-0.000 000 000 742 147 676 646 696 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 696 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
-0.000 000 000 742 147 676 646 696 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:

|-0.000 000 000 742 147 676 646 696 5| = 0.000 000 000 742 147 676 646 696 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: 0.000 000 000 742 147 676 646 696 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) 0.000 000 000 742 147 676 646 696 5 × 2 = 0 + 0.000 000 001 484 295 353 293 393;
  • 2) 0.000 000 001 484 295 353 293 393 × 2 = 0 + 0.000 000 002 968 590 706 586 786;
  • 3) 0.000 000 002 968 590 706 586 786 × 2 = 0 + 0.000 000 005 937 181 413 173 572;
  • 4) 0.000 000 005 937 181 413 173 572 × 2 = 0 + 0.000 000 011 874 362 826 347 144;
  • 5) 0.000 000 011 874 362 826 347 144 × 2 = 0 + 0.000 000 023 748 725 652 694 288;
  • 6) 0.000 000 023 748 725 652 694 288 × 2 = 0 + 0.000 000 047 497 451 305 388 576;
  • 7) 0.000 000 047 497 451 305 388 576 × 2 = 0 + 0.000 000 094 994 902 610 777 152;
  • 8) 0.000 000 094 994 902 610 777 152 × 2 = 0 + 0.000 000 189 989 805 221 554 304;
  • 9) 0.000 000 189 989 805 221 554 304 × 2 = 0 + 0.000 000 379 979 610 443 108 608;
  • 10) 0.000 000 379 979 610 443 108 608 × 2 = 0 + 0.000 000 759 959 220 886 217 216;
  • 11) 0.000 000 759 959 220 886 217 216 × 2 = 0 + 0.000 001 519 918 441 772 434 432;
  • 12) 0.000 001 519 918 441 772 434 432 × 2 = 0 + 0.000 003 039 836 883 544 868 864;
  • 13) 0.000 003 039 836 883 544 868 864 × 2 = 0 + 0.000 006 079 673 767 089 737 728;
  • 14) 0.000 006 079 673 767 089 737 728 × 2 = 0 + 0.000 012 159 347 534 179 475 456;
  • 15) 0.000 012 159 347 534 179 475 456 × 2 = 0 + 0.000 024 318 695 068 358 950 912;
  • 16) 0.000 024 318 695 068 358 950 912 × 2 = 0 + 0.000 048 637 390 136 717 901 824;
  • 17) 0.000 048 637 390 136 717 901 824 × 2 = 0 + 0.000 097 274 780 273 435 803 648;
  • 18) 0.000 097 274 780 273 435 803 648 × 2 = 0 + 0.000 194 549 560 546 871 607 296;
  • 19) 0.000 194 549 560 546 871 607 296 × 2 = 0 + 0.000 389 099 121 093 743 214 592;
  • 20) 0.000 389 099 121 093 743 214 592 × 2 = 0 + 0.000 778 198 242 187 486 429 184;
  • 21) 0.000 778 198 242 187 486 429 184 × 2 = 0 + 0.001 556 396 484 374 972 858 368;
  • 22) 0.001 556 396 484 374 972 858 368 × 2 = 0 + 0.003 112 792 968 749 945 716 736;
  • 23) 0.003 112 792 968 749 945 716 736 × 2 = 0 + 0.006 225 585 937 499 891 433 472;
  • 24) 0.006 225 585 937 499 891 433 472 × 2 = 0 + 0.012 451 171 874 999 782 866 944;
  • 25) 0.012 451 171 874 999 782 866 944 × 2 = 0 + 0.024 902 343 749 999 565 733 888;
  • 26) 0.024 902 343 749 999 565 733 888 × 2 = 0 + 0.049 804 687 499 999 131 467 776;
  • 27) 0.049 804 687 499 999 131 467 776 × 2 = 0 + 0.099 609 374 999 998 262 935 552;
  • 28) 0.099 609 374 999 998 262 935 552 × 2 = 0 + 0.199 218 749 999 996 525 871 104;
  • 29) 0.199 218 749 999 996 525 871 104 × 2 = 0 + 0.398 437 499 999 993 051 742 208;
  • 30) 0.398 437 499 999 993 051 742 208 × 2 = 0 + 0.796 874 999 999 986 103 484 416;
  • 31) 0.796 874 999 999 986 103 484 416 × 2 = 1 + 0.593 749 999 999 972 206 968 832;
  • 32) 0.593 749 999 999 972 206 968 832 × 2 = 1 + 0.187 499 999 999 944 413 937 664;
  • 33) 0.187 499 999 999 944 413 937 664 × 2 = 0 + 0.374 999 999 999 888 827 875 328;
  • 34) 0.374 999 999 999 888 827 875 328 × 2 = 0 + 0.749 999 999 999 777 655 750 656;
  • 35) 0.749 999 999 999 777 655 750 656 × 2 = 1 + 0.499 999 999 999 555 311 501 312;
  • 36) 0.499 999 999 999 555 311 501 312 × 2 = 0 + 0.999 999 999 999 110 623 002 624;
  • 37) 0.999 999 999 999 110 623 002 624 × 2 = 1 + 0.999 999 999 998 221 246 005 248;
  • 38) 0.999 999 999 998 221 246 005 248 × 2 = 1 + 0.999 999 999 996 442 492 010 496;
  • 39) 0.999 999 999 996 442 492 010 496 × 2 = 1 + 0.999 999 999 992 884 984 020 992;
  • 40) 0.999 999 999 992 884 984 020 992 × 2 = 1 + 0.999 999 999 985 769 968 041 984;
  • 41) 0.999 999 999 985 769 968 041 984 × 2 = 1 + 0.999 999 999 971 539 936 083 968;
  • 42) 0.999 999 999 971 539 936 083 968 × 2 = 1 + 0.999 999 999 943 079 872 167 936;
  • 43) 0.999 999 999 943 079 872 167 936 × 2 = 1 + 0.999 999 999 886 159 744 335 872;
  • 44) 0.999 999 999 886 159 744 335 872 × 2 = 1 + 0.999 999 999 772 319 488 671 744;
  • 45) 0.999 999 999 772 319 488 671 744 × 2 = 1 + 0.999 999 999 544 638 977 343 488;
  • 46) 0.999 999 999 544 638 977 343 488 × 2 = 1 + 0.999 999 999 089 277 954 686 976;
  • 47) 0.999 999 999 089 277 954 686 976 × 2 = 1 + 0.999 999 998 178 555 909 373 952;
  • 48) 0.999 999 998 178 555 909 373 952 × 2 = 1 + 0.999 999 996 357 111 818 747 904;
  • 49) 0.999 999 996 357 111 818 747 904 × 2 = 1 + 0.999 999 992 714 223 637 495 808;
  • 50) 0.999 999 992 714 223 637 495 808 × 2 = 1 + 0.999 999 985 428 447 274 991 616;
  • 51) 0.999 999 985 428 447 274 991 616 × 2 = 1 + 0.999 999 970 856 894 549 983 232;
  • 52) 0.999 999 970 856 894 549 983 232 × 2 = 1 + 0.999 999 941 713 789 099 966 464;
  • 53) 0.999 999 941 713 789 099 966 464 × 2 = 1 + 0.999 999 883 427 578 199 932 928;
  • 54) 0.999 999 883 427 578 199 932 928 × 2 = 1 + 0.999 999 766 855 156 399 865 856;

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:


0.000 000 000 742 147 676 646 696 5(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 696 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:


0.000 000 000 742 147 676 646 696 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 -0.000 000 000 742 147 676 646 696 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