-0.000 000 000 742 147 681 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 681 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
-0.000 000 000 742 147 681 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:

|-0.000 000 000 742 147 681 7| = 0.000 000 000 742 147 681 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: 0.000 000 000 742 147 681 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) 0.000 000 000 742 147 681 7 × 2 = 0 + 0.000 000 001 484 295 363 4;
  • 2) 0.000 000 001 484 295 363 4 × 2 = 0 + 0.000 000 002 968 590 726 8;
  • 3) 0.000 000 002 968 590 726 8 × 2 = 0 + 0.000 000 005 937 181 453 6;
  • 4) 0.000 000 005 937 181 453 6 × 2 = 0 + 0.000 000 011 874 362 907 2;
  • 5) 0.000 000 011 874 362 907 2 × 2 = 0 + 0.000 000 023 748 725 814 4;
  • 6) 0.000 000 023 748 725 814 4 × 2 = 0 + 0.000 000 047 497 451 628 8;
  • 7) 0.000 000 047 497 451 628 8 × 2 = 0 + 0.000 000 094 994 903 257 6;
  • 8) 0.000 000 094 994 903 257 6 × 2 = 0 + 0.000 000 189 989 806 515 2;
  • 9) 0.000 000 189 989 806 515 2 × 2 = 0 + 0.000 000 379 979 613 030 4;
  • 10) 0.000 000 379 979 613 030 4 × 2 = 0 + 0.000 000 759 959 226 060 8;
  • 11) 0.000 000 759 959 226 060 8 × 2 = 0 + 0.000 001 519 918 452 121 6;
  • 12) 0.000 001 519 918 452 121 6 × 2 = 0 + 0.000 003 039 836 904 243 2;
  • 13) 0.000 003 039 836 904 243 2 × 2 = 0 + 0.000 006 079 673 808 486 4;
  • 14) 0.000 006 079 673 808 486 4 × 2 = 0 + 0.000 012 159 347 616 972 8;
  • 15) 0.000 012 159 347 616 972 8 × 2 = 0 + 0.000 024 318 695 233 945 6;
  • 16) 0.000 024 318 695 233 945 6 × 2 = 0 + 0.000 048 637 390 467 891 2;
  • 17) 0.000 048 637 390 467 891 2 × 2 = 0 + 0.000 097 274 780 935 782 4;
  • 18) 0.000 097 274 780 935 782 4 × 2 = 0 + 0.000 194 549 561 871 564 8;
  • 19) 0.000 194 549 561 871 564 8 × 2 = 0 + 0.000 389 099 123 743 129 6;
  • 20) 0.000 389 099 123 743 129 6 × 2 = 0 + 0.000 778 198 247 486 259 2;
  • 21) 0.000 778 198 247 486 259 2 × 2 = 0 + 0.001 556 396 494 972 518 4;
  • 22) 0.001 556 396 494 972 518 4 × 2 = 0 + 0.003 112 792 989 945 036 8;
  • 23) 0.003 112 792 989 945 036 8 × 2 = 0 + 0.006 225 585 979 890 073 6;
  • 24) 0.006 225 585 979 890 073 6 × 2 = 0 + 0.012 451 171 959 780 147 2;
  • 25) 0.012 451 171 959 780 147 2 × 2 = 0 + 0.024 902 343 919 560 294 4;
  • 26) 0.024 902 343 919 560 294 4 × 2 = 0 + 0.049 804 687 839 120 588 8;
  • 27) 0.049 804 687 839 120 588 8 × 2 = 0 + 0.099 609 375 678 241 177 6;
  • 28) 0.099 609 375 678 241 177 6 × 2 = 0 + 0.199 218 751 356 482 355 2;
  • 29) 0.199 218 751 356 482 355 2 × 2 = 0 + 0.398 437 502 712 964 710 4;
  • 30) 0.398 437 502 712 964 710 4 × 2 = 0 + 0.796 875 005 425 929 420 8;
  • 31) 0.796 875 005 425 929 420 8 × 2 = 1 + 0.593 750 010 851 858 841 6;
  • 32) 0.593 750 010 851 858 841 6 × 2 = 1 + 0.187 500 021 703 717 683 2;
  • 33) 0.187 500 021 703 717 683 2 × 2 = 0 + 0.375 000 043 407 435 366 4;
  • 34) 0.375 000 043 407 435 366 4 × 2 = 0 + 0.750 000 086 814 870 732 8;
  • 35) 0.750 000 086 814 870 732 8 × 2 = 1 + 0.500 000 173 629 741 465 6;
  • 36) 0.500 000 173 629 741 465 6 × 2 = 1 + 0.000 000 347 259 482 931 2;
  • 37) 0.000 000 347 259 482 931 2 × 2 = 0 + 0.000 000 694 518 965 862 4;
  • 38) 0.000 000 694 518 965 862 4 × 2 = 0 + 0.000 001 389 037 931 724 8;
  • 39) 0.000 001 389 037 931 724 8 × 2 = 0 + 0.000 002 778 075 863 449 6;
  • 40) 0.000 002 778 075 863 449 6 × 2 = 0 + 0.000 005 556 151 726 899 2;
  • 41) 0.000 005 556 151 726 899 2 × 2 = 0 + 0.000 011 112 303 453 798 4;
  • 42) 0.000 011 112 303 453 798 4 × 2 = 0 + 0.000 022 224 606 907 596 8;
  • 43) 0.000 022 224 606 907 596 8 × 2 = 0 + 0.000 044 449 213 815 193 6;
  • 44) 0.000 044 449 213 815 193 6 × 2 = 0 + 0.000 088 898 427 630 387 2;
  • 45) 0.000 088 898 427 630 387 2 × 2 = 0 + 0.000 177 796 855 260 774 4;
  • 46) 0.000 177 796 855 260 774 4 × 2 = 0 + 0.000 355 593 710 521 548 8;
  • 47) 0.000 355 593 710 521 548 8 × 2 = 0 + 0.000 711 187 421 043 097 6;
  • 48) 0.000 711 187 421 043 097 6 × 2 = 0 + 0.001 422 374 842 086 195 2;
  • 49) 0.001 422 374 842 086 195 2 × 2 = 0 + 0.002 844 749 684 172 390 4;
  • 50) 0.002 844 749 684 172 390 4 × 2 = 0 + 0.005 689 499 368 344 780 8;
  • 51) 0.005 689 499 368 344 780 8 × 2 = 0 + 0.011 378 998 736 689 561 6;
  • 52) 0.011 378 998 736 689 561 6 × 2 = 0 + 0.022 757 997 473 379 123 2;
  • 53) 0.022 757 997 473 379 123 2 × 2 = 0 + 0.045 515 994 946 758 246 4;
  • 54) 0.045 515 994 946 758 246 4 × 2 = 0 + 0.091 031 989 893 516 492 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:


0.000 000 000 742 147 681 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 681 7(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:


0.000 000 000 742 147 681 7(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 -0.000 000 000 742 147 681 7 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