-0.000 000 000 742 147 676 635 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 635 7(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
-0.000 000 000 742 147 676 635 7(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:

|-0.000 000 000 742 147 676 635 7| = 0.000 000 000 742 147 676 635 7


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: 0.000 000 000 742 147 676 635 7.

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) 0.000 000 000 742 147 676 635 7 × 2 = 0 + 0.000 000 001 484 295 353 271 4;
  • 2) 0.000 000 001 484 295 353 271 4 × 2 = 0 + 0.000 000 002 968 590 706 542 8;
  • 3) 0.000 000 002 968 590 706 542 8 × 2 = 0 + 0.000 000 005 937 181 413 085 6;
  • 4) 0.000 000 005 937 181 413 085 6 × 2 = 0 + 0.000 000 011 874 362 826 171 2;
  • 5) 0.000 000 011 874 362 826 171 2 × 2 = 0 + 0.000 000 023 748 725 652 342 4;
  • 6) 0.000 000 023 748 725 652 342 4 × 2 = 0 + 0.000 000 047 497 451 304 684 8;
  • 7) 0.000 000 047 497 451 304 684 8 × 2 = 0 + 0.000 000 094 994 902 609 369 6;
  • 8) 0.000 000 094 994 902 609 369 6 × 2 = 0 + 0.000 000 189 989 805 218 739 2;
  • 9) 0.000 000 189 989 805 218 739 2 × 2 = 0 + 0.000 000 379 979 610 437 478 4;
  • 10) 0.000 000 379 979 610 437 478 4 × 2 = 0 + 0.000 000 759 959 220 874 956 8;
  • 11) 0.000 000 759 959 220 874 956 8 × 2 = 0 + 0.000 001 519 918 441 749 913 6;
  • 12) 0.000 001 519 918 441 749 913 6 × 2 = 0 + 0.000 003 039 836 883 499 827 2;
  • 13) 0.000 003 039 836 883 499 827 2 × 2 = 0 + 0.000 006 079 673 766 999 654 4;
  • 14) 0.000 006 079 673 766 999 654 4 × 2 = 0 + 0.000 012 159 347 533 999 308 8;
  • 15) 0.000 012 159 347 533 999 308 8 × 2 = 0 + 0.000 024 318 695 067 998 617 6;
  • 16) 0.000 024 318 695 067 998 617 6 × 2 = 0 + 0.000 048 637 390 135 997 235 2;
  • 17) 0.000 048 637 390 135 997 235 2 × 2 = 0 + 0.000 097 274 780 271 994 470 4;
  • 18) 0.000 097 274 780 271 994 470 4 × 2 = 0 + 0.000 194 549 560 543 988 940 8;
  • 19) 0.000 194 549 560 543 988 940 8 × 2 = 0 + 0.000 389 099 121 087 977 881 6;
  • 20) 0.000 389 099 121 087 977 881 6 × 2 = 0 + 0.000 778 198 242 175 955 763 2;
  • 21) 0.000 778 198 242 175 955 763 2 × 2 = 0 + 0.001 556 396 484 351 911 526 4;
  • 22) 0.001 556 396 484 351 911 526 4 × 2 = 0 + 0.003 112 792 968 703 823 052 8;
  • 23) 0.003 112 792 968 703 823 052 8 × 2 = 0 + 0.006 225 585 937 407 646 105 6;
  • 24) 0.006 225 585 937 407 646 105 6 × 2 = 0 + 0.012 451 171 874 815 292 211 2;
  • 25) 0.012 451 171 874 815 292 211 2 × 2 = 0 + 0.024 902 343 749 630 584 422 4;
  • 26) 0.024 902 343 749 630 584 422 4 × 2 = 0 + 0.049 804 687 499 261 168 844 8;
  • 27) 0.049 804 687 499 261 168 844 8 × 2 = 0 + 0.099 609 374 998 522 337 689 6;
  • 28) 0.099 609 374 998 522 337 689 6 × 2 = 0 + 0.199 218 749 997 044 675 379 2;
  • 29) 0.199 218 749 997 044 675 379 2 × 2 = 0 + 0.398 437 499 994 089 350 758 4;
  • 30) 0.398 437 499 994 089 350 758 4 × 2 = 0 + 0.796 874 999 988 178 701 516 8;
  • 31) 0.796 874 999 988 178 701 516 8 × 2 = 1 + 0.593 749 999 976 357 403 033 6;
  • 32) 0.593 749 999 976 357 403 033 6 × 2 = 1 + 0.187 499 999 952 714 806 067 2;
  • 33) 0.187 499 999 952 714 806 067 2 × 2 = 0 + 0.374 999 999 905 429 612 134 4;
  • 34) 0.374 999 999 905 429 612 134 4 × 2 = 0 + 0.749 999 999 810 859 224 268 8;
  • 35) 0.749 999 999 810 859 224 268 8 × 2 = 1 + 0.499 999 999 621 718 448 537 6;
  • 36) 0.499 999 999 621 718 448 537 6 × 2 = 0 + 0.999 999 999 243 436 897 075 2;
  • 37) 0.999 999 999 243 436 897 075 2 × 2 = 1 + 0.999 999 998 486 873 794 150 4;
  • 38) 0.999 999 998 486 873 794 150 4 × 2 = 1 + 0.999 999 996 973 747 588 300 8;
  • 39) 0.999 999 996 973 747 588 300 8 × 2 = 1 + 0.999 999 993 947 495 176 601 6;
  • 40) 0.999 999 993 947 495 176 601 6 × 2 = 1 + 0.999 999 987 894 990 353 203 2;
  • 41) 0.999 999 987 894 990 353 203 2 × 2 = 1 + 0.999 999 975 789 980 706 406 4;
  • 42) 0.999 999 975 789 980 706 406 4 × 2 = 1 + 0.999 999 951 579 961 412 812 8;
  • 43) 0.999 999 951 579 961 412 812 8 × 2 = 1 + 0.999 999 903 159 922 825 625 6;
  • 44) 0.999 999 903 159 922 825 625 6 × 2 = 1 + 0.999 999 806 319 845 651 251 2;
  • 45) 0.999 999 806 319 845 651 251 2 × 2 = 1 + 0.999 999 612 639 691 302 502 4;
  • 46) 0.999 999 612 639 691 302 502 4 × 2 = 1 + 0.999 999 225 279 382 605 004 8;
  • 47) 0.999 999 225 279 382 605 004 8 × 2 = 1 + 0.999 998 450 558 765 210 009 6;
  • 48) 0.999 998 450 558 765 210 009 6 × 2 = 1 + 0.999 996 901 117 530 420 019 2;
  • 49) 0.999 996 901 117 530 420 019 2 × 2 = 1 + 0.999 993 802 235 060 840 038 4;
  • 50) 0.999 993 802 235 060 840 038 4 × 2 = 1 + 0.999 987 604 470 121 680 076 8;
  • 51) 0.999 987 604 470 121 680 076 8 × 2 = 1 + 0.999 975 208 940 243 360 153 6;
  • 52) 0.999 975 208 940 243 360 153 6 × 2 = 1 + 0.999 950 417 880 486 720 307 2;
  • 53) 0.999 950 417 880 486 720 307 2 × 2 = 1 + 0.999 900 835 760 973 440 614 4;
  • 54) 0.999 900 835 760 973 440 614 4 × 2 = 1 + 0.999 801 671 521 946 881 228 8;

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:


0.000 000 000 742 147 676 635 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 635 7(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:


0.000 000 000 742 147 676 635 7(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 -0.000 000 000 742 147 676 635 7 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