-0.000 000 000 742 147 384 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 384(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
-0.000 000 000 742 147 384(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:

|-0.000 000 000 742 147 384| = 0.000 000 000 742 147 384


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: 0.000 000 000 742 147 384.

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) 0.000 000 000 742 147 384 × 2 = 0 + 0.000 000 001 484 294 768;
  • 2) 0.000 000 001 484 294 768 × 2 = 0 + 0.000 000 002 968 589 536;
  • 3) 0.000 000 002 968 589 536 × 2 = 0 + 0.000 000 005 937 179 072;
  • 4) 0.000 000 005 937 179 072 × 2 = 0 + 0.000 000 011 874 358 144;
  • 5) 0.000 000 011 874 358 144 × 2 = 0 + 0.000 000 023 748 716 288;
  • 6) 0.000 000 023 748 716 288 × 2 = 0 + 0.000 000 047 497 432 576;
  • 7) 0.000 000 047 497 432 576 × 2 = 0 + 0.000 000 094 994 865 152;
  • 8) 0.000 000 094 994 865 152 × 2 = 0 + 0.000 000 189 989 730 304;
  • 9) 0.000 000 189 989 730 304 × 2 = 0 + 0.000 000 379 979 460 608;
  • 10) 0.000 000 379 979 460 608 × 2 = 0 + 0.000 000 759 958 921 216;
  • 11) 0.000 000 759 958 921 216 × 2 = 0 + 0.000 001 519 917 842 432;
  • 12) 0.000 001 519 917 842 432 × 2 = 0 + 0.000 003 039 835 684 864;
  • 13) 0.000 003 039 835 684 864 × 2 = 0 + 0.000 006 079 671 369 728;
  • 14) 0.000 006 079 671 369 728 × 2 = 0 + 0.000 012 159 342 739 456;
  • 15) 0.000 012 159 342 739 456 × 2 = 0 + 0.000 024 318 685 478 912;
  • 16) 0.000 024 318 685 478 912 × 2 = 0 + 0.000 048 637 370 957 824;
  • 17) 0.000 048 637 370 957 824 × 2 = 0 + 0.000 097 274 741 915 648;
  • 18) 0.000 097 274 741 915 648 × 2 = 0 + 0.000 194 549 483 831 296;
  • 19) 0.000 194 549 483 831 296 × 2 = 0 + 0.000 389 098 967 662 592;
  • 20) 0.000 389 098 967 662 592 × 2 = 0 + 0.000 778 197 935 325 184;
  • 21) 0.000 778 197 935 325 184 × 2 = 0 + 0.001 556 395 870 650 368;
  • 22) 0.001 556 395 870 650 368 × 2 = 0 + 0.003 112 791 741 300 736;
  • 23) 0.003 112 791 741 300 736 × 2 = 0 + 0.006 225 583 482 601 472;
  • 24) 0.006 225 583 482 601 472 × 2 = 0 + 0.012 451 166 965 202 944;
  • 25) 0.012 451 166 965 202 944 × 2 = 0 + 0.024 902 333 930 405 888;
  • 26) 0.024 902 333 930 405 888 × 2 = 0 + 0.049 804 667 860 811 776;
  • 27) 0.049 804 667 860 811 776 × 2 = 0 + 0.099 609 335 721 623 552;
  • 28) 0.099 609 335 721 623 552 × 2 = 0 + 0.199 218 671 443 247 104;
  • 29) 0.199 218 671 443 247 104 × 2 = 0 + 0.398 437 342 886 494 208;
  • 30) 0.398 437 342 886 494 208 × 2 = 0 + 0.796 874 685 772 988 416;
  • 31) 0.796 874 685 772 988 416 × 2 = 1 + 0.593 749 371 545 976 832;
  • 32) 0.593 749 371 545 976 832 × 2 = 1 + 0.187 498 743 091 953 664;
  • 33) 0.187 498 743 091 953 664 × 2 = 0 + 0.374 997 486 183 907 328;
  • 34) 0.374 997 486 183 907 328 × 2 = 0 + 0.749 994 972 367 814 656;
  • 35) 0.749 994 972 367 814 656 × 2 = 1 + 0.499 989 944 735 629 312;
  • 36) 0.499 989 944 735 629 312 × 2 = 0 + 0.999 979 889 471 258 624;
  • 37) 0.999 979 889 471 258 624 × 2 = 1 + 0.999 959 778 942 517 248;
  • 38) 0.999 959 778 942 517 248 × 2 = 1 + 0.999 919 557 885 034 496;
  • 39) 0.999 919 557 885 034 496 × 2 = 1 + 0.999 839 115 770 068 992;
  • 40) 0.999 839 115 770 068 992 × 2 = 1 + 0.999 678 231 540 137 984;
  • 41) 0.999 678 231 540 137 984 × 2 = 1 + 0.999 356 463 080 275 968;
  • 42) 0.999 356 463 080 275 968 × 2 = 1 + 0.998 712 926 160 551 936;
  • 43) 0.998 712 926 160 551 936 × 2 = 1 + 0.997 425 852 321 103 872;
  • 44) 0.997 425 852 321 103 872 × 2 = 1 + 0.994 851 704 642 207 744;
  • 45) 0.994 851 704 642 207 744 × 2 = 1 + 0.989 703 409 284 415 488;
  • 46) 0.989 703 409 284 415 488 × 2 = 1 + 0.979 406 818 568 830 976;
  • 47) 0.979 406 818 568 830 976 × 2 = 1 + 0.958 813 637 137 661 952;
  • 48) 0.958 813 637 137 661 952 × 2 = 1 + 0.917 627 274 275 323 904;
  • 49) 0.917 627 274 275 323 904 × 2 = 1 + 0.835 254 548 550 647 808;
  • 50) 0.835 254 548 550 647 808 × 2 = 1 + 0.670 509 097 101 295 616;
  • 51) 0.670 509 097 101 295 616 × 2 = 1 + 0.341 018 194 202 591 232;
  • 52) 0.341 018 194 202 591 232 × 2 = 0 + 0.682 036 388 405 182 464;
  • 53) 0.682 036 388 405 182 464 × 2 = 1 + 0.364 072 776 810 364 928;
  • 54) 0.364 072 776 810 364 928 × 2 = 0 + 0.728 145 553 620 729 856;

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:


0.000 000 000 742 147 384(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 384(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(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:


0.000 000 000 742 147 384(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 10(2) × 20 =

1.1001 0111 1111 1111 1111 010(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 010


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 1010 =

100 1011 1111 1111 1111 1010


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 1010


Decimal number -0.000 000 000 742 147 384 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1010

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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