-0.000 000 000 742 147 676 617 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 617 3(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
-0.000 000 000 742 147 676 617 3(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:

|-0.000 000 000 742 147 676 617 3| = 0.000 000 000 742 147 676 617 3


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: 0.000 000 000 742 147 676 617 3.

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) 0.000 000 000 742 147 676 617 3 × 2 = 0 + 0.000 000 001 484 295 353 234 6;
  • 2) 0.000 000 001 484 295 353 234 6 × 2 = 0 + 0.000 000 002 968 590 706 469 2;
  • 3) 0.000 000 002 968 590 706 469 2 × 2 = 0 + 0.000 000 005 937 181 412 938 4;
  • 4) 0.000 000 005 937 181 412 938 4 × 2 = 0 + 0.000 000 011 874 362 825 876 8;
  • 5) 0.000 000 011 874 362 825 876 8 × 2 = 0 + 0.000 000 023 748 725 651 753 6;
  • 6) 0.000 000 023 748 725 651 753 6 × 2 = 0 + 0.000 000 047 497 451 303 507 2;
  • 7) 0.000 000 047 497 451 303 507 2 × 2 = 0 + 0.000 000 094 994 902 607 014 4;
  • 8) 0.000 000 094 994 902 607 014 4 × 2 = 0 + 0.000 000 189 989 805 214 028 8;
  • 9) 0.000 000 189 989 805 214 028 8 × 2 = 0 + 0.000 000 379 979 610 428 057 6;
  • 10) 0.000 000 379 979 610 428 057 6 × 2 = 0 + 0.000 000 759 959 220 856 115 2;
  • 11) 0.000 000 759 959 220 856 115 2 × 2 = 0 + 0.000 001 519 918 441 712 230 4;
  • 12) 0.000 001 519 918 441 712 230 4 × 2 = 0 + 0.000 003 039 836 883 424 460 8;
  • 13) 0.000 003 039 836 883 424 460 8 × 2 = 0 + 0.000 006 079 673 766 848 921 6;
  • 14) 0.000 006 079 673 766 848 921 6 × 2 = 0 + 0.000 012 159 347 533 697 843 2;
  • 15) 0.000 012 159 347 533 697 843 2 × 2 = 0 + 0.000 024 318 695 067 395 686 4;
  • 16) 0.000 024 318 695 067 395 686 4 × 2 = 0 + 0.000 048 637 390 134 791 372 8;
  • 17) 0.000 048 637 390 134 791 372 8 × 2 = 0 + 0.000 097 274 780 269 582 745 6;
  • 18) 0.000 097 274 780 269 582 745 6 × 2 = 0 + 0.000 194 549 560 539 165 491 2;
  • 19) 0.000 194 549 560 539 165 491 2 × 2 = 0 + 0.000 389 099 121 078 330 982 4;
  • 20) 0.000 389 099 121 078 330 982 4 × 2 = 0 + 0.000 778 198 242 156 661 964 8;
  • 21) 0.000 778 198 242 156 661 964 8 × 2 = 0 + 0.001 556 396 484 313 323 929 6;
  • 22) 0.001 556 396 484 313 323 929 6 × 2 = 0 + 0.003 112 792 968 626 647 859 2;
  • 23) 0.003 112 792 968 626 647 859 2 × 2 = 0 + 0.006 225 585 937 253 295 718 4;
  • 24) 0.006 225 585 937 253 295 718 4 × 2 = 0 + 0.012 451 171 874 506 591 436 8;
  • 25) 0.012 451 171 874 506 591 436 8 × 2 = 0 + 0.024 902 343 749 013 182 873 6;
  • 26) 0.024 902 343 749 013 182 873 6 × 2 = 0 + 0.049 804 687 498 026 365 747 2;
  • 27) 0.049 804 687 498 026 365 747 2 × 2 = 0 + 0.099 609 374 996 052 731 494 4;
  • 28) 0.099 609 374 996 052 731 494 4 × 2 = 0 + 0.199 218 749 992 105 462 988 8;
  • 29) 0.199 218 749 992 105 462 988 8 × 2 = 0 + 0.398 437 499 984 210 925 977 6;
  • 30) 0.398 437 499 984 210 925 977 6 × 2 = 0 + 0.796 874 999 968 421 851 955 2;
  • 31) 0.796 874 999 968 421 851 955 2 × 2 = 1 + 0.593 749 999 936 843 703 910 4;
  • 32) 0.593 749 999 936 843 703 910 4 × 2 = 1 + 0.187 499 999 873 687 407 820 8;
  • 33) 0.187 499 999 873 687 407 820 8 × 2 = 0 + 0.374 999 999 747 374 815 641 6;
  • 34) 0.374 999 999 747 374 815 641 6 × 2 = 0 + 0.749 999 999 494 749 631 283 2;
  • 35) 0.749 999 999 494 749 631 283 2 × 2 = 1 + 0.499 999 998 989 499 262 566 4;
  • 36) 0.499 999 998 989 499 262 566 4 × 2 = 0 + 0.999 999 997 978 998 525 132 8;
  • 37) 0.999 999 997 978 998 525 132 8 × 2 = 1 + 0.999 999 995 957 997 050 265 6;
  • 38) 0.999 999 995 957 997 050 265 6 × 2 = 1 + 0.999 999 991 915 994 100 531 2;
  • 39) 0.999 999 991 915 994 100 531 2 × 2 = 1 + 0.999 999 983 831 988 201 062 4;
  • 40) 0.999 999 983 831 988 201 062 4 × 2 = 1 + 0.999 999 967 663 976 402 124 8;
  • 41) 0.999 999 967 663 976 402 124 8 × 2 = 1 + 0.999 999 935 327 952 804 249 6;
  • 42) 0.999 999 935 327 952 804 249 6 × 2 = 1 + 0.999 999 870 655 905 608 499 2;
  • 43) 0.999 999 870 655 905 608 499 2 × 2 = 1 + 0.999 999 741 311 811 216 998 4;
  • 44) 0.999 999 741 311 811 216 998 4 × 2 = 1 + 0.999 999 482 623 622 433 996 8;
  • 45) 0.999 999 482 623 622 433 996 8 × 2 = 1 + 0.999 998 965 247 244 867 993 6;
  • 46) 0.999 998 965 247 244 867 993 6 × 2 = 1 + 0.999 997 930 494 489 735 987 2;
  • 47) 0.999 997 930 494 489 735 987 2 × 2 = 1 + 0.999 995 860 988 979 471 974 4;
  • 48) 0.999 995 860 988 979 471 974 4 × 2 = 1 + 0.999 991 721 977 958 943 948 8;
  • 49) 0.999 991 721 977 958 943 948 8 × 2 = 1 + 0.999 983 443 955 917 887 897 6;
  • 50) 0.999 983 443 955 917 887 897 6 × 2 = 1 + 0.999 966 887 911 835 775 795 2;
  • 51) 0.999 966 887 911 835 775 795 2 × 2 = 1 + 0.999 933 775 823 671 551 590 4;
  • 52) 0.999 933 775 823 671 551 590 4 × 2 = 1 + 0.999 867 551 647 343 103 180 8;
  • 53) 0.999 867 551 647 343 103 180 8 × 2 = 1 + 0.999 735 103 294 686 206 361 6;
  • 54) 0.999 735 103 294 686 206 361 6 × 2 = 1 + 0.999 470 206 589 372 412 723 2;

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:


0.000 000 000 742 147 676 617 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 617 3(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:


0.000 000 000 742 147 676 617 3(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 -0.000 000 000 742 147 676 617 3 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