-0.000 000 000 742 147 676 646 680 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 680 5(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
-0.000 000 000 742 147 676 646 680 5(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:

|-0.000 000 000 742 147 676 646 680 5| = 0.000 000 000 742 147 676 646 680 5


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: 0.000 000 000 742 147 676 646 680 5.

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) 0.000 000 000 742 147 676 646 680 5 × 2 = 0 + 0.000 000 001 484 295 353 293 361;
  • 2) 0.000 000 001 484 295 353 293 361 × 2 = 0 + 0.000 000 002 968 590 706 586 722;
  • 3) 0.000 000 002 968 590 706 586 722 × 2 = 0 + 0.000 000 005 937 181 413 173 444;
  • 4) 0.000 000 005 937 181 413 173 444 × 2 = 0 + 0.000 000 011 874 362 826 346 888;
  • 5) 0.000 000 011 874 362 826 346 888 × 2 = 0 + 0.000 000 023 748 725 652 693 776;
  • 6) 0.000 000 023 748 725 652 693 776 × 2 = 0 + 0.000 000 047 497 451 305 387 552;
  • 7) 0.000 000 047 497 451 305 387 552 × 2 = 0 + 0.000 000 094 994 902 610 775 104;
  • 8) 0.000 000 094 994 902 610 775 104 × 2 = 0 + 0.000 000 189 989 805 221 550 208;
  • 9) 0.000 000 189 989 805 221 550 208 × 2 = 0 + 0.000 000 379 979 610 443 100 416;
  • 10) 0.000 000 379 979 610 443 100 416 × 2 = 0 + 0.000 000 759 959 220 886 200 832;
  • 11) 0.000 000 759 959 220 886 200 832 × 2 = 0 + 0.000 001 519 918 441 772 401 664;
  • 12) 0.000 001 519 918 441 772 401 664 × 2 = 0 + 0.000 003 039 836 883 544 803 328;
  • 13) 0.000 003 039 836 883 544 803 328 × 2 = 0 + 0.000 006 079 673 767 089 606 656;
  • 14) 0.000 006 079 673 767 089 606 656 × 2 = 0 + 0.000 012 159 347 534 179 213 312;
  • 15) 0.000 012 159 347 534 179 213 312 × 2 = 0 + 0.000 024 318 695 068 358 426 624;
  • 16) 0.000 024 318 695 068 358 426 624 × 2 = 0 + 0.000 048 637 390 136 716 853 248;
  • 17) 0.000 048 637 390 136 716 853 248 × 2 = 0 + 0.000 097 274 780 273 433 706 496;
  • 18) 0.000 097 274 780 273 433 706 496 × 2 = 0 + 0.000 194 549 560 546 867 412 992;
  • 19) 0.000 194 549 560 546 867 412 992 × 2 = 0 + 0.000 389 099 121 093 734 825 984;
  • 20) 0.000 389 099 121 093 734 825 984 × 2 = 0 + 0.000 778 198 242 187 469 651 968;
  • 21) 0.000 778 198 242 187 469 651 968 × 2 = 0 + 0.001 556 396 484 374 939 303 936;
  • 22) 0.001 556 396 484 374 939 303 936 × 2 = 0 + 0.003 112 792 968 749 878 607 872;
  • 23) 0.003 112 792 968 749 878 607 872 × 2 = 0 + 0.006 225 585 937 499 757 215 744;
  • 24) 0.006 225 585 937 499 757 215 744 × 2 = 0 + 0.012 451 171 874 999 514 431 488;
  • 25) 0.012 451 171 874 999 514 431 488 × 2 = 0 + 0.024 902 343 749 999 028 862 976;
  • 26) 0.024 902 343 749 999 028 862 976 × 2 = 0 + 0.049 804 687 499 998 057 725 952;
  • 27) 0.049 804 687 499 998 057 725 952 × 2 = 0 + 0.099 609 374 999 996 115 451 904;
  • 28) 0.099 609 374 999 996 115 451 904 × 2 = 0 + 0.199 218 749 999 992 230 903 808;
  • 29) 0.199 218 749 999 992 230 903 808 × 2 = 0 + 0.398 437 499 999 984 461 807 616;
  • 30) 0.398 437 499 999 984 461 807 616 × 2 = 0 + 0.796 874 999 999 968 923 615 232;
  • 31) 0.796 874 999 999 968 923 615 232 × 2 = 1 + 0.593 749 999 999 937 847 230 464;
  • 32) 0.593 749 999 999 937 847 230 464 × 2 = 1 + 0.187 499 999 999 875 694 460 928;
  • 33) 0.187 499 999 999 875 694 460 928 × 2 = 0 + 0.374 999 999 999 751 388 921 856;
  • 34) 0.374 999 999 999 751 388 921 856 × 2 = 0 + 0.749 999 999 999 502 777 843 712;
  • 35) 0.749 999 999 999 502 777 843 712 × 2 = 1 + 0.499 999 999 999 005 555 687 424;
  • 36) 0.499 999 999 999 005 555 687 424 × 2 = 0 + 0.999 999 999 998 011 111 374 848;
  • 37) 0.999 999 999 998 011 111 374 848 × 2 = 1 + 0.999 999 999 996 022 222 749 696;
  • 38) 0.999 999 999 996 022 222 749 696 × 2 = 1 + 0.999 999 999 992 044 445 499 392;
  • 39) 0.999 999 999 992 044 445 499 392 × 2 = 1 + 0.999 999 999 984 088 890 998 784;
  • 40) 0.999 999 999 984 088 890 998 784 × 2 = 1 + 0.999 999 999 968 177 781 997 568;
  • 41) 0.999 999 999 968 177 781 997 568 × 2 = 1 + 0.999 999 999 936 355 563 995 136;
  • 42) 0.999 999 999 936 355 563 995 136 × 2 = 1 + 0.999 999 999 872 711 127 990 272;
  • 43) 0.999 999 999 872 711 127 990 272 × 2 = 1 + 0.999 999 999 745 422 255 980 544;
  • 44) 0.999 999 999 745 422 255 980 544 × 2 = 1 + 0.999 999 999 490 844 511 961 088;
  • 45) 0.999 999 999 490 844 511 961 088 × 2 = 1 + 0.999 999 998 981 689 023 922 176;
  • 46) 0.999 999 998 981 689 023 922 176 × 2 = 1 + 0.999 999 997 963 378 047 844 352;
  • 47) 0.999 999 997 963 378 047 844 352 × 2 = 1 + 0.999 999 995 926 756 095 688 704;
  • 48) 0.999 999 995 926 756 095 688 704 × 2 = 1 + 0.999 999 991 853 512 191 377 408;
  • 49) 0.999 999 991 853 512 191 377 408 × 2 = 1 + 0.999 999 983 707 024 382 754 816;
  • 50) 0.999 999 983 707 024 382 754 816 × 2 = 1 + 0.999 999 967 414 048 765 509 632;
  • 51) 0.999 999 967 414 048 765 509 632 × 2 = 1 + 0.999 999 934 828 097 531 019 264;
  • 52) 0.999 999 934 828 097 531 019 264 × 2 = 1 + 0.999 999 869 656 195 062 038 528;
  • 53) 0.999 999 869 656 195 062 038 528 × 2 = 1 + 0.999 999 739 312 390 124 077 056;
  • 54) 0.999 999 739 312 390 124 077 056 × 2 = 1 + 0.999 999 478 624 780 248 154 112;

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:


0.000 000 000 742 147 676 646 680 5(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 680 5(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:


0.000 000 000 742 147 676 646 680 5(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 -0.000 000 000 742 147 676 646 680 5 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