-0.000 000 000 742 147 676 649 15 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 649 15(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
-0.000 000 000 742 147 676 649 15(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:

|-0.000 000 000 742 147 676 649 15| = 0.000 000 000 742 147 676 649 15


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: 0.000 000 000 742 147 676 649 15.

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) 0.000 000 000 742 147 676 649 15 × 2 = 0 + 0.000 000 001 484 295 353 298 3;
  • 2) 0.000 000 001 484 295 353 298 3 × 2 = 0 + 0.000 000 002 968 590 706 596 6;
  • 3) 0.000 000 002 968 590 706 596 6 × 2 = 0 + 0.000 000 005 937 181 413 193 2;
  • 4) 0.000 000 005 937 181 413 193 2 × 2 = 0 + 0.000 000 011 874 362 826 386 4;
  • 5) 0.000 000 011 874 362 826 386 4 × 2 = 0 + 0.000 000 023 748 725 652 772 8;
  • 6) 0.000 000 023 748 725 652 772 8 × 2 = 0 + 0.000 000 047 497 451 305 545 6;
  • 7) 0.000 000 047 497 451 305 545 6 × 2 = 0 + 0.000 000 094 994 902 611 091 2;
  • 8) 0.000 000 094 994 902 611 091 2 × 2 = 0 + 0.000 000 189 989 805 222 182 4;
  • 9) 0.000 000 189 989 805 222 182 4 × 2 = 0 + 0.000 000 379 979 610 444 364 8;
  • 10) 0.000 000 379 979 610 444 364 8 × 2 = 0 + 0.000 000 759 959 220 888 729 6;
  • 11) 0.000 000 759 959 220 888 729 6 × 2 = 0 + 0.000 001 519 918 441 777 459 2;
  • 12) 0.000 001 519 918 441 777 459 2 × 2 = 0 + 0.000 003 039 836 883 554 918 4;
  • 13) 0.000 003 039 836 883 554 918 4 × 2 = 0 + 0.000 006 079 673 767 109 836 8;
  • 14) 0.000 006 079 673 767 109 836 8 × 2 = 0 + 0.000 012 159 347 534 219 673 6;
  • 15) 0.000 012 159 347 534 219 673 6 × 2 = 0 + 0.000 024 318 695 068 439 347 2;
  • 16) 0.000 024 318 695 068 439 347 2 × 2 = 0 + 0.000 048 637 390 136 878 694 4;
  • 17) 0.000 048 637 390 136 878 694 4 × 2 = 0 + 0.000 097 274 780 273 757 388 8;
  • 18) 0.000 097 274 780 273 757 388 8 × 2 = 0 + 0.000 194 549 560 547 514 777 6;
  • 19) 0.000 194 549 560 547 514 777 6 × 2 = 0 + 0.000 389 099 121 095 029 555 2;
  • 20) 0.000 389 099 121 095 029 555 2 × 2 = 0 + 0.000 778 198 242 190 059 110 4;
  • 21) 0.000 778 198 242 190 059 110 4 × 2 = 0 + 0.001 556 396 484 380 118 220 8;
  • 22) 0.001 556 396 484 380 118 220 8 × 2 = 0 + 0.003 112 792 968 760 236 441 6;
  • 23) 0.003 112 792 968 760 236 441 6 × 2 = 0 + 0.006 225 585 937 520 472 883 2;
  • 24) 0.006 225 585 937 520 472 883 2 × 2 = 0 + 0.012 451 171 875 040 945 766 4;
  • 25) 0.012 451 171 875 040 945 766 4 × 2 = 0 + 0.024 902 343 750 081 891 532 8;
  • 26) 0.024 902 343 750 081 891 532 8 × 2 = 0 + 0.049 804 687 500 163 783 065 6;
  • 27) 0.049 804 687 500 163 783 065 6 × 2 = 0 + 0.099 609 375 000 327 566 131 2;
  • 28) 0.099 609 375 000 327 566 131 2 × 2 = 0 + 0.199 218 750 000 655 132 262 4;
  • 29) 0.199 218 750 000 655 132 262 4 × 2 = 0 + 0.398 437 500 001 310 264 524 8;
  • 30) 0.398 437 500 001 310 264 524 8 × 2 = 0 + 0.796 875 000 002 620 529 049 6;
  • 31) 0.796 875 000 002 620 529 049 6 × 2 = 1 + 0.593 750 000 005 241 058 099 2;
  • 32) 0.593 750 000 005 241 058 099 2 × 2 = 1 + 0.187 500 000 010 482 116 198 4;
  • 33) 0.187 500 000 010 482 116 198 4 × 2 = 0 + 0.375 000 000 020 964 232 396 8;
  • 34) 0.375 000 000 020 964 232 396 8 × 2 = 0 + 0.750 000 000 041 928 464 793 6;
  • 35) 0.750 000 000 041 928 464 793 6 × 2 = 1 + 0.500 000 000 083 856 929 587 2;
  • 36) 0.500 000 000 083 856 929 587 2 × 2 = 1 + 0.000 000 000 167 713 859 174 4;
  • 37) 0.000 000 000 167 713 859 174 4 × 2 = 0 + 0.000 000 000 335 427 718 348 8;
  • 38) 0.000 000 000 335 427 718 348 8 × 2 = 0 + 0.000 000 000 670 855 436 697 6;
  • 39) 0.000 000 000 670 855 436 697 6 × 2 = 0 + 0.000 000 001 341 710 873 395 2;
  • 40) 0.000 000 001 341 710 873 395 2 × 2 = 0 + 0.000 000 002 683 421 746 790 4;
  • 41) 0.000 000 002 683 421 746 790 4 × 2 = 0 + 0.000 000 005 366 843 493 580 8;
  • 42) 0.000 000 005 366 843 493 580 8 × 2 = 0 + 0.000 000 010 733 686 987 161 6;
  • 43) 0.000 000 010 733 686 987 161 6 × 2 = 0 + 0.000 000 021 467 373 974 323 2;
  • 44) 0.000 000 021 467 373 974 323 2 × 2 = 0 + 0.000 000 042 934 747 948 646 4;
  • 45) 0.000 000 042 934 747 948 646 4 × 2 = 0 + 0.000 000 085 869 495 897 292 8;
  • 46) 0.000 000 085 869 495 897 292 8 × 2 = 0 + 0.000 000 171 738 991 794 585 6;
  • 47) 0.000 000 171 738 991 794 585 6 × 2 = 0 + 0.000 000 343 477 983 589 171 2;
  • 48) 0.000 000 343 477 983 589 171 2 × 2 = 0 + 0.000 000 686 955 967 178 342 4;
  • 49) 0.000 000 686 955 967 178 342 4 × 2 = 0 + 0.000 001 373 911 934 356 684 8;
  • 50) 0.000 001 373 911 934 356 684 8 × 2 = 0 + 0.000 002 747 823 868 713 369 6;
  • 51) 0.000 002 747 823 868 713 369 6 × 2 = 0 + 0.000 005 495 647 737 426 739 2;
  • 52) 0.000 005 495 647 737 426 739 2 × 2 = 0 + 0.000 010 991 295 474 853 478 4;
  • 53) 0.000 010 991 295 474 853 478 4 × 2 = 0 + 0.000 021 982 590 949 706 956 8;
  • 54) 0.000 021 982 590 949 706 956 8 × 2 = 0 + 0.000 043 965 181 899 413 913 6;

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:


0.000 000 000 742 147 676 649 15(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 649 15(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 649 15(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 649 15 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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