-0.000 000 000 742 147 676 479 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 479(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
-0.000 000 000 742 147 676 479(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:

|-0.000 000 000 742 147 676 479| = 0.000 000 000 742 147 676 479


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: 0.000 000 000 742 147 676 479.

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) 0.000 000 000 742 147 676 479 × 2 = 0 + 0.000 000 001 484 295 352 958;
  • 2) 0.000 000 001 484 295 352 958 × 2 = 0 + 0.000 000 002 968 590 705 916;
  • 3) 0.000 000 002 968 590 705 916 × 2 = 0 + 0.000 000 005 937 181 411 832;
  • 4) 0.000 000 005 937 181 411 832 × 2 = 0 + 0.000 000 011 874 362 823 664;
  • 5) 0.000 000 011 874 362 823 664 × 2 = 0 + 0.000 000 023 748 725 647 328;
  • 6) 0.000 000 023 748 725 647 328 × 2 = 0 + 0.000 000 047 497 451 294 656;
  • 7) 0.000 000 047 497 451 294 656 × 2 = 0 + 0.000 000 094 994 902 589 312;
  • 8) 0.000 000 094 994 902 589 312 × 2 = 0 + 0.000 000 189 989 805 178 624;
  • 9) 0.000 000 189 989 805 178 624 × 2 = 0 + 0.000 000 379 979 610 357 248;
  • 10) 0.000 000 379 979 610 357 248 × 2 = 0 + 0.000 000 759 959 220 714 496;
  • 11) 0.000 000 759 959 220 714 496 × 2 = 0 + 0.000 001 519 918 441 428 992;
  • 12) 0.000 001 519 918 441 428 992 × 2 = 0 + 0.000 003 039 836 882 857 984;
  • 13) 0.000 003 039 836 882 857 984 × 2 = 0 + 0.000 006 079 673 765 715 968;
  • 14) 0.000 006 079 673 765 715 968 × 2 = 0 + 0.000 012 159 347 531 431 936;
  • 15) 0.000 012 159 347 531 431 936 × 2 = 0 + 0.000 024 318 695 062 863 872;
  • 16) 0.000 024 318 695 062 863 872 × 2 = 0 + 0.000 048 637 390 125 727 744;
  • 17) 0.000 048 637 390 125 727 744 × 2 = 0 + 0.000 097 274 780 251 455 488;
  • 18) 0.000 097 274 780 251 455 488 × 2 = 0 + 0.000 194 549 560 502 910 976;
  • 19) 0.000 194 549 560 502 910 976 × 2 = 0 + 0.000 389 099 121 005 821 952;
  • 20) 0.000 389 099 121 005 821 952 × 2 = 0 + 0.000 778 198 242 011 643 904;
  • 21) 0.000 778 198 242 011 643 904 × 2 = 0 + 0.001 556 396 484 023 287 808;
  • 22) 0.001 556 396 484 023 287 808 × 2 = 0 + 0.003 112 792 968 046 575 616;
  • 23) 0.003 112 792 968 046 575 616 × 2 = 0 + 0.006 225 585 936 093 151 232;
  • 24) 0.006 225 585 936 093 151 232 × 2 = 0 + 0.012 451 171 872 186 302 464;
  • 25) 0.012 451 171 872 186 302 464 × 2 = 0 + 0.024 902 343 744 372 604 928;
  • 26) 0.024 902 343 744 372 604 928 × 2 = 0 + 0.049 804 687 488 745 209 856;
  • 27) 0.049 804 687 488 745 209 856 × 2 = 0 + 0.099 609 374 977 490 419 712;
  • 28) 0.099 609 374 977 490 419 712 × 2 = 0 + 0.199 218 749 954 980 839 424;
  • 29) 0.199 218 749 954 980 839 424 × 2 = 0 + 0.398 437 499 909 961 678 848;
  • 30) 0.398 437 499 909 961 678 848 × 2 = 0 + 0.796 874 999 819 923 357 696;
  • 31) 0.796 874 999 819 923 357 696 × 2 = 1 + 0.593 749 999 639 846 715 392;
  • 32) 0.593 749 999 639 846 715 392 × 2 = 1 + 0.187 499 999 279 693 430 784;
  • 33) 0.187 499 999 279 693 430 784 × 2 = 0 + 0.374 999 998 559 386 861 568;
  • 34) 0.374 999 998 559 386 861 568 × 2 = 0 + 0.749 999 997 118 773 723 136;
  • 35) 0.749 999 997 118 773 723 136 × 2 = 1 + 0.499 999 994 237 547 446 272;
  • 36) 0.499 999 994 237 547 446 272 × 2 = 0 + 0.999 999 988 475 094 892 544;
  • 37) 0.999 999 988 475 094 892 544 × 2 = 1 + 0.999 999 976 950 189 785 088;
  • 38) 0.999 999 976 950 189 785 088 × 2 = 1 + 0.999 999 953 900 379 570 176;
  • 39) 0.999 999 953 900 379 570 176 × 2 = 1 + 0.999 999 907 800 759 140 352;
  • 40) 0.999 999 907 800 759 140 352 × 2 = 1 + 0.999 999 815 601 518 280 704;
  • 41) 0.999 999 815 601 518 280 704 × 2 = 1 + 0.999 999 631 203 036 561 408;
  • 42) 0.999 999 631 203 036 561 408 × 2 = 1 + 0.999 999 262 406 073 122 816;
  • 43) 0.999 999 262 406 073 122 816 × 2 = 1 + 0.999 998 524 812 146 245 632;
  • 44) 0.999 998 524 812 146 245 632 × 2 = 1 + 0.999 997 049 624 292 491 264;
  • 45) 0.999 997 049 624 292 491 264 × 2 = 1 + 0.999 994 099 248 584 982 528;
  • 46) 0.999 994 099 248 584 982 528 × 2 = 1 + 0.999 988 198 497 169 965 056;
  • 47) 0.999 988 198 497 169 965 056 × 2 = 1 + 0.999 976 396 994 339 930 112;
  • 48) 0.999 976 396 994 339 930 112 × 2 = 1 + 0.999 952 793 988 679 860 224;
  • 49) 0.999 952 793 988 679 860 224 × 2 = 1 + 0.999 905 587 977 359 720 448;
  • 50) 0.999 905 587 977 359 720 448 × 2 = 1 + 0.999 811 175 954 719 440 896;
  • 51) 0.999 811 175 954 719 440 896 × 2 = 1 + 0.999 622 351 909 438 881 792;
  • 52) 0.999 622 351 909 438 881 792 × 2 = 1 + 0.999 244 703 818 877 763 584;
  • 53) 0.999 244 703 818 877 763 584 × 2 = 1 + 0.998 489 407 637 755 527 168;
  • 54) 0.998 489 407 637 755 527 168 × 2 = 1 + 0.996 978 815 275 511 054 336;

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:


0.000 000 000 742 147 676 479(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 479(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:


0.000 000 000 742 147 676 479(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 -0.000 000 000 742 147 676 479 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