-0.000 000 000 742 147 676 646 688 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 688 1(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
-0.000 000 000 742 147 676 646 688 1(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:

|-0.000 000 000 742 147 676 646 688 1| = 0.000 000 000 742 147 676 646 688 1


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: 0.000 000 000 742 147 676 646 688 1.

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) 0.000 000 000 742 147 676 646 688 1 × 2 = 0 + 0.000 000 001 484 295 353 293 376 2;
  • 2) 0.000 000 001 484 295 353 293 376 2 × 2 = 0 + 0.000 000 002 968 590 706 586 752 4;
  • 3) 0.000 000 002 968 590 706 586 752 4 × 2 = 0 + 0.000 000 005 937 181 413 173 504 8;
  • 4) 0.000 000 005 937 181 413 173 504 8 × 2 = 0 + 0.000 000 011 874 362 826 347 009 6;
  • 5) 0.000 000 011 874 362 826 347 009 6 × 2 = 0 + 0.000 000 023 748 725 652 694 019 2;
  • 6) 0.000 000 023 748 725 652 694 019 2 × 2 = 0 + 0.000 000 047 497 451 305 388 038 4;
  • 7) 0.000 000 047 497 451 305 388 038 4 × 2 = 0 + 0.000 000 094 994 902 610 776 076 8;
  • 8) 0.000 000 094 994 902 610 776 076 8 × 2 = 0 + 0.000 000 189 989 805 221 552 153 6;
  • 9) 0.000 000 189 989 805 221 552 153 6 × 2 = 0 + 0.000 000 379 979 610 443 104 307 2;
  • 10) 0.000 000 379 979 610 443 104 307 2 × 2 = 0 + 0.000 000 759 959 220 886 208 614 4;
  • 11) 0.000 000 759 959 220 886 208 614 4 × 2 = 0 + 0.000 001 519 918 441 772 417 228 8;
  • 12) 0.000 001 519 918 441 772 417 228 8 × 2 = 0 + 0.000 003 039 836 883 544 834 457 6;
  • 13) 0.000 003 039 836 883 544 834 457 6 × 2 = 0 + 0.000 006 079 673 767 089 668 915 2;
  • 14) 0.000 006 079 673 767 089 668 915 2 × 2 = 0 + 0.000 012 159 347 534 179 337 830 4;
  • 15) 0.000 012 159 347 534 179 337 830 4 × 2 = 0 + 0.000 024 318 695 068 358 675 660 8;
  • 16) 0.000 024 318 695 068 358 675 660 8 × 2 = 0 + 0.000 048 637 390 136 717 351 321 6;
  • 17) 0.000 048 637 390 136 717 351 321 6 × 2 = 0 + 0.000 097 274 780 273 434 702 643 2;
  • 18) 0.000 097 274 780 273 434 702 643 2 × 2 = 0 + 0.000 194 549 560 546 869 405 286 4;
  • 19) 0.000 194 549 560 546 869 405 286 4 × 2 = 0 + 0.000 389 099 121 093 738 810 572 8;
  • 20) 0.000 389 099 121 093 738 810 572 8 × 2 = 0 + 0.000 778 198 242 187 477 621 145 6;
  • 21) 0.000 778 198 242 187 477 621 145 6 × 2 = 0 + 0.001 556 396 484 374 955 242 291 2;
  • 22) 0.001 556 396 484 374 955 242 291 2 × 2 = 0 + 0.003 112 792 968 749 910 484 582 4;
  • 23) 0.003 112 792 968 749 910 484 582 4 × 2 = 0 + 0.006 225 585 937 499 820 969 164 8;
  • 24) 0.006 225 585 937 499 820 969 164 8 × 2 = 0 + 0.012 451 171 874 999 641 938 329 6;
  • 25) 0.012 451 171 874 999 641 938 329 6 × 2 = 0 + 0.024 902 343 749 999 283 876 659 2;
  • 26) 0.024 902 343 749 999 283 876 659 2 × 2 = 0 + 0.049 804 687 499 998 567 753 318 4;
  • 27) 0.049 804 687 499 998 567 753 318 4 × 2 = 0 + 0.099 609 374 999 997 135 506 636 8;
  • 28) 0.099 609 374 999 997 135 506 636 8 × 2 = 0 + 0.199 218 749 999 994 271 013 273 6;
  • 29) 0.199 218 749 999 994 271 013 273 6 × 2 = 0 + 0.398 437 499 999 988 542 026 547 2;
  • 30) 0.398 437 499 999 988 542 026 547 2 × 2 = 0 + 0.796 874 999 999 977 084 053 094 4;
  • 31) 0.796 874 999 999 977 084 053 094 4 × 2 = 1 + 0.593 749 999 999 954 168 106 188 8;
  • 32) 0.593 749 999 999 954 168 106 188 8 × 2 = 1 + 0.187 499 999 999 908 336 212 377 6;
  • 33) 0.187 499 999 999 908 336 212 377 6 × 2 = 0 + 0.374 999 999 999 816 672 424 755 2;
  • 34) 0.374 999 999 999 816 672 424 755 2 × 2 = 0 + 0.749 999 999 999 633 344 849 510 4;
  • 35) 0.749 999 999 999 633 344 849 510 4 × 2 = 1 + 0.499 999 999 999 266 689 699 020 8;
  • 36) 0.499 999 999 999 266 689 699 020 8 × 2 = 0 + 0.999 999 999 998 533 379 398 041 6;
  • 37) 0.999 999 999 998 533 379 398 041 6 × 2 = 1 + 0.999 999 999 997 066 758 796 083 2;
  • 38) 0.999 999 999 997 066 758 796 083 2 × 2 = 1 + 0.999 999 999 994 133 517 592 166 4;
  • 39) 0.999 999 999 994 133 517 592 166 4 × 2 = 1 + 0.999 999 999 988 267 035 184 332 8;
  • 40) 0.999 999 999 988 267 035 184 332 8 × 2 = 1 + 0.999 999 999 976 534 070 368 665 6;
  • 41) 0.999 999 999 976 534 070 368 665 6 × 2 = 1 + 0.999 999 999 953 068 140 737 331 2;
  • 42) 0.999 999 999 953 068 140 737 331 2 × 2 = 1 + 0.999 999 999 906 136 281 474 662 4;
  • 43) 0.999 999 999 906 136 281 474 662 4 × 2 = 1 + 0.999 999 999 812 272 562 949 324 8;
  • 44) 0.999 999 999 812 272 562 949 324 8 × 2 = 1 + 0.999 999 999 624 545 125 898 649 6;
  • 45) 0.999 999 999 624 545 125 898 649 6 × 2 = 1 + 0.999 999 999 249 090 251 797 299 2;
  • 46) 0.999 999 999 249 090 251 797 299 2 × 2 = 1 + 0.999 999 998 498 180 503 594 598 4;
  • 47) 0.999 999 998 498 180 503 594 598 4 × 2 = 1 + 0.999 999 996 996 361 007 189 196 8;
  • 48) 0.999 999 996 996 361 007 189 196 8 × 2 = 1 + 0.999 999 993 992 722 014 378 393 6;
  • 49) 0.999 999 993 992 722 014 378 393 6 × 2 = 1 + 0.999 999 987 985 444 028 756 787 2;
  • 50) 0.999 999 987 985 444 028 756 787 2 × 2 = 1 + 0.999 999 975 970 888 057 513 574 4;
  • 51) 0.999 999 975 970 888 057 513 574 4 × 2 = 1 + 0.999 999 951 941 776 115 027 148 8;
  • 52) 0.999 999 951 941 776 115 027 148 8 × 2 = 1 + 0.999 999 903 883 552 230 054 297 6;
  • 53) 0.999 999 903 883 552 230 054 297 6 × 2 = 1 + 0.999 999 807 767 104 460 108 595 2;
  • 54) 0.999 999 807 767 104 460 108 595 2 × 2 = 1 + 0.999 999 615 534 208 920 217 190 4;

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:


0.000 000 000 742 147 676 646 688 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 688 1(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:


0.000 000 000 742 147 676 646 688 1(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 -0.000 000 000 742 147 676 646 688 1 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