-0.000 000 000 742 147 423 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 423(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
-0.000 000 000 742 147 423(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:

|-0.000 000 000 742 147 423| = 0.000 000 000 742 147 423


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: 0.000 000 000 742 147 423.

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) 0.000 000 000 742 147 423 × 2 = 0 + 0.000 000 001 484 294 846;
  • 2) 0.000 000 001 484 294 846 × 2 = 0 + 0.000 000 002 968 589 692;
  • 3) 0.000 000 002 968 589 692 × 2 = 0 + 0.000 000 005 937 179 384;
  • 4) 0.000 000 005 937 179 384 × 2 = 0 + 0.000 000 011 874 358 768;
  • 5) 0.000 000 011 874 358 768 × 2 = 0 + 0.000 000 023 748 717 536;
  • 6) 0.000 000 023 748 717 536 × 2 = 0 + 0.000 000 047 497 435 072;
  • 7) 0.000 000 047 497 435 072 × 2 = 0 + 0.000 000 094 994 870 144;
  • 8) 0.000 000 094 994 870 144 × 2 = 0 + 0.000 000 189 989 740 288;
  • 9) 0.000 000 189 989 740 288 × 2 = 0 + 0.000 000 379 979 480 576;
  • 10) 0.000 000 379 979 480 576 × 2 = 0 + 0.000 000 759 958 961 152;
  • 11) 0.000 000 759 958 961 152 × 2 = 0 + 0.000 001 519 917 922 304;
  • 12) 0.000 001 519 917 922 304 × 2 = 0 + 0.000 003 039 835 844 608;
  • 13) 0.000 003 039 835 844 608 × 2 = 0 + 0.000 006 079 671 689 216;
  • 14) 0.000 006 079 671 689 216 × 2 = 0 + 0.000 012 159 343 378 432;
  • 15) 0.000 012 159 343 378 432 × 2 = 0 + 0.000 024 318 686 756 864;
  • 16) 0.000 024 318 686 756 864 × 2 = 0 + 0.000 048 637 373 513 728;
  • 17) 0.000 048 637 373 513 728 × 2 = 0 + 0.000 097 274 747 027 456;
  • 18) 0.000 097 274 747 027 456 × 2 = 0 + 0.000 194 549 494 054 912;
  • 19) 0.000 194 549 494 054 912 × 2 = 0 + 0.000 389 098 988 109 824;
  • 20) 0.000 389 098 988 109 824 × 2 = 0 + 0.000 778 197 976 219 648;
  • 21) 0.000 778 197 976 219 648 × 2 = 0 + 0.001 556 395 952 439 296;
  • 22) 0.001 556 395 952 439 296 × 2 = 0 + 0.003 112 791 904 878 592;
  • 23) 0.003 112 791 904 878 592 × 2 = 0 + 0.006 225 583 809 757 184;
  • 24) 0.006 225 583 809 757 184 × 2 = 0 + 0.012 451 167 619 514 368;
  • 25) 0.012 451 167 619 514 368 × 2 = 0 + 0.024 902 335 239 028 736;
  • 26) 0.024 902 335 239 028 736 × 2 = 0 + 0.049 804 670 478 057 472;
  • 27) 0.049 804 670 478 057 472 × 2 = 0 + 0.099 609 340 956 114 944;
  • 28) 0.099 609 340 956 114 944 × 2 = 0 + 0.199 218 681 912 229 888;
  • 29) 0.199 218 681 912 229 888 × 2 = 0 + 0.398 437 363 824 459 776;
  • 30) 0.398 437 363 824 459 776 × 2 = 0 + 0.796 874 727 648 919 552;
  • 31) 0.796 874 727 648 919 552 × 2 = 1 + 0.593 749 455 297 839 104;
  • 32) 0.593 749 455 297 839 104 × 2 = 1 + 0.187 498 910 595 678 208;
  • 33) 0.187 498 910 595 678 208 × 2 = 0 + 0.374 997 821 191 356 416;
  • 34) 0.374 997 821 191 356 416 × 2 = 0 + 0.749 995 642 382 712 832;
  • 35) 0.749 995 642 382 712 832 × 2 = 1 + 0.499 991 284 765 425 664;
  • 36) 0.499 991 284 765 425 664 × 2 = 0 + 0.999 982 569 530 851 328;
  • 37) 0.999 982 569 530 851 328 × 2 = 1 + 0.999 965 139 061 702 656;
  • 38) 0.999 965 139 061 702 656 × 2 = 1 + 0.999 930 278 123 405 312;
  • 39) 0.999 930 278 123 405 312 × 2 = 1 + 0.999 860 556 246 810 624;
  • 40) 0.999 860 556 246 810 624 × 2 = 1 + 0.999 721 112 493 621 248;
  • 41) 0.999 721 112 493 621 248 × 2 = 1 + 0.999 442 224 987 242 496;
  • 42) 0.999 442 224 987 242 496 × 2 = 1 + 0.998 884 449 974 484 992;
  • 43) 0.998 884 449 974 484 992 × 2 = 1 + 0.997 768 899 948 969 984;
  • 44) 0.997 768 899 948 969 984 × 2 = 1 + 0.995 537 799 897 939 968;
  • 45) 0.995 537 799 897 939 968 × 2 = 1 + 0.991 075 599 795 879 936;
  • 46) 0.991 075 599 795 879 936 × 2 = 1 + 0.982 151 199 591 759 872;
  • 47) 0.982 151 199 591 759 872 × 2 = 1 + 0.964 302 399 183 519 744;
  • 48) 0.964 302 399 183 519 744 × 2 = 1 + 0.928 604 798 367 039 488;
  • 49) 0.928 604 798 367 039 488 × 2 = 1 + 0.857 209 596 734 078 976;
  • 50) 0.857 209 596 734 078 976 × 2 = 1 + 0.714 419 193 468 157 952;
  • 51) 0.714 419 193 468 157 952 × 2 = 1 + 0.428 838 386 936 315 904;
  • 52) 0.428 838 386 936 315 904 × 2 = 0 + 0.857 676 773 872 631 808;
  • 53) 0.857 676 773 872 631 808 × 2 = 1 + 0.715 353 547 745 263 616;
  • 54) 0.715 353 547 745 263 616 × 2 = 1 + 0.430 707 095 490 527 232;

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:


0.000 000 000 742 147 423(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 423(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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:


0.000 000 000 742 147 423(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) × 20 =

1.1001 0111 1111 1111 1111 011(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 011


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 1011 =

100 1011 1111 1111 1111 1011


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 1011


Decimal number -0.000 000 000 742 147 423 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1011

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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