-0.000 000 000 742 147 676 646 714 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 714 3(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
-0.000 000 000 742 147 676 646 714 3(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:

|-0.000 000 000 742 147 676 646 714 3| = 0.000 000 000 742 147 676 646 714 3


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: 0.000 000 000 742 147 676 646 714 3.

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) 0.000 000 000 742 147 676 646 714 3 × 2 = 0 + 0.000 000 001 484 295 353 293 428 6;
  • 2) 0.000 000 001 484 295 353 293 428 6 × 2 = 0 + 0.000 000 002 968 590 706 586 857 2;
  • 3) 0.000 000 002 968 590 706 586 857 2 × 2 = 0 + 0.000 000 005 937 181 413 173 714 4;
  • 4) 0.000 000 005 937 181 413 173 714 4 × 2 = 0 + 0.000 000 011 874 362 826 347 428 8;
  • 5) 0.000 000 011 874 362 826 347 428 8 × 2 = 0 + 0.000 000 023 748 725 652 694 857 6;
  • 6) 0.000 000 023 748 725 652 694 857 6 × 2 = 0 + 0.000 000 047 497 451 305 389 715 2;
  • 7) 0.000 000 047 497 451 305 389 715 2 × 2 = 0 + 0.000 000 094 994 902 610 779 430 4;
  • 8) 0.000 000 094 994 902 610 779 430 4 × 2 = 0 + 0.000 000 189 989 805 221 558 860 8;
  • 9) 0.000 000 189 989 805 221 558 860 8 × 2 = 0 + 0.000 000 379 979 610 443 117 721 6;
  • 10) 0.000 000 379 979 610 443 117 721 6 × 2 = 0 + 0.000 000 759 959 220 886 235 443 2;
  • 11) 0.000 000 759 959 220 886 235 443 2 × 2 = 0 + 0.000 001 519 918 441 772 470 886 4;
  • 12) 0.000 001 519 918 441 772 470 886 4 × 2 = 0 + 0.000 003 039 836 883 544 941 772 8;
  • 13) 0.000 003 039 836 883 544 941 772 8 × 2 = 0 + 0.000 006 079 673 767 089 883 545 6;
  • 14) 0.000 006 079 673 767 089 883 545 6 × 2 = 0 + 0.000 012 159 347 534 179 767 091 2;
  • 15) 0.000 012 159 347 534 179 767 091 2 × 2 = 0 + 0.000 024 318 695 068 359 534 182 4;
  • 16) 0.000 024 318 695 068 359 534 182 4 × 2 = 0 + 0.000 048 637 390 136 719 068 364 8;
  • 17) 0.000 048 637 390 136 719 068 364 8 × 2 = 0 + 0.000 097 274 780 273 438 136 729 6;
  • 18) 0.000 097 274 780 273 438 136 729 6 × 2 = 0 + 0.000 194 549 560 546 876 273 459 2;
  • 19) 0.000 194 549 560 546 876 273 459 2 × 2 = 0 + 0.000 389 099 121 093 752 546 918 4;
  • 20) 0.000 389 099 121 093 752 546 918 4 × 2 = 0 + 0.000 778 198 242 187 505 093 836 8;
  • 21) 0.000 778 198 242 187 505 093 836 8 × 2 = 0 + 0.001 556 396 484 375 010 187 673 6;
  • 22) 0.001 556 396 484 375 010 187 673 6 × 2 = 0 + 0.003 112 792 968 750 020 375 347 2;
  • 23) 0.003 112 792 968 750 020 375 347 2 × 2 = 0 + 0.006 225 585 937 500 040 750 694 4;
  • 24) 0.006 225 585 937 500 040 750 694 4 × 2 = 0 + 0.012 451 171 875 000 081 501 388 8;
  • 25) 0.012 451 171 875 000 081 501 388 8 × 2 = 0 + 0.024 902 343 750 000 163 002 777 6;
  • 26) 0.024 902 343 750 000 163 002 777 6 × 2 = 0 + 0.049 804 687 500 000 326 005 555 2;
  • 27) 0.049 804 687 500 000 326 005 555 2 × 2 = 0 + 0.099 609 375 000 000 652 011 110 4;
  • 28) 0.099 609 375 000 000 652 011 110 4 × 2 = 0 + 0.199 218 750 000 001 304 022 220 8;
  • 29) 0.199 218 750 000 001 304 022 220 8 × 2 = 0 + 0.398 437 500 000 002 608 044 441 6;
  • 30) 0.398 437 500 000 002 608 044 441 6 × 2 = 0 + 0.796 875 000 000 005 216 088 883 2;
  • 31) 0.796 875 000 000 005 216 088 883 2 × 2 = 1 + 0.593 750 000 000 010 432 177 766 4;
  • 32) 0.593 750 000 000 010 432 177 766 4 × 2 = 1 + 0.187 500 000 000 020 864 355 532 8;
  • 33) 0.187 500 000 000 020 864 355 532 8 × 2 = 0 + 0.375 000 000 000 041 728 711 065 6;
  • 34) 0.375 000 000 000 041 728 711 065 6 × 2 = 0 + 0.750 000 000 000 083 457 422 131 2;
  • 35) 0.750 000 000 000 083 457 422 131 2 × 2 = 1 + 0.500 000 000 000 166 914 844 262 4;
  • 36) 0.500 000 000 000 166 914 844 262 4 × 2 = 1 + 0.000 000 000 000 333 829 688 524 8;
  • 37) 0.000 000 000 000 333 829 688 524 8 × 2 = 0 + 0.000 000 000 000 667 659 377 049 6;
  • 38) 0.000 000 000 000 667 659 377 049 6 × 2 = 0 + 0.000 000 000 001 335 318 754 099 2;
  • 39) 0.000 000 000 001 335 318 754 099 2 × 2 = 0 + 0.000 000 000 002 670 637 508 198 4;
  • 40) 0.000 000 000 002 670 637 508 198 4 × 2 = 0 + 0.000 000 000 005 341 275 016 396 8;
  • 41) 0.000 000 000 005 341 275 016 396 8 × 2 = 0 + 0.000 000 000 010 682 550 032 793 6;
  • 42) 0.000 000 000 010 682 550 032 793 6 × 2 = 0 + 0.000 000 000 021 365 100 065 587 2;
  • 43) 0.000 000 000 021 365 100 065 587 2 × 2 = 0 + 0.000 000 000 042 730 200 131 174 4;
  • 44) 0.000 000 000 042 730 200 131 174 4 × 2 = 0 + 0.000 000 000 085 460 400 262 348 8;
  • 45) 0.000 000 000 085 460 400 262 348 8 × 2 = 0 + 0.000 000 000 170 920 800 524 697 6;
  • 46) 0.000 000 000 170 920 800 524 697 6 × 2 = 0 + 0.000 000 000 341 841 601 049 395 2;
  • 47) 0.000 000 000 341 841 601 049 395 2 × 2 = 0 + 0.000 000 000 683 683 202 098 790 4;
  • 48) 0.000 000 000 683 683 202 098 790 4 × 2 = 0 + 0.000 000 001 367 366 404 197 580 8;
  • 49) 0.000 000 001 367 366 404 197 580 8 × 2 = 0 + 0.000 000 002 734 732 808 395 161 6;
  • 50) 0.000 000 002 734 732 808 395 161 6 × 2 = 0 + 0.000 000 005 469 465 616 790 323 2;
  • 51) 0.000 000 005 469 465 616 790 323 2 × 2 = 0 + 0.000 000 010 938 931 233 580 646 4;
  • 52) 0.000 000 010 938 931 233 580 646 4 × 2 = 0 + 0.000 000 021 877 862 467 161 292 8;
  • 53) 0.000 000 021 877 862 467 161 292 8 × 2 = 0 + 0.000 000 043 755 724 934 322 585 6;
  • 54) 0.000 000 043 755 724 934 322 585 6 × 2 = 0 + 0.000 000 087 511 449 868 645 171 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:


0.000 000 000 742 147 676 646 714 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 714 3(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:


0.000 000 000 742 147 676 646 714 3(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 -0.000 000 000 742 147 676 646 714 3 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