-0.000 000 000 742 147 483 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 483(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
-0.000 000 000 742 147 483(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:

|-0.000 000 000 742 147 483| = 0.000 000 000 742 147 483


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: 0.000 000 000 742 147 483.

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) 0.000 000 000 742 147 483 × 2 = 0 + 0.000 000 001 484 294 966;
  • 2) 0.000 000 001 484 294 966 × 2 = 0 + 0.000 000 002 968 589 932;
  • 3) 0.000 000 002 968 589 932 × 2 = 0 + 0.000 000 005 937 179 864;
  • 4) 0.000 000 005 937 179 864 × 2 = 0 + 0.000 000 011 874 359 728;
  • 5) 0.000 000 011 874 359 728 × 2 = 0 + 0.000 000 023 748 719 456;
  • 6) 0.000 000 023 748 719 456 × 2 = 0 + 0.000 000 047 497 438 912;
  • 7) 0.000 000 047 497 438 912 × 2 = 0 + 0.000 000 094 994 877 824;
  • 8) 0.000 000 094 994 877 824 × 2 = 0 + 0.000 000 189 989 755 648;
  • 9) 0.000 000 189 989 755 648 × 2 = 0 + 0.000 000 379 979 511 296;
  • 10) 0.000 000 379 979 511 296 × 2 = 0 + 0.000 000 759 959 022 592;
  • 11) 0.000 000 759 959 022 592 × 2 = 0 + 0.000 001 519 918 045 184;
  • 12) 0.000 001 519 918 045 184 × 2 = 0 + 0.000 003 039 836 090 368;
  • 13) 0.000 003 039 836 090 368 × 2 = 0 + 0.000 006 079 672 180 736;
  • 14) 0.000 006 079 672 180 736 × 2 = 0 + 0.000 012 159 344 361 472;
  • 15) 0.000 012 159 344 361 472 × 2 = 0 + 0.000 024 318 688 722 944;
  • 16) 0.000 024 318 688 722 944 × 2 = 0 + 0.000 048 637 377 445 888;
  • 17) 0.000 048 637 377 445 888 × 2 = 0 + 0.000 097 274 754 891 776;
  • 18) 0.000 097 274 754 891 776 × 2 = 0 + 0.000 194 549 509 783 552;
  • 19) 0.000 194 549 509 783 552 × 2 = 0 + 0.000 389 099 019 567 104;
  • 20) 0.000 389 099 019 567 104 × 2 = 0 + 0.000 778 198 039 134 208;
  • 21) 0.000 778 198 039 134 208 × 2 = 0 + 0.001 556 396 078 268 416;
  • 22) 0.001 556 396 078 268 416 × 2 = 0 + 0.003 112 792 156 536 832;
  • 23) 0.003 112 792 156 536 832 × 2 = 0 + 0.006 225 584 313 073 664;
  • 24) 0.006 225 584 313 073 664 × 2 = 0 + 0.012 451 168 626 147 328;
  • 25) 0.012 451 168 626 147 328 × 2 = 0 + 0.024 902 337 252 294 656;
  • 26) 0.024 902 337 252 294 656 × 2 = 0 + 0.049 804 674 504 589 312;
  • 27) 0.049 804 674 504 589 312 × 2 = 0 + 0.099 609 349 009 178 624;
  • 28) 0.099 609 349 009 178 624 × 2 = 0 + 0.199 218 698 018 357 248;
  • 29) 0.199 218 698 018 357 248 × 2 = 0 + 0.398 437 396 036 714 496;
  • 30) 0.398 437 396 036 714 496 × 2 = 0 + 0.796 874 792 073 428 992;
  • 31) 0.796 874 792 073 428 992 × 2 = 1 + 0.593 749 584 146 857 984;
  • 32) 0.593 749 584 146 857 984 × 2 = 1 + 0.187 499 168 293 715 968;
  • 33) 0.187 499 168 293 715 968 × 2 = 0 + 0.374 998 336 587 431 936;
  • 34) 0.374 998 336 587 431 936 × 2 = 0 + 0.749 996 673 174 863 872;
  • 35) 0.749 996 673 174 863 872 × 2 = 1 + 0.499 993 346 349 727 744;
  • 36) 0.499 993 346 349 727 744 × 2 = 0 + 0.999 986 692 699 455 488;
  • 37) 0.999 986 692 699 455 488 × 2 = 1 + 0.999 973 385 398 910 976;
  • 38) 0.999 973 385 398 910 976 × 2 = 1 + 0.999 946 770 797 821 952;
  • 39) 0.999 946 770 797 821 952 × 2 = 1 + 0.999 893 541 595 643 904;
  • 40) 0.999 893 541 595 643 904 × 2 = 1 + 0.999 787 083 191 287 808;
  • 41) 0.999 787 083 191 287 808 × 2 = 1 + 0.999 574 166 382 575 616;
  • 42) 0.999 574 166 382 575 616 × 2 = 1 + 0.999 148 332 765 151 232;
  • 43) 0.999 148 332 765 151 232 × 2 = 1 + 0.998 296 665 530 302 464;
  • 44) 0.998 296 665 530 302 464 × 2 = 1 + 0.996 593 331 060 604 928;
  • 45) 0.996 593 331 060 604 928 × 2 = 1 + 0.993 186 662 121 209 856;
  • 46) 0.993 186 662 121 209 856 × 2 = 1 + 0.986 373 324 242 419 712;
  • 47) 0.986 373 324 242 419 712 × 2 = 1 + 0.972 746 648 484 839 424;
  • 48) 0.972 746 648 484 839 424 × 2 = 1 + 0.945 493 296 969 678 848;
  • 49) 0.945 493 296 969 678 848 × 2 = 1 + 0.890 986 593 939 357 696;
  • 50) 0.890 986 593 939 357 696 × 2 = 1 + 0.781 973 187 878 715 392;
  • 51) 0.781 973 187 878 715 392 × 2 = 1 + 0.563 946 375 757 430 784;
  • 52) 0.563 946 375 757 430 784 × 2 = 1 + 0.127 892 751 514 861 568;
  • 53) 0.127 892 751 514 861 568 × 2 = 0 + 0.255 785 503 029 723 136;
  • 54) 0.255 785 503 029 723 136 × 2 = 0 + 0.511 571 006 059 446 272;

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:


0.000 000 000 742 147 483(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 483(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(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:


0.000 000 000 742 147 483(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 100(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 100


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 1100 =

100 1011 1111 1111 1111 1100


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 1100


Decimal number -0.000 000 000 742 147 483 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1100

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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