 0.187 499 999 967 317 694 873 6 × 2 = 0 + 0.374 999 999 934 635 389 747 2;
  • 34) 0.374 999 999 934 635 389 747 2 × 2 = 0 + 0.749 999 999 869 270 779 494 4;
  • 35) 0.749 999 999 869 270 779 494 4 × 2 = 1 + 0.499 999 999 738 541 558 988 8;
  • 36) 0.499 999 999 738 541 558 988 8 × 2 = 0 + 0.999 999 999 477 083 117 977 6;
  • 37) 0.999 999 999 477 083 117 977 6 × 2 = 1 + 0.999 999 998 954 166 235 955 2;
  • 38) 0.999 999 998 954 166 235 955 2 × 2 = 1 + 0.999 999 997 908 332 471 910 4;
  • 39) 0.999 999 997 908 332 471 910 4 × 2 = 1 + 0.999 999 995 816 664 943 820 8;
  • 40) 0.999 999 995 816 664 943 820 8 × 2 = 1 + 0.999 999 991 633 329 887 641 6;
  • 41) 0.999 999 991 633 329 887 641 6 × 2 = 1 + 0.999 999 983 266 659 775 283 2;
  • 42) 0.999 999 983 266 659 775 283 2 × 2 = 1 + 0.999 999 966 533 319 550 566 4;
  • 43) 0.999 999 966 533 319 550 566 4 × 2 = 1 + 0.999 999 933 066 639 101 132 8;
  • 44) 0.999 999 933 066 639 101 132 8 × 2 = 1 + 0.999 999 866 133 278 202 265 6;
  • 45) 0.999 999 866 133 278 202 265 6 × 2 = 1 + 0.999 999 732 266 556 404 531 2;
  • 46) 0.999 999 732 266 556 404 531 2 × 2 = 1 + 0.999 999 464 533 112 809 062 4;
  • 47) 0.999 999 464 533 112 809 062 4 × 2 = 1 + 0.999 998 929 066 225 618 124 8;
  • 48) 0.999 998 929 066 225 618 124 8 × 2 = 1 + 0.999 997 858 132 451 236 249 6;
  • 49) 0.999 997 858 132 451 236 249 6 × 2 = 1 + 0.999 995 716 264 902 472 499 2;
  • 50) 0.999 995 716 264 902 472 499 2 × 2 = 1 + 0.999 991 432 529 804 944 998 4;
  • 51) 0.999 991 432 529 804 944 998 4 × 2 = 1 + 0.999 982 865 059 609 889 996 8;
  • 52) 0.999 982 865 059 609 889 996 8 × 2 = 1 + 0.999 965 730 119 219 779 993 6;
  • 53) 0.999 965 730 119 219 779 993 6 × 2 = 1 + 0.999 931 460 238 439 559 987 2;
  • 54) 0.999 931 460 238 439 559 987 2 × 2 = 1 + 0.999 862 920 476 879 119 974 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:


0.000 000 000 742 147 676 639 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 639 1(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:


0.000 000 000 742 147 676 639 1(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 -0.000 000 000 742 147 676 639 1 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