-0.000 000 000 742 147 676 361 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 361(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
-0.000 000 000 742 147 676 361(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:

|-0.000 000 000 742 147 676 361| = 0.000 000 000 742 147 676 361


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: 0.000 000 000 742 147 676 361.

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) 0.000 000 000 742 147 676 361 × 2 = 0 + 0.000 000 001 484 295 352 722;
  • 2) 0.000 000 001 484 295 352 722 × 2 = 0 + 0.000 000 002 968 590 705 444;
  • 3) 0.000 000 002 968 590 705 444 × 2 = 0 + 0.000 000 005 937 181 410 888;
  • 4) 0.000 000 005 937 181 410 888 × 2 = 0 + 0.000 000 011 874 362 821 776;
  • 5) 0.000 000 011 874 362 821 776 × 2 = 0 + 0.000 000 023 748 725 643 552;
  • 6) 0.000 000 023 748 725 643 552 × 2 = 0 + 0.000 000 047 497 451 287 104;
  • 7) 0.000 000 047 497 451 287 104 × 2 = 0 + 0.000 000 094 994 902 574 208;
  • 8) 0.000 000 094 994 902 574 208 × 2 = 0 + 0.000 000 189 989 805 148 416;
  • 9) 0.000 000 189 989 805 148 416 × 2 = 0 + 0.000 000 379 979 610 296 832;
  • 10) 0.000 000 379 979 610 296 832 × 2 = 0 + 0.000 000 759 959 220 593 664;
  • 11) 0.000 000 759 959 220 593 664 × 2 = 0 + 0.000 001 519 918 441 187 328;
  • 12) 0.000 001 519 918 441 187 328 × 2 = 0 + 0.000 003 039 836 882 374 656;
  • 13) 0.000 003 039 836 882 374 656 × 2 = 0 + 0.000 006 079 673 764 749 312;
  • 14) 0.000 006 079 673 764 749 312 × 2 = 0 + 0.000 012 159 347 529 498 624;
  • 15) 0.000 012 159 347 529 498 624 × 2 = 0 + 0.000 024 318 695 058 997 248;
  • 16) 0.000 024 318 695 058 997 248 × 2 = 0 + 0.000 048 637 390 117 994 496;
  • 17) 0.000 048 637 390 117 994 496 × 2 = 0 + 0.000 097 274 780 235 988 992;
  • 18) 0.000 097 274 780 235 988 992 × 2 = 0 + 0.000 194 549 560 471 977 984;
  • 19) 0.000 194 549 560 471 977 984 × 2 = 0 + 0.000 389 099 120 943 955 968;
  • 20) 0.000 389 099 120 943 955 968 × 2 = 0 + 0.000 778 198 241 887 911 936;
  • 21) 0.000 778 198 241 887 911 936 × 2 = 0 + 0.001 556 396 483 775 823 872;
  • 22) 0.001 556 396 483 775 823 872 × 2 = 0 + 0.003 112 792 967 551 647 744;
  • 23) 0.003 112 792 967 551 647 744 × 2 = 0 + 0.006 225 585 935 103 295 488;
  • 24) 0.006 225 585 935 103 295 488 × 2 = 0 + 0.012 451 171 870 206 590 976;
  • 25) 0.012 451 171 870 206 590 976 × 2 = 0 + 0.024 902 343 740 413 181 952;
  • 26) 0.024 902 343 740 413 181 952 × 2 = 0 + 0.049 804 687 480 826 363 904;
  • 27) 0.049 804 687 480 826 363 904 × 2 = 0 + 0.099 609 374 961 652 727 808;
  • 28) 0.099 609 374 961 652 727 808 × 2 = 0 + 0.199 218 749 923 305 455 616;
  • 29) 0.199 218 749 923 305 455 616 × 2 = 0 + 0.398 437 499 846 610 911 232;
  • 30) 0.398 437 499 846 610 911 232 × 2 = 0 + 0.796 874 999 693 221 822 464;
  • 31) 0.796 874 999 693 221 822 464 × 2 = 1 + 0.593 749 999 386 443 644 928;
  • 32) 0.593 749 999 386 443 644 928 × 2 = 1 + 0.187 499 998 772 887 289 856;
  • 33) 0.187 499 998 772 887 289 856 × 2 = 0 + 0.374 999 997 545 774 579 712;
  • 34) 0.374 999 997 545 774 579 712 × 2 = 0 + 0.749 999 995 091 549 159 424;
  • 35) 0.749 999 995 091 549 159 424 × 2 = 1 + 0.499 999 990 183 098 318 848;
  • 36) 0.499 999 990 183 098 318 848 × 2 = 0 + 0.999 999 980 366 196 637 696;
  • 37) 0.999 999 980 366 196 637 696 × 2 = 1 + 0.999 999 960 732 393 275 392;
  • 38) 0.999 999 960 732 393 275 392 × 2 = 1 + 0.999 999 921 464 786 550 784;
  • 39) 0.999 999 921 464 786 550 784 × 2 = 1 + 0.999 999 842 929 573 101 568;
  • 40) 0.999 999 842 929 573 101 568 × 2 = 1 + 0.999 999 685 859 146 203 136;
  • 41) 0.999 999 685 859 146 203 136 × 2 = 1 + 0.999 999 371 718 292 406 272;
  • 42) 0.999 999 371 718 292 406 272 × 2 = 1 + 0.999 998 743 436 584 812 544;
  • 43) 0.999 998 743 436 584 812 544 × 2 = 1 + 0.999 997 486 873 169 625 088;
  • 44) 0.999 997 486 873 169 625 088 × 2 = 1 + 0.999 994 973 746 339 250 176;
  • 45) 0.999 994 973 746 339 250 176 × 2 = 1 + 0.999 989 947 492 678 500 352;
  • 46) 0.999 989 947 492 678 500 352 × 2 = 1 + 0.999 979 894 985 357 000 704;
  • 47) 0.999 979 894 985 357 000 704 × 2 = 1 + 0.999 959 789 970 714 001 408;
  • 48) 0.999 959 789 970 714 001 408 × 2 = 1 + 0.999 919 579 941 428 002 816;
  • 49) 0.999 919 579 941 428 002 816 × 2 = 1 + 0.999 839 159 882 856 005 632;
  • 50) 0.999 839 159 882 856 005 632 × 2 = 1 + 0.999 678 319 765 712 011 264;
  • 51) 0.999 678 319 765 712 011 264 × 2 = 1 + 0.999 356 639 531 424 022 528;
  • 52) 0.999 356 639 531 424 022 528 × 2 = 1 + 0.998 713 279 062 848 045 056;
  • 53) 0.998 713 279 062 848 045 056 × 2 = 1 + 0.997 426 558 125 696 090 112;
  • 54) 0.997 426 558 125 696 090 112 × 2 = 1 + 0.994 853 116 251 392 180 224;

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:


0.000 000 000 742 147 676 361(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 361(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:


0.000 000 000 742 147 676 361(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 -0.000 000 000 742 147 676 361 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