-0.000 000 000 742 147 676 646 421 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 421(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
-0.000 000 000 742 147 676 646 421(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:

|-0.000 000 000 742 147 676 646 421| = 0.000 000 000 742 147 676 646 421


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: 0.000 000 000 742 147 676 646 421.

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) 0.000 000 000 742 147 676 646 421 × 2 = 0 + 0.000 000 001 484 295 353 292 842;
  • 2) 0.000 000 001 484 295 353 292 842 × 2 = 0 + 0.000 000 002 968 590 706 585 684;
  • 3) 0.000 000 002 968 590 706 585 684 × 2 = 0 + 0.000 000 005 937 181 413 171 368;
  • 4) 0.000 000 005 937 181 413 171 368 × 2 = 0 + 0.000 000 011 874 362 826 342 736;
  • 5) 0.000 000 011 874 362 826 342 736 × 2 = 0 + 0.000 000 023 748 725 652 685 472;
  • 6) 0.000 000 023 748 725 652 685 472 × 2 = 0 + 0.000 000 047 497 451 305 370 944;
  • 7) 0.000 000 047 497 451 305 370 944 × 2 = 0 + 0.000 000 094 994 902 610 741 888;
  • 8) 0.000 000 094 994 902 610 741 888 × 2 = 0 + 0.000 000 189 989 805 221 483 776;
  • 9) 0.000 000 189 989 805 221 483 776 × 2 = 0 + 0.000 000 379 979 610 442 967 552;
  • 10) 0.000 000 379 979 610 442 967 552 × 2 = 0 + 0.000 000 759 959 220 885 935 104;
  • 11) 0.000 000 759 959 220 885 935 104 × 2 = 0 + 0.000 001 519 918 441 771 870 208;
  • 12) 0.000 001 519 918 441 771 870 208 × 2 = 0 + 0.000 003 039 836 883 543 740 416;
  • 13) 0.000 003 039 836 883 543 740 416 × 2 = 0 + 0.000 006 079 673 767 087 480 832;
  • 14) 0.000 006 079 673 767 087 480 832 × 2 = 0 + 0.000 012 159 347 534 174 961 664;
  • 15) 0.000 012 159 347 534 174 961 664 × 2 = 0 + 0.000 024 318 695 068 349 923 328;
  • 16) 0.000 024 318 695 068 349 923 328 × 2 = 0 + 0.000 048 637 390 136 699 846 656;
  • 17) 0.000 048 637 390 136 699 846 656 × 2 = 0 + 0.000 097 274 780 273 399 693 312;
  • 18) 0.000 097 274 780 273 399 693 312 × 2 = 0 + 0.000 194 549 560 546 799 386 624;
  • 19) 0.000 194 549 560 546 799 386 624 × 2 = 0 + 0.000 389 099 121 093 598 773 248;
  • 20) 0.000 389 099 121 093 598 773 248 × 2 = 0 + 0.000 778 198 242 187 197 546 496;
  • 21) 0.000 778 198 242 187 197 546 496 × 2 = 0 + 0.001 556 396 484 374 395 092 992;
  • 22) 0.001 556 396 484 374 395 092 992 × 2 = 0 + 0.003 112 792 968 748 790 185 984;
  • 23) 0.003 112 792 968 748 790 185 984 × 2 = 0 + 0.006 225 585 937 497 580 371 968;
  • 24) 0.006 225 585 937 497 580 371 968 × 2 = 0 + 0.012 451 171 874 995 160 743 936;
  • 25) 0.012 451 171 874 995 160 743 936 × 2 = 0 + 0.024 902 343 749 990 321 487 872;
  • 26) 0.024 902 343 749 990 321 487 872 × 2 = 0 + 0.049 804 687 499 980 642 975 744;
  • 27) 0.049 804 687 499 980 642 975 744 × 2 = 0 + 0.099 609 374 999 961 285 951 488;
  • 28) 0.099 609 374 999 961 285 951 488 × 2 = 0 + 0.199 218 749 999 922 571 902 976;
  • 29) 0.199 218 749 999 922 571 902 976 × 2 = 0 + 0.398 437 499 999 845 143 805 952;
  • 30) 0.398 437 499 999 845 143 805 952 × 2 = 0 + 0.796 874 999 999 690 287 611 904;
  • 31) 0.796 874 999 999 690 287 611 904 × 2 = 1 + 0.593 749 999 999 380 575 223 808;
  • 32) 0.593 749 999 999 380 575 223 808 × 2 = 1 + 0.187 499 999 998 761 150 447 616;
  • 33) 0.187 499 999 998 761 150 447 616 × 2 = 0 + 0.374 999 999 997 522 300 895 232;
  • 34) 0.374 999 999 997 522 300 895 232 × 2 = 0 + 0.749 999 999 995 044 601 790 464;
  • 35) 0.749 999 999 995 044 601 790 464 × 2 = 1 + 0.499 999 999 990 089 203 580 928;
  • 36) 0.499 999 999 990 089 203 580 928 × 2 = 0 + 0.999 999 999 980 178 407 161 856;
  • 37) 0.999 999 999 980 178 407 161 856 × 2 = 1 + 0.999 999 999 960 356 814 323 712;
  • 38) 0.999 999 999 960 356 814 323 712 × 2 = 1 + 0.999 999 999 920 713 628 647 424;
  • 39) 0.999 999 999 920 713 628 647 424 × 2 = 1 + 0.999 999 999 841 427 257 294 848;
  • 40) 0.999 999 999 841 427 257 294 848 × 2 = 1 + 0.999 999 999 682 854 514 589 696;
  • 41) 0.999 999 999 682 854 514 589 696 × 2 = 1 + 0.999 999 999 365 709 029 179 392;
  • 42) 0.999 999 999 365 709 029 179 392 × 2 = 1 + 0.999 999 998 731 418 058 358 784;
  • 43) 0.999 999 998 731 418 058 358 784 × 2 = 1 + 0.999 999 997 462 836 116 717 568;
  • 44) 0.999 999 997 462 836 116 717 568 × 2 = 1 + 0.999 999 994 925 672 233 435 136;
  • 45) 0.999 999 994 925 672 233 435 136 × 2 = 1 + 0.999 999 989 851 344 466 870 272;
  • 46) 0.999 999 989 851 344 466 870 272 × 2 = 1 + 0.999 999 979 702 688 933 740 544;
  • 47) 0.999 999 979 702 688 933 740 544 × 2 = 1 + 0.999 999 959 405 377 867 481 088;
  • 48) 0.999 999 959 405 377 867 481 088 × 2 = 1 + 0.999 999 918 810 755 734 962 176;
  • 49) 0.999 999 918 810 755 734 962 176 × 2 = 1 + 0.999 999 837 621 511 469 924 352;
  • 50) 0.999 999 837 621 511 469 924 352 × 2 = 1 + 0.999 999 675 243 022 939 848 704;
  • 51) 0.999 999 675 243 022 939 848 704 × 2 = 1 + 0.999 999 350 486 045 879 697 408;
  • 52) 0.999 999 350 486 045 879 697 408 × 2 = 1 + 0.999 998 700 972 091 759 394 816;
  • 53) 0.999 998 700 972 091 759 394 816 × 2 = 1 + 0.999 997 401 944 183 518 789 632;
  • 54) 0.999 997 401 944 183 518 789 632 × 2 = 1 + 0.999 994 803 888 367 037 579 264;

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:


0.000 000 000 742 147 676 646 421(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 421(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:


0.000 000 000 742 147 676 646 421(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 -0.000 000 000 742 147 676 646 421 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