-0.000 000 000 742 147 676 646 681 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 681 4(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
-0.000 000 000 742 147 676 646 681 4(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:

|-0.000 000 000 742 147 676 646 681 4| = 0.000 000 000 742 147 676 646 681 4


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: 0.000 000 000 742 147 676 646 681 4.

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) 0.000 000 000 742 147 676 646 681 4 × 2 = 0 + 0.000 000 001 484 295 353 293 362 8;
  • 2) 0.000 000 001 484 295 353 293 362 8 × 2 = 0 + 0.000 000 002 968 590 706 586 725 6;
  • 3) 0.000 000 002 968 590 706 586 725 6 × 2 = 0 + 0.000 000 005 937 181 413 173 451 2;
  • 4) 0.000 000 005 937 181 413 173 451 2 × 2 = 0 + 0.000 000 011 874 362 826 346 902 4;
  • 5) 0.000 000 011 874 362 826 346 902 4 × 2 = 0 + 0.000 000 023 748 725 652 693 804 8;
  • 6) 0.000 000 023 748 725 652 693 804 8 × 2 = 0 + 0.000 000 047 497 451 305 387 609 6;
  • 7) 0.000 000 047 497 451 305 387 609 6 × 2 = 0 + 0.000 000 094 994 902 610 775 219 2;
  • 8) 0.000 000 094 994 902 610 775 219 2 × 2 = 0 + 0.000 000 189 989 805 221 550 438 4;
  • 9) 0.000 000 189 989 805 221 550 438 4 × 2 = 0 + 0.000 000 379 979 610 443 100 876 8;
  • 10) 0.000 000 379 979 610 443 100 876 8 × 2 = 0 + 0.000 000 759 959 220 886 201 753 6;
  • 11) 0.000 000 759 959 220 886 201 753 6 × 2 = 0 + 0.000 001 519 918 441 772 403 507 2;
  • 12) 0.000 001 519 918 441 772 403 507 2 × 2 = 0 + 0.000 003 039 836 883 544 807 014 4;
  • 13) 0.000 003 039 836 883 544 807 014 4 × 2 = 0 + 0.000 006 079 673 767 089 614 028 8;
  • 14) 0.000 006 079 673 767 089 614 028 8 × 2 = 0 + 0.000 012 159 347 534 179 228 057 6;
  • 15) 0.000 012 159 347 534 179 228 057 6 × 2 = 0 + 0.000 024 318 695 068 358 456 115 2;
  • 16) 0.000 024 318 695 068 358 456 115 2 × 2 = 0 + 0.000 048 637 390 136 716 912 230 4;
  • 17) 0.000 048 637 390 136 716 912 230 4 × 2 = 0 + 0.000 097 274 780 273 433 824 460 8;
  • 18) 0.000 097 274 780 273 433 824 460 8 × 2 = 0 + 0.000 194 549 560 546 867 648 921 6;
  • 19) 0.000 194 549 560 546 867 648 921 6 × 2 = 0 + 0.000 389 099 121 093 735 297 843 2;
  • 20) 0.000 389 099 121 093 735 297 843 2 × 2 = 0 + 0.000 778 198 242 187 470 595 686 4;
  • 21) 0.000 778 198 242 187 470 595 686 4 × 2 = 0 + 0.001 556 396 484 374 941 191 372 8;
  • 22) 0.001 556 396 484 374 941 191 372 8 × 2 = 0 + 0.003 112 792 968 749 882 382 745 6;
  • 23) 0.003 112 792 968 749 882 382 745 6 × 2 = 0 + 0.006 225 585 937 499 764 765 491 2;
  • 24) 0.006 225 585 937 499 764 765 491 2 × 2 = 0 + 0.012 451 171 874 999 529 530 982 4;
  • 25) 0.012 451 171 874 999 529 530 982 4 × 2 = 0 + 0.024 902 343 749 999 059 061 964 8;
  • 26) 0.024 902 343 749 999 059 061 964 8 × 2 = 0 + 0.049 804 687 499 998 118 123 929 6;
  • 27) 0.049 804 687 499 998 118 123 929 6 × 2 = 0 + 0.099 609 374 999 996 236 247 859 2;
  • 28) 0.099 609 374 999 996 236 247 859 2 × 2 = 0 + 0.199 218 749 999 992 472 495 718 4;
  • 29) 0.199 218 749 999 992 472 495 718 4 × 2 = 0 + 0.398 437 499 999 984 944 991 436 8;
  • 30) 0.398 437 499 999 984 944 991 436 8 × 2 = 0 + 0.796 874 999 999 969 889 982 873 6;
  • 31) 0.796 874 999 999 969 889 982 873 6 × 2 = 1 + 0.593 749 999 999 939 779 965 747 2;
  • 32) 0.593 749 999 999 939 779 965 747 2 × 2 = 1 + 0.187 499 999 999 879 559 931 494 4;
  • 33) 0.187 499 999 999 879 559 931 494 4 × 2 = 0 + 0.374 999 999 999 759 119 862 988 8;
  • 34) 0.374 999 999 999 759 119 862 988 8 × 2 = 0 + 0.749 999 999 999 518 239 725 977 6;
  • 35) 0.749 999 999 999 518 239 725 977 6 × 2 = 1 + 0.499 999 999 999 036 479 451 955 2;
  • 36) 0.499 999 999 999 036 479 451 955 2 × 2 = 0 + 0.999 999 999 998 072 958 903 910 4;
  • 37) 0.999 999 999 998 072 958 903 910 4 × 2 = 1 + 0.999 999 999 996 145 917 807 820 8;
  • 38) 0.999 999 999 996 145 917 807 820 8 × 2 = 1 + 0.999 999 999 992 291 835 615 641 6;
  • 39) 0.999 999 999 992 291 835 615 641 6 × 2 = 1 + 0.999 999 999 984 583 671 231 283 2;
  • 40) 0.999 999 999 984 583 671 231 283 2 × 2 = 1 + 0.999 999 999 969 167 342 462 566 4;
  • 41) 0.999 999 999 969 167 342 462 566 4 × 2 = 1 + 0.999 999 999 938 334 684 925 132 8;
  • 42) 0.999 999 999 938 334 684 925 132 8 × 2 = 1 + 0.999 999 999 876 669 369 850 265 6;
  • 43) 0.999 999 999 876 669 369 850 265 6 × 2 = 1 + 0.999 999 999 753 338 739 700 531 2;
  • 44) 0.999 999 999 753 338 739 700 531 2 × 2 = 1 + 0.999 999 999 506 677 479 401 062 4;
  • 45) 0.999 999 999 506 677 479 401 062 4 × 2 = 1 + 0.999 999 999 013 354 958 802 124 8;
  • 46) 0.999 999 999 013 354 958 802 124 8 × 2 = 1 + 0.999 999 998 026 709 917 604 249 6;
  • 47) 0.999 999 998 026 709 917 604 249 6 × 2 = 1 + 0.999 999 996 053 419 835 208 499 2;
  • 48) 0.999 999 996 053 419 835 208 499 2 × 2 = 1 + 0.999 999 992 106 839 670 416 998 4;
  • 49) 0.999 999 992 106 839 670 416 998 4 × 2 = 1 + 0.999 999 984 213 679 340 833 996 8;
  • 50) 0.999 999 984 213 679 340 833 996 8 × 2 = 1 + 0.999 999 968 427 358 681 667 993 6;
  • 51) 0.999 999 968 427 358 681 667 993 6 × 2 = 1 + 0.999 999 936 854 717 363 335 987 2;
  • 52) 0.999 999 936 854 717 363 335 987 2 × 2 = 1 + 0.999 999 873 709 434 726 671 974 4;
  • 53) 0.999 999 873 709 434 726 671 974 4 × 2 = 1 + 0.999 999 747 418 869 453 343 948 8;
  • 54) 0.999 999 747 418 869 453 343 948 8 × 2 = 1 + 0.999 999 494 837 738 906 687 897 6;

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:


0.000 000 000 742 147 676 646 681 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 681 4(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:


0.000 000 000 742 147 676 646 681 4(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 -0.000 000 000 742 147 676 646 681 4 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