-0.000 000 000 742 147 676 616 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 616 7(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
-0.000 000 000 742 147 676 616 7(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:

|-0.000 000 000 742 147 676 616 7| = 0.000 000 000 742 147 676 616 7


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: 0.000 000 000 742 147 676 616 7.

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) 0.000 000 000 742 147 676 616 7 × 2 = 0 + 0.000 000 001 484 295 353 233 4;
  • 2) 0.000 000 001 484 295 353 233 4 × 2 = 0 + 0.000 000 002 968 590 706 466 8;
  • 3) 0.000 000 002 968 590 706 466 8 × 2 = 0 + 0.000 000 005 937 181 412 933 6;
  • 4) 0.000 000 005 937 181 412 933 6 × 2 = 0 + 0.000 000 011 874 362 825 867 2;
  • 5) 0.000 000 011 874 362 825 867 2 × 2 = 0 + 0.000 000 023 748 725 651 734 4;
  • 6) 0.000 000 023 748 725 651 734 4 × 2 = 0 + 0.000 000 047 497 451 303 468 8;
  • 7) 0.000 000 047 497 451 303 468 8 × 2 = 0 + 0.000 000 094 994 902 606 937 6;
  • 8) 0.000 000 094 994 902 606 937 6 × 2 = 0 + 0.000 000 189 989 805 213 875 2;
  • 9) 0.000 000 189 989 805 213 875 2 × 2 = 0 + 0.000 000 379 979 610 427 750 4;
  • 10) 0.000 000 379 979 610 427 750 4 × 2 = 0 + 0.000 000 759 959 220 855 500 8;
  • 11) 0.000 000 759 959 220 855 500 8 × 2 = 0 + 0.000 001 519 918 441 711 001 6;
  • 12) 0.000 001 519 918 441 711 001 6 × 2 = 0 + 0.000 003 039 836 883 422 003 2;
  • 13) 0.000 003 039 836 883 422 003 2 × 2 = 0 + 0.000 006 079 673 766 844 006 4;
  • 14) 0.000 006 079 673 766 844 006 4 × 2 = 0 + 0.000 012 159 347 533 688 012 8;
  • 15) 0.000 012 159 347 533 688 012 8 × 2 = 0 + 0.000 024 318 695 067 376 025 6;
  • 16) 0.000 024 318 695 067 376 025 6 × 2 = 0 + 0.000 048 637 390 134 752 051 2;
  • 17) 0.000 048 637 390 134 752 051 2 × 2 = 0 + 0.000 097 274 780 269 504 102 4;
  • 18) 0.000 097 274 780 269 504 102 4 × 2 = 0 + 0.000 194 549 560 539 008 204 8;
  • 19) 0.000 194 549 560 539 008 204 8 × 2 = 0 + 0.000 389 099 121 078 016 409 6;
  • 20) 0.000 389 099 121 078 016 409 6 × 2 = 0 + 0.000 778 198 242 156 032 819 2;
  • 21) 0.000 778 198 242 156 032 819 2 × 2 = 0 + 0.001 556 396 484 312 065 638 4;
  • 22) 0.001 556 396 484 312 065 638 4 × 2 = 0 + 0.003 112 792 968 624 131 276 8;
  • 23) 0.003 112 792 968 624 131 276 8 × 2 = 0 + 0.006 225 585 937 248 262 553 6;
  • 24) 0.006 225 585 937 248 262 553 6 × 2 = 0 + 0.012 451 171 874 496 525 107 2;
  • 25) 0.012 451 171 874 496 525 107 2 × 2 = 0 + 0.024 902 343 748 993 050 214 4;
  • 26) 0.024 902 343 748 993 050 214 4 × 2 = 0 + 0.049 804 687 497 986 100 428 8;
  • 27) 0.049 804 687 497 986 100 428 8 × 2 = 0 + 0.099 609 374 995 972 200 857 6;
  • 28) 0.099 609 374 995 972 200 857 6 × 2 = 0 + 0.199 218 749 991 944 401 715 2;
  • 29) 0.199 218 749 991 944 401 715 2 × 2 = 0 + 0.398 437 499 983 888 803 430 4;
  • 30) 0.398 437 499 983 888 803 430 4 × 2 = 0 + 0.796 874 999 967 777 606 860 8;
  • 31) 0.796 874 999 967 777 606 860 8 × 2 = 1 + 0.593 749 999 935 555 213 721 6;
  • 32) 0.593 749 999 935 555 213 721 6 × 2 = 1 + 0.187 499 999 871 110 427 443 2;
  • 33) 0.187 499 999 871 110 427 443 2 × 2 = 0 + 0.374 999 999 742 220 854 886 4;
  • 34) 0.374 999 999 742 220 854 886 4 × 2 = 0 + 0.749 999 999 484 441 709 772 8;
  • 35) 0.749 999 999 484 441 709 772 8 × 2 = 1 + 0.499 999 998 968 883 419 545 6;
  • 36) 0.499 999 998 968 883 419 545 6 × 2 = 0 + 0.999 999 997 937 766 839 091 2;
  • 37) 0.999 999 997 937 766 839 091 2 × 2 = 1 + 0.999 999 995 875 533 678 182 4;
  • 38) 0.999 999 995 875 533 678 182 4 × 2 = 1 + 0.999 999 991 751 067 356 364 8;
  • 39) 0.999 999 991 751 067 356 364 8 × 2 = 1 + 0.999 999 983 502 134 712 729 6;
  • 40) 0.999 999 983 502 134 712 729 6 × 2 = 1 + 0.999 999 967 004 269 425 459 2;
  • 41) 0.999 999 967 004 269 425 459 2 × 2 = 1 + 0.999 999 934 008 538 850 918 4;
  • 42) 0.999 999 934 008 538 850 918 4 × 2 = 1 + 0.999 999 868 017 077 701 836 8;
  • 43) 0.999 999 868 017 077 701 836 8 × 2 = 1 + 0.999 999 736 034 155 403 673 6;
  • 44) 0.999 999 736 034 155 403 673 6 × 2 = 1 + 0.999 999 472 068 310 807 347 2;
  • 45) 0.999 999 472 068 310 807 347 2 × 2 = 1 + 0.999 998 944 136 621 614 694 4;
  • 46) 0.999 998 944 136 621 614 694 4 × 2 = 1 + 0.999 997 888 273 243 229 388 8;
  • 47) 0.999 997 888 273 243 229 388 8 × 2 = 1 + 0.999 995 776 546 486 458 777 6;
  • 48) 0.999 995 776 546 486 458 777 6 × 2 = 1 + 0.999 991 553 092 972 917 555 2;
  • 49) 0.999 991 553 092 972 917 555 2 × 2 = 1 + 0.999 983 106 185 945 835 110 4;
  • 50) 0.999 983 106 185 945 835 110 4 × 2 = 1 + 0.999 966 212 371 891 670 220 8;
  • 51) 0.999 966 212 371 891 670 220 8 × 2 = 1 + 0.999 932 424 743 783 340 441 6;
  • 52) 0.999 932 424 743 783 340 441 6 × 2 = 1 + 0.999 864 849 487 566 680 883 2;
  • 53) 0.999 864 849 487 566 680 883 2 × 2 = 1 + 0.999 729 698 975 133 361 766 4;
  • 54) 0.999 729 698 975 133 361 766 4 × 2 = 1 + 0.999 459 397 950 266 723 532 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:


0.000 000 000 742 147 676 616 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 616 7(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:


0.000 000 000 742 147 676 616 7(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 -0.000 000 000 742 147 676 616 7 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