-0.000 000 000 742 147 682 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 682 4(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
-0.000 000 000 742 147 682 4(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:

|-0.000 000 000 742 147 682 4| = 0.000 000 000 742 147 682 4


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: 0.000 000 000 742 147 682 4.

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) 0.000 000 000 742 147 682 4 × 2 = 0 + 0.000 000 001 484 295 364 8;
  • 2) 0.000 000 001 484 295 364 8 × 2 = 0 + 0.000 000 002 968 590 729 6;
  • 3) 0.000 000 002 968 590 729 6 × 2 = 0 + 0.000 000 005 937 181 459 2;
  • 4) 0.000 000 005 937 181 459 2 × 2 = 0 + 0.000 000 011 874 362 918 4;
  • 5) 0.000 000 011 874 362 918 4 × 2 = 0 + 0.000 000 023 748 725 836 8;
  • 6) 0.000 000 023 748 725 836 8 × 2 = 0 + 0.000 000 047 497 451 673 6;
  • 7) 0.000 000 047 497 451 673 6 × 2 = 0 + 0.000 000 094 994 903 347 2;
  • 8) 0.000 000 094 994 903 347 2 × 2 = 0 + 0.000 000 189 989 806 694 4;
  • 9) 0.000 000 189 989 806 694 4 × 2 = 0 + 0.000 000 379 979 613 388 8;
  • 10) 0.000 000 379 979 613 388 8 × 2 = 0 + 0.000 000 759 959 226 777 6;
  • 11) 0.000 000 759 959 226 777 6 × 2 = 0 + 0.000 001 519 918 453 555 2;
  • 12) 0.000 001 519 918 453 555 2 × 2 = 0 + 0.000 003 039 836 907 110 4;
  • 13) 0.000 003 039 836 907 110 4 × 2 = 0 + 0.000 006 079 673 814 220 8;
  • 14) 0.000 006 079 673 814 220 8 × 2 = 0 + 0.000 012 159 347 628 441 6;
  • 15) 0.000 012 159 347 628 441 6 × 2 = 0 + 0.000 024 318 695 256 883 2;
  • 16) 0.000 024 318 695 256 883 2 × 2 = 0 + 0.000 048 637 390 513 766 4;
  • 17) 0.000 048 637 390 513 766 4 × 2 = 0 + 0.000 097 274 781 027 532 8;
  • 18) 0.000 097 274 781 027 532 8 × 2 = 0 + 0.000 194 549 562 055 065 6;
  • 19) 0.000 194 549 562 055 065 6 × 2 = 0 + 0.000 389 099 124 110 131 2;
  • 20) 0.000 389 099 124 110 131 2 × 2 = 0 + 0.000 778 198 248 220 262 4;
  • 21) 0.000 778 198 248 220 262 4 × 2 = 0 + 0.001 556 396 496 440 524 8;
  • 22) 0.001 556 396 496 440 524 8 × 2 = 0 + 0.003 112 792 992 881 049 6;
  • 23) 0.003 112 792 992 881 049 6 × 2 = 0 + 0.006 225 585 985 762 099 2;
  • 24) 0.006 225 585 985 762 099 2 × 2 = 0 + 0.012 451 171 971 524 198 4;
  • 25) 0.012 451 171 971 524 198 4 × 2 = 0 + 0.024 902 343 943 048 396 8;
  • 26) 0.024 902 343 943 048 396 8 × 2 = 0 + 0.049 804 687 886 096 793 6;
  • 27) 0.049 804 687 886 096 793 6 × 2 = 0 + 0.099 609 375 772 193 587 2;
  • 28) 0.099 609 375 772 193 587 2 × 2 = 0 + 0.199 218 751 544 387 174 4;
  • 29) 0.199 218 751 544 387 174 4 × 2 = 0 + 0.398 437 503 088 774 348 8;
  • 30) 0.398 437 503 088 774 348 8 × 2 = 0 + 0.796 875 006 177 548 697 6;
  • 31) 0.796 875 006 177 548 697 6 × 2 = 1 + 0.593 750 012 355 097 395 2;
  • 32) 0.593 750 012 355 097 395 2 × 2 = 1 + 0.187 500 024 710 194 790 4;
  • 33) 0.187 500 024 710 194 790 4 × 2 = 0 + 0.375 000 049 420 389 580 8;
  • 34) 0.375 000 049 420 389 580 8 × 2 = 0 + 0.750 000 098 840 779 161 6;
  • 35) 0.750 000 098 840 779 161 6 × 2 = 1 + 0.500 000 197 681 558 323 2;
  • 36) 0.500 000 197 681 558 323 2 × 2 = 1 + 0.000 000 395 363 116 646 4;
  • 37) 0.000 000 395 363 116 646 4 × 2 = 0 + 0.000 000 790 726 233 292 8;
  • 38) 0.000 000 790 726 233 292 8 × 2 = 0 + 0.000 001 581 452 466 585 6;
  • 39) 0.000 001 581 452 466 585 6 × 2 = 0 + 0.000 003 162 904 933 171 2;
  • 40) 0.000 003 162 904 933 171 2 × 2 = 0 + 0.000 006 325 809 866 342 4;
  • 41) 0.000 006 325 809 866 342 4 × 2 = 0 + 0.000 012 651 619 732 684 8;
  • 42) 0.000 012 651 619 732 684 8 × 2 = 0 + 0.000 025 303 239 465 369 6;
  • 43) 0.000 025 303 239 465 369 6 × 2 = 0 + 0.000 050 606 478 930 739 2;
  • 44) 0.000 050 606 478 930 739 2 × 2 = 0 + 0.000 101 212 957 861 478 4;
  • 45) 0.000 101 212 957 861 478 4 × 2 = 0 + 0.000 202 425 915 722 956 8;
  • 46) 0.000 202 425 915 722 956 8 × 2 = 0 + 0.000 404 851 831 445 913 6;
  • 47) 0.000 404 851 831 445 913 6 × 2 = 0 + 0.000 809 703 662 891 827 2;
  • 48) 0.000 809 703 662 891 827 2 × 2 = 0 + 0.001 619 407 325 783 654 4;
  • 49) 0.001 619 407 325 783 654 4 × 2 = 0 + 0.003 238 814 651 567 308 8;
  • 50) 0.003 238 814 651 567 308 8 × 2 = 0 + 0.006 477 629 303 134 617 6;
  • 51) 0.006 477 629 303 134 617 6 × 2 = 0 + 0.012 955 258 606 269 235 2;
  • 52) 0.012 955 258 606 269 235 2 × 2 = 0 + 0.025 910 517 212 538 470 4;
  • 53) 0.025 910 517 212 538 470 4 × 2 = 0 + 0.051 821 034 425 076 940 8;
  • 54) 0.051 821 034 425 076 940 8 × 2 = 0 + 0.103 642 068 850 153 881 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:


0.000 000 000 742 147 682 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 682 4(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:


0.000 000 000 742 147 682 4(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 -0.000 000 000 742 147 682 4 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