-0.000 000 000 742 147 676 646 706 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 706 3(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
-0.000 000 000 742 147 676 646 706 3(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:

|-0.000 000 000 742 147 676 646 706 3| = 0.000 000 000 742 147 676 646 706 3


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: 0.000 000 000 742 147 676 646 706 3.

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) 0.000 000 000 742 147 676 646 706 3 × 2 = 0 + 0.000 000 001 484 295 353 293 412 6;
  • 2) 0.000 000 001 484 295 353 293 412 6 × 2 = 0 + 0.000 000 002 968 590 706 586 825 2;
  • 3) 0.000 000 002 968 590 706 586 825 2 × 2 = 0 + 0.000 000 005 937 181 413 173 650 4;
  • 4) 0.000 000 005 937 181 413 173 650 4 × 2 = 0 + 0.000 000 011 874 362 826 347 300 8;
  • 5) 0.000 000 011 874 362 826 347 300 8 × 2 = 0 + 0.000 000 023 748 725 652 694 601 6;
  • 6) 0.000 000 023 748 725 652 694 601 6 × 2 = 0 + 0.000 000 047 497 451 305 389 203 2;
  • 7) 0.000 000 047 497 451 305 389 203 2 × 2 = 0 + 0.000 000 094 994 902 610 778 406 4;
  • 8) 0.000 000 094 994 902 610 778 406 4 × 2 = 0 + 0.000 000 189 989 805 221 556 812 8;
  • 9) 0.000 000 189 989 805 221 556 812 8 × 2 = 0 + 0.000 000 379 979 610 443 113 625 6;
  • 10) 0.000 000 379 979 610 443 113 625 6 × 2 = 0 + 0.000 000 759 959 220 886 227 251 2;
  • 11) 0.000 000 759 959 220 886 227 251 2 × 2 = 0 + 0.000 001 519 918 441 772 454 502 4;
  • 12) 0.000 001 519 918 441 772 454 502 4 × 2 = 0 + 0.000 003 039 836 883 544 909 004 8;
  • 13) 0.000 003 039 836 883 544 909 004 8 × 2 = 0 + 0.000 006 079 673 767 089 818 009 6;
  • 14) 0.000 006 079 673 767 089 818 009 6 × 2 = 0 + 0.000 012 159 347 534 179 636 019 2;
  • 15) 0.000 012 159 347 534 179 636 019 2 × 2 = 0 + 0.000 024 318 695 068 359 272 038 4;
  • 16) 0.000 024 318 695 068 359 272 038 4 × 2 = 0 + 0.000 048 637 390 136 718 544 076 8;
  • 17) 0.000 048 637 390 136 718 544 076 8 × 2 = 0 + 0.000 097 274 780 273 437 088 153 6;
  • 18) 0.000 097 274 780 273 437 088 153 6 × 2 = 0 + 0.000 194 549 560 546 874 176 307 2;
  • 19) 0.000 194 549 560 546 874 176 307 2 × 2 = 0 + 0.000 389 099 121 093 748 352 614 4;
  • 20) 0.000 389 099 121 093 748 352 614 4 × 2 = 0 + 0.000 778 198 242 187 496 705 228 8;
  • 21) 0.000 778 198 242 187 496 705 228 8 × 2 = 0 + 0.001 556 396 484 374 993 410 457 6;
  • 22) 0.001 556 396 484 374 993 410 457 6 × 2 = 0 + 0.003 112 792 968 749 986 820 915 2;
  • 23) 0.003 112 792 968 749 986 820 915 2 × 2 = 0 + 0.006 225 585 937 499 973 641 830 4;
  • 24) 0.006 225 585 937 499 973 641 830 4 × 2 = 0 + 0.012 451 171 874 999 947 283 660 8;
  • 25) 0.012 451 171 874 999 947 283 660 8 × 2 = 0 + 0.024 902 343 749 999 894 567 321 6;
  • 26) 0.024 902 343 749 999 894 567 321 6 × 2 = 0 + 0.049 804 687 499 999 789 134 643 2;
  • 27) 0.049 804 687 499 999 789 134 643 2 × 2 = 0 + 0.099 609 374 999 999 578 269 286 4;
  • 28) 0.099 609 374 999 999 578 269 286 4 × 2 = 0 + 0.199 218 749 999 999 156 538 572 8;
  • 29) 0.199 218 749 999 999 156 538 572 8 × 2 = 0 + 0.398 437 499 999 998 313 077 145 6;
  • 30) 0.398 437 499 999 998 313 077 145 6 × 2 = 0 + 0.796 874 999 999 996 626 154 291 2;
  • 31) 0.796 874 999 999 996 626 154 291 2 × 2 = 1 + 0.593 749 999 999 993 252 308 582 4;
  • 32) 0.593 749 999 999 993 252 308 582 4 × 2 = 1 + 0.187 499 999 999 986 504 617 164 8;
  • 33) 0.187 499 999 999 986 504 617 164 8 × 2 = 0 + 0.374 999 999 999 973 009 234 329 6;
  • 34) 0.374 999 999 999 973 009 234 329 6 × 2 = 0 + 0.749 999 999 999 946 018 468 659 2;
  • 35) 0.749 999 999 999 946 018 468 659 2 × 2 = 1 + 0.499 999 999 999 892 036 937 318 4;
  • 36) 0.499 999 999 999 892 036 937 318 4 × 2 = 0 + 0.999 999 999 999 784 073 874 636 8;
  • 37) 0.999 999 999 999 784 073 874 636 8 × 2 = 1 + 0.999 999 999 999 568 147 749 273 6;
  • 38) 0.999 999 999 999 568 147 749 273 6 × 2 = 1 + 0.999 999 999 999 136 295 498 547 2;
  • 39) 0.999 999 999 999 136 295 498 547 2 × 2 = 1 + 0.999 999 999 998 272 590 997 094 4;
  • 40) 0.999 999 999 998 272 590 997 094 4 × 2 = 1 + 0.999 999 999 996 545 181 994 188 8;
  • 41) 0.999 999 999 996 545 181 994 188 8 × 2 = 1 + 0.999 999 999 993 090 363 988 377 6;
  • 42) 0.999 999 999 993 090 363 988 377 6 × 2 = 1 + 0.999 999 999 986 180 727 976 755 2;
  • 43) 0.999 999 999 986 180 727 976 755 2 × 2 = 1 + 0.999 999 999 972 361 455 953 510 4;
  • 44) 0.999 999 999 972 361 455 953 510 4 × 2 = 1 + 0.999 999 999 944 722 911 907 020 8;
  • 45) 0.999 999 999 944 722 911 907 020 8 × 2 = 1 + 0.999 999 999 889 445 823 814 041 6;
  • 46) 0.999 999 999 889 445 823 814 041 6 × 2 = 1 + 0.999 999 999 778 891 647 628 083 2;
  • 47) 0.999 999 999 778 891 647 628 083 2 × 2 = 1 + 0.999 999 999 557 783 295 256 166 4;
  • 48) 0.999 999 999 557 783 295 256 166 4 × 2 = 1 + 0.999 999 999 115 566 590 512 332 8;
  • 49) 0.999 999 999 115 566 590 512 332 8 × 2 = 1 + 0.999 999 998 231 133 181 024 665 6;
  • 50) 0.999 999 998 231 133 181 024 665 6 × 2 = 1 + 0.999 999 996 462 266 362 049 331 2;
  • 51) 0.999 999 996 462 266 362 049 331 2 × 2 = 1 + 0.999 999 992 924 532 724 098 662 4;
  • 52) 0.999 999 992 924 532 724 098 662 4 × 2 = 1 + 0.999 999 985 849 065 448 197 324 8;
  • 53) 0.999 999 985 849 065 448 197 324 8 × 2 = 1 + 0.999 999 971 698 130 896 394 649 6;
  • 54) 0.999 999 971 698 130 896 394 649 6 × 2 = 1 + 0.999 999 943 396 261 792 789 299 2;

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:


0.000 000 000 742 147 676 646 706 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 706 3(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:


0.000 000 000 742 147 676 646 706 3(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 -0.000 000 000 742 147 676 646 706 3 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