-0.000 000 000 742 147 676 646 708 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 708 1(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
-0.000 000 000 742 147 676 646 708 1(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:

|-0.000 000 000 742 147 676 646 708 1| = 0.000 000 000 742 147 676 646 708 1


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: 0.000 000 000 742 147 676 646 708 1.

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) 0.000 000 000 742 147 676 646 708 1 × 2 = 0 + 0.000 000 001 484 295 353 293 416 2;
  • 2) 0.000 000 001 484 295 353 293 416 2 × 2 = 0 + 0.000 000 002 968 590 706 586 832 4;
  • 3) 0.000 000 002 968 590 706 586 832 4 × 2 = 0 + 0.000 000 005 937 181 413 173 664 8;
  • 4) 0.000 000 005 937 181 413 173 664 8 × 2 = 0 + 0.000 000 011 874 362 826 347 329 6;
  • 5) 0.000 000 011 874 362 826 347 329 6 × 2 = 0 + 0.000 000 023 748 725 652 694 659 2;
  • 6) 0.000 000 023 748 725 652 694 659 2 × 2 = 0 + 0.000 000 047 497 451 305 389 318 4;
  • 7) 0.000 000 047 497 451 305 389 318 4 × 2 = 0 + 0.000 000 094 994 902 610 778 636 8;
  • 8) 0.000 000 094 994 902 610 778 636 8 × 2 = 0 + 0.000 000 189 989 805 221 557 273 6;
  • 9) 0.000 000 189 989 805 221 557 273 6 × 2 = 0 + 0.000 000 379 979 610 443 114 547 2;
  • 10) 0.000 000 379 979 610 443 114 547 2 × 2 = 0 + 0.000 000 759 959 220 886 229 094 4;
  • 11) 0.000 000 759 959 220 886 229 094 4 × 2 = 0 + 0.000 001 519 918 441 772 458 188 8;
  • 12) 0.000 001 519 918 441 772 458 188 8 × 2 = 0 + 0.000 003 039 836 883 544 916 377 6;
  • 13) 0.000 003 039 836 883 544 916 377 6 × 2 = 0 + 0.000 006 079 673 767 089 832 755 2;
  • 14) 0.000 006 079 673 767 089 832 755 2 × 2 = 0 + 0.000 012 159 347 534 179 665 510 4;
  • 15) 0.000 012 159 347 534 179 665 510 4 × 2 = 0 + 0.000 024 318 695 068 359 331 020 8;
  • 16) 0.000 024 318 695 068 359 331 020 8 × 2 = 0 + 0.000 048 637 390 136 718 662 041 6;
  • 17) 0.000 048 637 390 136 718 662 041 6 × 2 = 0 + 0.000 097 274 780 273 437 324 083 2;
  • 18) 0.000 097 274 780 273 437 324 083 2 × 2 = 0 + 0.000 194 549 560 546 874 648 166 4;
  • 19) 0.000 194 549 560 546 874 648 166 4 × 2 = 0 + 0.000 389 099 121 093 749 296 332 8;
  • 20) 0.000 389 099 121 093 749 296 332 8 × 2 = 0 + 0.000 778 198 242 187 498 592 665 6;
  • 21) 0.000 778 198 242 187 498 592 665 6 × 2 = 0 + 0.001 556 396 484 374 997 185 331 2;
  • 22) 0.001 556 396 484 374 997 185 331 2 × 2 = 0 + 0.003 112 792 968 749 994 370 662 4;
  • 23) 0.003 112 792 968 749 994 370 662 4 × 2 = 0 + 0.006 225 585 937 499 988 741 324 8;
  • 24) 0.006 225 585 937 499 988 741 324 8 × 2 = 0 + 0.012 451 171 874 999 977 482 649 6;
  • 25) 0.012 451 171 874 999 977 482 649 6 × 2 = 0 + 0.024 902 343 749 999 954 965 299 2;
  • 26) 0.024 902 343 749 999 954 965 299 2 × 2 = 0 + 0.049 804 687 499 999 909 930 598 4;
  • 27) 0.049 804 687 499 999 909 930 598 4 × 2 = 0 + 0.099 609 374 999 999 819 861 196 8;
  • 28) 0.099 609 374 999 999 819 861 196 8 × 2 = 0 + 0.199 218 749 999 999 639 722 393 6;
  • 29) 0.199 218 749 999 999 639 722 393 6 × 2 = 0 + 0.398 437 499 999 999 279 444 787 2;
  • 30) 0.398 437 499 999 999 279 444 787 2 × 2 = 0 + 0.796 874 999 999 998 558 889 574 4;
  • 31) 0.796 874 999 999 998 558 889 574 4 × 2 = 1 + 0.593 749 999 999 997 117 779 148 8;
  • 32) 0.593 749 999 999 997 117 779 148 8 × 2 = 1 + 0.187 499 999 999 994 235 558 297 6;
  • 33) 0.187 499 999 999 994 235 558 297 6 × 2 = 0 + 0.374 999 999 999 988 471 116 595 2;
  • 34) 0.374 999 999 999 988 471 116 595 2 × 2 = 0 + 0.749 999 999 999 976 942 233 190 4;
  • 35) 0.749 999 999 999 976 942 233 190 4 × 2 = 1 + 0.499 999 999 999 953 884 466 380 8;
  • 36) 0.499 999 999 999 953 884 466 380 8 × 2 = 0 + 0.999 999 999 999 907 768 932 761 6;
  • 37) 0.999 999 999 999 907 768 932 761 6 × 2 = 1 + 0.999 999 999 999 815 537 865 523 2;
  • 38) 0.999 999 999 999 815 537 865 523 2 × 2 = 1 + 0.999 999 999 999 631 075 731 046 4;
  • 39) 0.999 999 999 999 631 075 731 046 4 × 2 = 1 + 0.999 999 999 999 262 151 462 092 8;
  • 40) 0.999 999 999 999 262 151 462 092 8 × 2 = 1 + 0.999 999 999 998 524 302 924 185 6;
  • 41) 0.999 999 999 998 524 302 924 185 6 × 2 = 1 + 0.999 999 999 997 048 605 848 371 2;
  • 42) 0.999 999 999 997 048 605 848 371 2 × 2 = 1 + 0.999 999 999 994 097 211 696 742 4;
  • 43) 0.999 999 999 994 097 211 696 742 4 × 2 = 1 + 0.999 999 999 988 194 423 393 484 8;
  • 44) 0.999 999 999 988 194 423 393 484 8 × 2 = 1 + 0.999 999 999 976 388 846 786 969 6;
  • 45) 0.999 999 999 976 388 846 786 969 6 × 2 = 1 + 0.999 999 999 952 777 693 573 939 2;
  • 46) 0.999 999 999 952 777 693 573 939 2 × 2 = 1 + 0.999 999 999 905 555 387 147 878 4;
  • 47) 0.999 999 999 905 555 387 147 878 4 × 2 = 1 + 0.999 999 999 811 110 774 295 756 8;
  • 48) 0.999 999 999 811 110 774 295 756 8 × 2 = 1 + 0.999 999 999 622 221 548 591 513 6;
  • 49) 0.999 999 999 622 221 548 591 513 6 × 2 = 1 + 0.999 999 999 244 443 097 183 027 2;
  • 50) 0.999 999 999 244 443 097 183 027 2 × 2 = 1 + 0.999 999 998 488 886 194 366 054 4;
  • 51) 0.999 999 998 488 886 194 366 054 4 × 2 = 1 + 0.999 999 996 977 772 388 732 108 8;
  • 52) 0.999 999 996 977 772 388 732 108 8 × 2 = 1 + 0.999 999 993 955 544 777 464 217 6;
  • 53) 0.999 999 993 955 544 777 464 217 6 × 2 = 1 + 0.999 999 987 911 089 554 928 435 2;
  • 54) 0.999 999 987 911 089 554 928 435 2 × 2 = 1 + 0.999 999 975 822 179 109 856 870 4;

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:


0.000 000 000 742 147 676 646 708 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 708 1(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:


0.000 000 000 742 147 676 646 708 1(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 -0.000 000 000 742 147 676 646 708 1 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