-0.000 000 000 742 147 676 229 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 229(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
-0.000 000 000 742 147 676 229(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:

|-0.000 000 000 742 147 676 229| = 0.000 000 000 742 147 676 229


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: 0.000 000 000 742 147 676 229.

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) 0.000 000 000 742 147 676 229 × 2 = 0 + 0.000 000 001 484 295 352 458;
  • 2) 0.000 000 001 484 295 352 458 × 2 = 0 + 0.000 000 002 968 590 704 916;
  • 3) 0.000 000 002 968 590 704 916 × 2 = 0 + 0.000 000 005 937 181 409 832;
  • 4) 0.000 000 005 937 181 409 832 × 2 = 0 + 0.000 000 011 874 362 819 664;
  • 5) 0.000 000 011 874 362 819 664 × 2 = 0 + 0.000 000 023 748 725 639 328;
  • 6) 0.000 000 023 748 725 639 328 × 2 = 0 + 0.000 000 047 497 451 278 656;
  • 7) 0.000 000 047 497 451 278 656 × 2 = 0 + 0.000 000 094 994 902 557 312;
  • 8) 0.000 000 094 994 902 557 312 × 2 = 0 + 0.000 000 189 989 805 114 624;
  • 9) 0.000 000 189 989 805 114 624 × 2 = 0 + 0.000 000 379 979 610 229 248;
  • 10) 0.000 000 379 979 610 229 248 × 2 = 0 + 0.000 000 759 959 220 458 496;
  • 11) 0.000 000 759 959 220 458 496 × 2 = 0 + 0.000 001 519 918 440 916 992;
  • 12) 0.000 001 519 918 440 916 992 × 2 = 0 + 0.000 003 039 836 881 833 984;
  • 13) 0.000 003 039 836 881 833 984 × 2 = 0 + 0.000 006 079 673 763 667 968;
  • 14) 0.000 006 079 673 763 667 968 × 2 = 0 + 0.000 012 159 347 527 335 936;
  • 15) 0.000 012 159 347 527 335 936 × 2 = 0 + 0.000 024 318 695 054 671 872;
  • 16) 0.000 024 318 695 054 671 872 × 2 = 0 + 0.000 048 637 390 109 343 744;
  • 17) 0.000 048 637 390 109 343 744 × 2 = 0 + 0.000 097 274 780 218 687 488;
  • 18) 0.000 097 274 780 218 687 488 × 2 = 0 + 0.000 194 549 560 437 374 976;
  • 19) 0.000 194 549 560 437 374 976 × 2 = 0 + 0.000 389 099 120 874 749 952;
  • 20) 0.000 389 099 120 874 749 952 × 2 = 0 + 0.000 778 198 241 749 499 904;
  • 21) 0.000 778 198 241 749 499 904 × 2 = 0 + 0.001 556 396 483 498 999 808;
  • 22) 0.001 556 396 483 498 999 808 × 2 = 0 + 0.003 112 792 966 997 999 616;
  • 23) 0.003 112 792 966 997 999 616 × 2 = 0 + 0.006 225 585 933 995 999 232;
  • 24) 0.006 225 585 933 995 999 232 × 2 = 0 + 0.012 451 171 867 991 998 464;
  • 25) 0.012 451 171 867 991 998 464 × 2 = 0 + 0.024 902 343 735 983 996 928;
  • 26) 0.024 902 343 735 983 996 928 × 2 = 0 + 0.049 804 687 471 967 993 856;
  • 27) 0.049 804 687 471 967 993 856 × 2 = 0 + 0.099 609 374 943 935 987 712;
  • 28) 0.099 609 374 943 935 987 712 × 2 = 0 + 0.199 218 749 887 871 975 424;
  • 29) 0.199 218 749 887 871 975 424 × 2 = 0 + 0.398 437 499 775 743 950 848;
  • 30) 0.398 437 499 775 743 950 848 × 2 = 0 + 0.796 874 999 551 487 901 696;
  • 31) 0.796 874 999 551 487 901 696 × 2 = 1 + 0.593 749 999 102 975 803 392;
  • 32) 0.593 749 999 102 975 803 392 × 2 = 1 + 0.187 499 998 205 951 606 784;
  • 33) 0.187 499 998 205 951 606 784 × 2 = 0 + 0.374 999 996 411 903 213 568;
  • 34) 0.374 999 996 411 903 213 568 × 2 = 0 + 0.749 999 992 823 806 427 136;
  • 35) 0.749 999 992 823 806 427 136 × 2 = 1 + 0.499 999 985 647 612 854 272;
  • 36) 0.499 999 985 647 612 854 272 × 2 = 0 + 0.999 999 971 295 225 708 544;
  • 37) 0.999 999 971 295 225 708 544 × 2 = 1 + 0.999 999 942 590 451 417 088;
  • 38) 0.999 999 942 590 451 417 088 × 2 = 1 + 0.999 999 885 180 902 834 176;
  • 39) 0.999 999 885 180 902 834 176 × 2 = 1 + 0.999 999 770 361 805 668 352;
  • 40) 0.999 999 770 361 805 668 352 × 2 = 1 + 0.999 999 540 723 611 336 704;
  • 41) 0.999 999 540 723 611 336 704 × 2 = 1 + 0.999 999 081 447 222 673 408;
  • 42) 0.999 999 081 447 222 673 408 × 2 = 1 + 0.999 998 162 894 445 346 816;
  • 43) 0.999 998 162 894 445 346 816 × 2 = 1 + 0.999 996 325 788 890 693 632;
  • 44) 0.999 996 325 788 890 693 632 × 2 = 1 + 0.999 992 651 577 781 387 264;
  • 45) 0.999 992 651 577 781 387 264 × 2 = 1 + 0.999 985 303 155 562 774 528;
  • 46) 0.999 985 303 155 562 774 528 × 2 = 1 + 0.999 970 606 311 125 549 056;
  • 47) 0.999 970 606 311 125 549 056 × 2 = 1 + 0.999 941 212 622 251 098 112;
  • 48) 0.999 941 212 622 251 098 112 × 2 = 1 + 0.999 882 425 244 502 196 224;
  • 49) 0.999 882 425 244 502 196 224 × 2 = 1 + 0.999 764 850 489 004 392 448;
  • 50) 0.999 764 850 489 004 392 448 × 2 = 1 + 0.999 529 700 978 008 784 896;
  • 51) 0.999 529 700 978 008 784 896 × 2 = 1 + 0.999 059 401 956 017 569 792;
  • 52) 0.999 059 401 956 017 569 792 × 2 = 1 + 0.998 118 803 912 035 139 584;
  • 53) 0.998 118 803 912 035 139 584 × 2 = 1 + 0.996 237 607 824 070 279 168;
  • 54) 0.996 237 607 824 070 279 168 × 2 = 1 + 0.992 475 215 648 140 558 336;

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:


0.000 000 000 742 147 676 229(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 229(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:


0.000 000 000 742 147 676 229(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 -0.000 000 000 742 147 676 229 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