-0.000 000 000 742 147 676 646 703 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 703 2(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
-0.000 000 000 742 147 676 646 703 2(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:

|-0.000 000 000 742 147 676 646 703 2| = 0.000 000 000 742 147 676 646 703 2


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: 0.000 000 000 742 147 676 646 703 2.

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) 0.000 000 000 742 147 676 646 703 2 × 2 = 0 + 0.000 000 001 484 295 353 293 406 4;
  • 2) 0.000 000 001 484 295 353 293 406 4 × 2 = 0 + 0.000 000 002 968 590 706 586 812 8;
  • 3) 0.000 000 002 968 590 706 586 812 8 × 2 = 0 + 0.000 000 005 937 181 413 173 625 6;
  • 4) 0.000 000 005 937 181 413 173 625 6 × 2 = 0 + 0.000 000 011 874 362 826 347 251 2;
  • 5) 0.000 000 011 874 362 826 347 251 2 × 2 = 0 + 0.000 000 023 748 725 652 694 502 4;
  • 6) 0.000 000 023 748 725 652 694 502 4 × 2 = 0 + 0.000 000 047 497 451 305 389 004 8;
  • 7) 0.000 000 047 497 451 305 389 004 8 × 2 = 0 + 0.000 000 094 994 902 610 778 009 6;
  • 8) 0.000 000 094 994 902 610 778 009 6 × 2 = 0 + 0.000 000 189 989 805 221 556 019 2;
  • 9) 0.000 000 189 989 805 221 556 019 2 × 2 = 0 + 0.000 000 379 979 610 443 112 038 4;
  • 10) 0.000 000 379 979 610 443 112 038 4 × 2 = 0 + 0.000 000 759 959 220 886 224 076 8;
  • 11) 0.000 000 759 959 220 886 224 076 8 × 2 = 0 + 0.000 001 519 918 441 772 448 153 6;
  • 12) 0.000 001 519 918 441 772 448 153 6 × 2 = 0 + 0.000 003 039 836 883 544 896 307 2;
  • 13) 0.000 003 039 836 883 544 896 307 2 × 2 = 0 + 0.000 006 079 673 767 089 792 614 4;
  • 14) 0.000 006 079 673 767 089 792 614 4 × 2 = 0 + 0.000 012 159 347 534 179 585 228 8;
  • 15) 0.000 012 159 347 534 179 585 228 8 × 2 = 0 + 0.000 024 318 695 068 359 170 457 6;
  • 16) 0.000 024 318 695 068 359 170 457 6 × 2 = 0 + 0.000 048 637 390 136 718 340 915 2;
  • 17) 0.000 048 637 390 136 718 340 915 2 × 2 = 0 + 0.000 097 274 780 273 436 681 830 4;
  • 18) 0.000 097 274 780 273 436 681 830 4 × 2 = 0 + 0.000 194 549 560 546 873 363 660 8;
  • 19) 0.000 194 549 560 546 873 363 660 8 × 2 = 0 + 0.000 389 099 121 093 746 727 321 6;
  • 20) 0.000 389 099 121 093 746 727 321 6 × 2 = 0 + 0.000 778 198 242 187 493 454 643 2;
  • 21) 0.000 778 198 242 187 493 454 643 2 × 2 = 0 + 0.001 556 396 484 374 986 909 286 4;
  • 22) 0.001 556 396 484 374 986 909 286 4 × 2 = 0 + 0.003 112 792 968 749 973 818 572 8;
  • 23) 0.003 112 792 968 749 973 818 572 8 × 2 = 0 + 0.006 225 585 937 499 947 637 145 6;
  • 24) 0.006 225 585 937 499 947 637 145 6 × 2 = 0 + 0.012 451 171 874 999 895 274 291 2;
  • 25) 0.012 451 171 874 999 895 274 291 2 × 2 = 0 + 0.024 902 343 749 999 790 548 582 4;
  • 26) 0.024 902 343 749 999 790 548 582 4 × 2 = 0 + 0.049 804 687 499 999 581 097 164 8;
  • 27) 0.049 804 687 499 999 581 097 164 8 × 2 = 0 + 0.099 609 374 999 999 162 194 329 6;
  • 28) 0.099 609 374 999 999 162 194 329 6 × 2 = 0 + 0.199 218 749 999 998 324 388 659 2;
  • 29) 0.199 218 749 999 998 324 388 659 2 × 2 = 0 + 0.398 437 499 999 996 648 777 318 4;
  • 30) 0.398 437 499 999 996 648 777 318 4 × 2 = 0 + 0.796 874 999 999 993 297 554 636 8;
  • 31) 0.796 874 999 999 993 297 554 636 8 × 2 = 1 + 0.593 749 999 999 986 595 109 273 6;
  • 32) 0.593 749 999 999 986 595 109 273 6 × 2 = 1 + 0.187 499 999 999 973 190 218 547 2;
  • 33) 0.187 499 999 999 973 190 218 547 2 × 2 = 0 + 0.374 999 999 999 946 380 437 094 4;
  • 34) 0.374 999 999 999 946 380 437 094 4 × 2 = 0 + 0.749 999 999 999 892 760 874 188 8;
  • 35) 0.749 999 999 999 892 760 874 188 8 × 2 = 1 + 0.499 999 999 999 785 521 748 377 6;
  • 36) 0.499 999 999 999 785 521 748 377 6 × 2 = 0 + 0.999 999 999 999 571 043 496 755 2;
  • 37) 0.999 999 999 999 571 043 496 755 2 × 2 = 1 + 0.999 999 999 999 142 086 993 510 4;
  • 38) 0.999 999 999 999 142 086 993 510 4 × 2 = 1 + 0.999 999 999 998 284 173 987 020 8;
  • 39) 0.999 999 999 998 284 173 987 020 8 × 2 = 1 + 0.999 999 999 996 568 347 974 041 6;
  • 40) 0.999 999 999 996 568 347 974 041 6 × 2 = 1 + 0.999 999 999 993 136 695 948 083 2;
  • 41) 0.999 999 999 993 136 695 948 083 2 × 2 = 1 + 0.999 999 999 986 273 391 896 166 4;
  • 42) 0.999 999 999 986 273 391 896 166 4 × 2 = 1 + 0.999 999 999 972 546 783 792 332 8;
  • 43) 0.999 999 999 972 546 783 792 332 8 × 2 = 1 + 0.999 999 999 945 093 567 584 665 6;
  • 44) 0.999 999 999 945 093 567 584 665 6 × 2 = 1 + 0.999 999 999 890 187 135 169 331 2;
  • 45) 0.999 999 999 890 187 135 169 331 2 × 2 = 1 + 0.999 999 999 780 374 270 338 662 4;
  • 46) 0.999 999 999 780 374 270 338 662 4 × 2 = 1 + 0.999 999 999 560 748 540 677 324 8;
  • 47) 0.999 999 999 560 748 540 677 324 8 × 2 = 1 + 0.999 999 999 121 497 081 354 649 6;
  • 48) 0.999 999 999 121 497 081 354 649 6 × 2 = 1 + 0.999 999 998 242 994 162 709 299 2;
  • 49) 0.999 999 998 242 994 162 709 299 2 × 2 = 1 + 0.999 999 996 485 988 325 418 598 4;
  • 50) 0.999 999 996 485 988 325 418 598 4 × 2 = 1 + 0.999 999 992 971 976 650 837 196 8;
  • 51) 0.999 999 992 971 976 650 837 196 8 × 2 = 1 + 0.999 999 985 943 953 301 674 393 6;
  • 52) 0.999 999 985 943 953 301 674 393 6 × 2 = 1 + 0.999 999 971 887 906 603 348 787 2;
  • 53) 0.999 999 971 887 906 603 348 787 2 × 2 = 1 + 0.999 999 943 775 813 206 697 574 4;
  • 54) 0.999 999 943 775 813 206 697 574 4 × 2 = 1 + 0.999 999 887 551 626 413 395 148 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:


0.000 000 000 742 147 676 646 703 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 703 2(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:


0.000 000 000 742 147 676 646 703 2(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 -0.000 000 000 742 147 676 646 703 2 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