-0.000 000 000 742 147 678 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 678 8(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
-0.000 000 000 742 147 678 8(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:

|-0.000 000 000 742 147 678 8| = 0.000 000 000 742 147 678 8


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: 0.000 000 000 742 147 678 8.

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) 0.000 000 000 742 147 678 8 × 2 = 0 + 0.000 000 001 484 295 357 6;
  • 2) 0.000 000 001 484 295 357 6 × 2 = 0 + 0.000 000 002 968 590 715 2;
  • 3) 0.000 000 002 968 590 715 2 × 2 = 0 + 0.000 000 005 937 181 430 4;
  • 4) 0.000 000 005 937 181 430 4 × 2 = 0 + 0.000 000 011 874 362 860 8;
  • 5) 0.000 000 011 874 362 860 8 × 2 = 0 + 0.000 000 023 748 725 721 6;
  • 6) 0.000 000 023 748 725 721 6 × 2 = 0 + 0.000 000 047 497 451 443 2;
  • 7) 0.000 000 047 497 451 443 2 × 2 = 0 + 0.000 000 094 994 902 886 4;
  • 8) 0.000 000 094 994 902 886 4 × 2 = 0 + 0.000 000 189 989 805 772 8;
  • 9) 0.000 000 189 989 805 772 8 × 2 = 0 + 0.000 000 379 979 611 545 6;
  • 10) 0.000 000 379 979 611 545 6 × 2 = 0 + 0.000 000 759 959 223 091 2;
  • 11) 0.000 000 759 959 223 091 2 × 2 = 0 + 0.000 001 519 918 446 182 4;
  • 12) 0.000 001 519 918 446 182 4 × 2 = 0 + 0.000 003 039 836 892 364 8;
  • 13) 0.000 003 039 836 892 364 8 × 2 = 0 + 0.000 006 079 673 784 729 6;
  • 14) 0.000 006 079 673 784 729 6 × 2 = 0 + 0.000 012 159 347 569 459 2;
  • 15) 0.000 012 159 347 569 459 2 × 2 = 0 + 0.000 024 318 695 138 918 4;
  • 16) 0.000 024 318 695 138 918 4 × 2 = 0 + 0.000 048 637 390 277 836 8;
  • 17) 0.000 048 637 390 277 836 8 × 2 = 0 + 0.000 097 274 780 555 673 6;
  • 18) 0.000 097 274 780 555 673 6 × 2 = 0 + 0.000 194 549 561 111 347 2;
  • 19) 0.000 194 549 561 111 347 2 × 2 = 0 + 0.000 389 099 122 222 694 4;
  • 20) 0.000 389 099 122 222 694 4 × 2 = 0 + 0.000 778 198 244 445 388 8;
  • 21) 0.000 778 198 244 445 388 8 × 2 = 0 + 0.001 556 396 488 890 777 6;
  • 22) 0.001 556 396 488 890 777 6 × 2 = 0 + 0.003 112 792 977 781 555 2;
  • 23) 0.003 112 792 977 781 555 2 × 2 = 0 + 0.006 225 585 955 563 110 4;
  • 24) 0.006 225 585 955 563 110 4 × 2 = 0 + 0.012 451 171 911 126 220 8;
  • 25) 0.012 451 171 911 126 220 8 × 2 = 0 + 0.024 902 343 822 252 441 6;
  • 26) 0.024 902 343 822 252 441 6 × 2 = 0 + 0.049 804 687 644 504 883 2;
  • 27) 0.049 804 687 644 504 883 2 × 2 = 0 + 0.099 609 375 289 009 766 4;
  • 28) 0.099 609 375 289 009 766 4 × 2 = 0 + 0.199 218 750 578 019 532 8;
  • 29) 0.199 218 750 578 019 532 8 × 2 = 0 + 0.398 437 501 156 039 065 6;
  • 30) 0.398 437 501 156 039 065 6 × 2 = 0 + 0.796 875 002 312 078 131 2;
  • 31) 0.796 875 002 312 078 131 2 × 2 = 1 + 0.593 750 004 624 156 262 4;
  • 32) 0.593 750 004 624 156 262 4 × 2 = 1 + 0.187 500 009 248 312 524 8;
  • 33) 0.187 500 009 248 312 524 8 × 2 = 0 + 0.375 000 018 496 625 049 6;
  • 34) 0.375 000 018 496 625 049 6 × 2 = 0 + 0.750 000 036 993 250 099 2;
  • 35) 0.750 000 036 993 250 099 2 × 2 = 1 + 0.500 000 073 986 500 198 4;
  • 36) 0.500 000 073 986 500 198 4 × 2 = 1 + 0.000 000 147 973 000 396 8;
  • 37) 0.000 000 147 973 000 396 8 × 2 = 0 + 0.000 000 295 946 000 793 6;
  • 38) 0.000 000 295 946 000 793 6 × 2 = 0 + 0.000 000 591 892 001 587 2;
  • 39) 0.000 000 591 892 001 587 2 × 2 = 0 + 0.000 001 183 784 003 174 4;
  • 40) 0.000 001 183 784 003 174 4 × 2 = 0 + 0.000 002 367 568 006 348 8;
  • 41) 0.000 002 367 568 006 348 8 × 2 = 0 + 0.000 004 735 136 012 697 6;
  • 42) 0.000 004 735 136 012 697 6 × 2 = 0 + 0.000 009 470 272 025 395 2;
  • 43) 0.000 009 470 272 025 395 2 × 2 = 0 + 0.000 018 940 544 050 790 4;
  • 44) 0.000 018 940 544 050 790 4 × 2 = 0 + 0.000 037 881 088 101 580 8;
  • 45) 0.000 037 881 088 101 580 8 × 2 = 0 + 0.000 075 762 176 203 161 6;
  • 46) 0.000 075 762 176 203 161 6 × 2 = 0 + 0.000 151 524 352 406 323 2;
  • 47) 0.000 151 524 352 406 323 2 × 2 = 0 + 0.000 303 048 704 812 646 4;
  • 48) 0.000 303 048 704 812 646 4 × 2 = 0 + 0.000 606 097 409 625 292 8;
  • 49) 0.000 606 097 409 625 292 8 × 2 = 0 + 0.001 212 194 819 250 585 6;
  • 50) 0.001 212 194 819 250 585 6 × 2 = 0 + 0.002 424 389 638 501 171 2;
  • 51) 0.002 424 389 638 501 171 2 × 2 = 0 + 0.004 848 779 277 002 342 4;
  • 52) 0.004 848 779 277 002 342 4 × 2 = 0 + 0.009 697 558 554 004 684 8;
  • 53) 0.009 697 558 554 004 684 8 × 2 = 0 + 0.019 395 117 108 009 369 6;
  • 54) 0.019 395 117 108 009 369 6 × 2 = 0 + 0.038 790 234 216 018 739 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:


0.000 000 000 742 147 678 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 678 8(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:


0.000 000 000 742 147 678 8(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 -0.000 000 000 742 147 678 8 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