-0.000 000 000 742 147 676 646 666 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 666(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
-0.000 000 000 742 147 676 646 666(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:

|-0.000 000 000 742 147 676 646 666| = 0.000 000 000 742 147 676 646 666


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: 0.000 000 000 742 147 676 646 666.

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) 0.000 000 000 742 147 676 646 666 × 2 = 0 + 0.000 000 001 484 295 353 293 332;
  • 2) 0.000 000 001 484 295 353 293 332 × 2 = 0 + 0.000 000 002 968 590 706 586 664;
  • 3) 0.000 000 002 968 590 706 586 664 × 2 = 0 + 0.000 000 005 937 181 413 173 328;
  • 4) 0.000 000 005 937 181 413 173 328 × 2 = 0 + 0.000 000 011 874 362 826 346 656;
  • 5) 0.000 000 011 874 362 826 346 656 × 2 = 0 + 0.000 000 023 748 725 652 693 312;
  • 6) 0.000 000 023 748 725 652 693 312 × 2 = 0 + 0.000 000 047 497 451 305 386 624;
  • 7) 0.000 000 047 497 451 305 386 624 × 2 = 0 + 0.000 000 094 994 902 610 773 248;
  • 8) 0.000 000 094 994 902 610 773 248 × 2 = 0 + 0.000 000 189 989 805 221 546 496;
  • 9) 0.000 000 189 989 805 221 546 496 × 2 = 0 + 0.000 000 379 979 610 443 092 992;
  • 10) 0.000 000 379 979 610 443 092 992 × 2 = 0 + 0.000 000 759 959 220 886 185 984;
  • 11) 0.000 000 759 959 220 886 185 984 × 2 = 0 + 0.000 001 519 918 441 772 371 968;
  • 12) 0.000 001 519 918 441 772 371 968 × 2 = 0 + 0.000 003 039 836 883 544 743 936;
  • 13) 0.000 003 039 836 883 544 743 936 × 2 = 0 + 0.000 006 079 673 767 089 487 872;
  • 14) 0.000 006 079 673 767 089 487 872 × 2 = 0 + 0.000 012 159 347 534 178 975 744;
  • 15) 0.000 012 159 347 534 178 975 744 × 2 = 0 + 0.000 024 318 695 068 357 951 488;
  • 16) 0.000 024 318 695 068 357 951 488 × 2 = 0 + 0.000 048 637 390 136 715 902 976;
  • 17) 0.000 048 637 390 136 715 902 976 × 2 = 0 + 0.000 097 274 780 273 431 805 952;
  • 18) 0.000 097 274 780 273 431 805 952 × 2 = 0 + 0.000 194 549 560 546 863 611 904;
  • 19) 0.000 194 549 560 546 863 611 904 × 2 = 0 + 0.000 389 099 121 093 727 223 808;
  • 20) 0.000 389 099 121 093 727 223 808 × 2 = 0 + 0.000 778 198 242 187 454 447 616;
  • 21) 0.000 778 198 242 187 454 447 616 × 2 = 0 + 0.001 556 396 484 374 908 895 232;
  • 22) 0.001 556 396 484 374 908 895 232 × 2 = 0 + 0.003 112 792 968 749 817 790 464;
  • 23) 0.003 112 792 968 749 817 790 464 × 2 = 0 + 0.006 225 585 937 499 635 580 928;
  • 24) 0.006 225 585 937 499 635 580 928 × 2 = 0 + 0.012 451 171 874 999 271 161 856;
  • 25) 0.012 451 171 874 999 271 161 856 × 2 = 0 + 0.024 902 343 749 998 542 323 712;
  • 26) 0.024 902 343 749 998 542 323 712 × 2 = 0 + 0.049 804 687 499 997 084 647 424;
  • 27) 0.049 804 687 499 997 084 647 424 × 2 = 0 + 0.099 609 374 999 994 169 294 848;
  • 28) 0.099 609 374 999 994 169 294 848 × 2 = 0 + 0.199 218 749 999 988 338 589 696;
  • 29) 0.199 218 749 999 988 338 589 696 × 2 = 0 + 0.398 437 499 999 976 677 179 392;
  • 30) 0.398 437 499 999 976 677 179 392 × 2 = 0 + 0.796 874 999 999 953 354 358 784;
  • 31) 0.796 874 999 999 953 354 358 784 × 2 = 1 + 0.593 749 999 999 906 708 717 568;
  • 32) 0.593 749 999 999 906 708 717 568 × 2 = 1 + 0.187 499 999 999 813 417 435 136;
  • 33) 0.187 499 999 999 813 417 435 136 × 2 = 0 + 0.374 999 999 999 626 834 870 272;
  • 34) 0.374 999 999 999 626 834 870 272 × 2 = 0 + 0.749 999 999 999 253 669 740 544;
  • 35) 0.749 999 999 999 253 669 740 544 × 2 = 1 + 0.499 999 999 998 507 339 481 088;
  • 36) 0.499 999 999 998 507 339 481 088 × 2 = 0 + 0.999 999 999 997 014 678 962 176;
  • 37) 0.999 999 999 997 014 678 962 176 × 2 = 1 + 0.999 999 999 994 029 357 924 352;
  • 38) 0.999 999 999 994 029 357 924 352 × 2 = 1 + 0.999 999 999 988 058 715 848 704;
  • 39) 0.999 999 999 988 058 715 848 704 × 2 = 1 + 0.999 999 999 976 117 431 697 408;
  • 40) 0.999 999 999 976 117 431 697 408 × 2 = 1 + 0.999 999 999 952 234 863 394 816;
  • 41) 0.999 999 999 952 234 863 394 816 × 2 = 1 + 0.999 999 999 904 469 726 789 632;
  • 42) 0.999 999 999 904 469 726 789 632 × 2 = 1 + 0.999 999 999 808 939 453 579 264;
  • 43) 0.999 999 999 808 939 453 579 264 × 2 = 1 + 0.999 999 999 617 878 907 158 528;
  • 44) 0.999 999 999 617 878 907 158 528 × 2 = 1 + 0.999 999 999 235 757 814 317 056;
  • 45) 0.999 999 999 235 757 814 317 056 × 2 = 1 + 0.999 999 998 471 515 628 634 112;
  • 46) 0.999 999 998 471 515 628 634 112 × 2 = 1 + 0.999 999 996 943 031 257 268 224;
  • 47) 0.999 999 996 943 031 257 268 224 × 2 = 1 + 0.999 999 993 886 062 514 536 448;
  • 48) 0.999 999 993 886 062 514 536 448 × 2 = 1 + 0.999 999 987 772 125 029 072 896;
  • 49) 0.999 999 987 772 125 029 072 896 × 2 = 1 + 0.999 999 975 544 250 058 145 792;
  • 50) 0.999 999 975 544 250 058 145 792 × 2 = 1 + 0.999 999 951 088 500 116 291 584;
  • 51) 0.999 999 951 088 500 116 291 584 × 2 = 1 + 0.999 999 902 177 000 232 583 168;
  • 52) 0.999 999 902 177 000 232 583 168 × 2 = 1 + 0.999 999 804 354 000 465 166 336;
  • 53) 0.999 999 804 354 000 465 166 336 × 2 = 1 + 0.999 999 608 708 000 930 332 672;
  • 54) 0.999 999 608 708 000 930 332 672 × 2 = 1 + 0.999 999 217 416 001 860 665 344;

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:


0.000 000 000 742 147 676 646 666(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 666(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:


0.000 000 000 742 147 676 646 666(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 -0.000 000 000 742 147 676 646 666 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