-0.000 000 000 742 147 676 554 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 554(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
-0.000 000 000 742 147 676 554(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:

|-0.000 000 000 742 147 676 554| = 0.000 000 000 742 147 676 554


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: 0.000 000 000 742 147 676 554.

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) 0.000 000 000 742 147 676 554 × 2 = 0 + 0.000 000 001 484 295 353 108;
  • 2) 0.000 000 001 484 295 353 108 × 2 = 0 + 0.000 000 002 968 590 706 216;
  • 3) 0.000 000 002 968 590 706 216 × 2 = 0 + 0.000 000 005 937 181 412 432;
  • 4) 0.000 000 005 937 181 412 432 × 2 = 0 + 0.000 000 011 874 362 824 864;
  • 5) 0.000 000 011 874 362 824 864 × 2 = 0 + 0.000 000 023 748 725 649 728;
  • 6) 0.000 000 023 748 725 649 728 × 2 = 0 + 0.000 000 047 497 451 299 456;
  • 7) 0.000 000 047 497 451 299 456 × 2 = 0 + 0.000 000 094 994 902 598 912;
  • 8) 0.000 000 094 994 902 598 912 × 2 = 0 + 0.000 000 189 989 805 197 824;
  • 9) 0.000 000 189 989 805 197 824 × 2 = 0 + 0.000 000 379 979 610 395 648;
  • 10) 0.000 000 379 979 610 395 648 × 2 = 0 + 0.000 000 759 959 220 791 296;
  • 11) 0.000 000 759 959 220 791 296 × 2 = 0 + 0.000 001 519 918 441 582 592;
  • 12) 0.000 001 519 918 441 582 592 × 2 = 0 + 0.000 003 039 836 883 165 184;
  • 13) 0.000 003 039 836 883 165 184 × 2 = 0 + 0.000 006 079 673 766 330 368;
  • 14) 0.000 006 079 673 766 330 368 × 2 = 0 + 0.000 012 159 347 532 660 736;
  • 15) 0.000 012 159 347 532 660 736 × 2 = 0 + 0.000 024 318 695 065 321 472;
  • 16) 0.000 024 318 695 065 321 472 × 2 = 0 + 0.000 048 637 390 130 642 944;
  • 17) 0.000 048 637 390 130 642 944 × 2 = 0 + 0.000 097 274 780 261 285 888;
  • 18) 0.000 097 274 780 261 285 888 × 2 = 0 + 0.000 194 549 560 522 571 776;
  • 19) 0.000 194 549 560 522 571 776 × 2 = 0 + 0.000 389 099 121 045 143 552;
  • 20) 0.000 389 099 121 045 143 552 × 2 = 0 + 0.000 778 198 242 090 287 104;
  • 21) 0.000 778 198 242 090 287 104 × 2 = 0 + 0.001 556 396 484 180 574 208;
  • 22) 0.001 556 396 484 180 574 208 × 2 = 0 + 0.003 112 792 968 361 148 416;
  • 23) 0.003 112 792 968 361 148 416 × 2 = 0 + 0.006 225 585 936 722 296 832;
  • 24) 0.006 225 585 936 722 296 832 × 2 = 0 + 0.012 451 171 873 444 593 664;
  • 25) 0.012 451 171 873 444 593 664 × 2 = 0 + 0.024 902 343 746 889 187 328;
  • 26) 0.024 902 343 746 889 187 328 × 2 = 0 + 0.049 804 687 493 778 374 656;
  • 27) 0.049 804 687 493 778 374 656 × 2 = 0 + 0.099 609 374 987 556 749 312;
  • 28) 0.099 609 374 987 556 749 312 × 2 = 0 + 0.199 218 749 975 113 498 624;
  • 29) 0.199 218 749 975 113 498 624 × 2 = 0 + 0.398 437 499 950 226 997 248;
  • 30) 0.398 437 499 950 226 997 248 × 2 = 0 + 0.796 874 999 900 453 994 496;
  • 31) 0.796 874 999 900 453 994 496 × 2 = 1 + 0.593 749 999 800 907 988 992;
  • 32) 0.593 749 999 800 907 988 992 × 2 = 1 + 0.187 499 999 601 815 977 984;
  • 33) 0.187 499 999 601 815 977 984 × 2 = 0 + 0.374 999 999 203 631 955 968;
  • 34) 0.374 999 999 203 631 955 968 × 2 = 0 + 0.749 999 998 407 263 911 936;
  • 35) 0.749 999 998 407 263 911 936 × 2 = 1 + 0.499 999 996 814 527 823 872;
  • 36) 0.499 999 996 814 527 823 872 × 2 = 0 + 0.999 999 993 629 055 647 744;
  • 37) 0.999 999 993 629 055 647 744 × 2 = 1 + 0.999 999 987 258 111 295 488;
  • 38) 0.999 999 987 258 111 295 488 × 2 = 1 + 0.999 999 974 516 222 590 976;
  • 39) 0.999 999 974 516 222 590 976 × 2 = 1 + 0.999 999 949 032 445 181 952;
  • 40) 0.999 999 949 032 445 181 952 × 2 = 1 + 0.999 999 898 064 890 363 904;
  • 41) 0.999 999 898 064 890 363 904 × 2 = 1 + 0.999 999 796 129 780 727 808;
  • 42) 0.999 999 796 129 780 727 808 × 2 = 1 + 0.999 999 592 259 561 455 616;
  • 43) 0.999 999 592 259 561 455 616 × 2 = 1 + 0.999 999 184 519 122 911 232;
  • 44) 0.999 999 184 519 122 911 232 × 2 = 1 + 0.999 998 369 038 245 822 464;
  • 45) 0.999 998 369 038 245 822 464 × 2 = 1 + 0.999 996 738 076 491 644 928;
  • 46) 0.999 996 738 076 491 644 928 × 2 = 1 + 0.999 993 476 152 983 289 856;
  • 47) 0.999 993 476 152 983 289 856 × 2 = 1 + 0.999 986 952 305 966 579 712;
  • 48) 0.999 986 952 305 966 579 712 × 2 = 1 + 0.999 973 904 611 933 159 424;
  • 49) 0.999 973 904 611 933 159 424 × 2 = 1 + 0.999 947 809 223 866 318 848;
  • 50) 0.999 947 809 223 866 318 848 × 2 = 1 + 0.999 895 618 447 732 637 696;
  • 51) 0.999 895 618 447 732 637 696 × 2 = 1 + 0.999 791 236 895 465 275 392;
  • 52) 0.999 791 236 895 465 275 392 × 2 = 1 + 0.999 582 473 790 930 550 784;
  • 53) 0.999 582 473 790 930 550 784 × 2 = 1 + 0.999 164 947 581 861 101 568;
  • 54) 0.999 164 947 581 861 101 568 × 2 = 1 + 0.998 329 895 163 722 203 136;

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:


0.000 000 000 742 147 676 554(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 554(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:


0.000 000 000 742 147 676 554(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 -0.000 000 000 742 147 676 554 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