-0.000 000 000 742 147 676 162 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 162(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
-0.000 000 000 742 147 676 162(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:

|-0.000 000 000 742 147 676 162| = 0.000 000 000 742 147 676 162


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: 0.000 000 000 742 147 676 162.

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) 0.000 000 000 742 147 676 162 × 2 = 0 + 0.000 000 001 484 295 352 324;
  • 2) 0.000 000 001 484 295 352 324 × 2 = 0 + 0.000 000 002 968 590 704 648;
  • 3) 0.000 000 002 968 590 704 648 × 2 = 0 + 0.000 000 005 937 181 409 296;
  • 4) 0.000 000 005 937 181 409 296 × 2 = 0 + 0.000 000 011 874 362 818 592;
  • 5) 0.000 000 011 874 362 818 592 × 2 = 0 + 0.000 000 023 748 725 637 184;
  • 6) 0.000 000 023 748 725 637 184 × 2 = 0 + 0.000 000 047 497 451 274 368;
  • 7) 0.000 000 047 497 451 274 368 × 2 = 0 + 0.000 000 094 994 902 548 736;
  • 8) 0.000 000 094 994 902 548 736 × 2 = 0 + 0.000 000 189 989 805 097 472;
  • 9) 0.000 000 189 989 805 097 472 × 2 = 0 + 0.000 000 379 979 610 194 944;
  • 10) 0.000 000 379 979 610 194 944 × 2 = 0 + 0.000 000 759 959 220 389 888;
  • 11) 0.000 000 759 959 220 389 888 × 2 = 0 + 0.000 001 519 918 440 779 776;
  • 12) 0.000 001 519 918 440 779 776 × 2 = 0 + 0.000 003 039 836 881 559 552;
  • 13) 0.000 003 039 836 881 559 552 × 2 = 0 + 0.000 006 079 673 763 119 104;
  • 14) 0.000 006 079 673 763 119 104 × 2 = 0 + 0.000 012 159 347 526 238 208;
  • 15) 0.000 012 159 347 526 238 208 × 2 = 0 + 0.000 024 318 695 052 476 416;
  • 16) 0.000 024 318 695 052 476 416 × 2 = 0 + 0.000 048 637 390 104 952 832;
  • 17) 0.000 048 637 390 104 952 832 × 2 = 0 + 0.000 097 274 780 209 905 664;
  • 18) 0.000 097 274 780 209 905 664 × 2 = 0 + 0.000 194 549 560 419 811 328;
  • 19) 0.000 194 549 560 419 811 328 × 2 = 0 + 0.000 389 099 120 839 622 656;
  • 20) 0.000 389 099 120 839 622 656 × 2 = 0 + 0.000 778 198 241 679 245 312;
  • 21) 0.000 778 198 241 679 245 312 × 2 = 0 + 0.001 556 396 483 358 490 624;
  • 22) 0.001 556 396 483 358 490 624 × 2 = 0 + 0.003 112 792 966 716 981 248;
  • 23) 0.003 112 792 966 716 981 248 × 2 = 0 + 0.006 225 585 933 433 962 496;
  • 24) 0.006 225 585 933 433 962 496 × 2 = 0 + 0.012 451 171 866 867 924 992;
  • 25) 0.012 451 171 866 867 924 992 × 2 = 0 + 0.024 902 343 733 735 849 984;
  • 26) 0.024 902 343 733 735 849 984 × 2 = 0 + 0.049 804 687 467 471 699 968;
  • 27) 0.049 804 687 467 471 699 968 × 2 = 0 + 0.099 609 374 934 943 399 936;
  • 28) 0.099 609 374 934 943 399 936 × 2 = 0 + 0.199 218 749 869 886 799 872;
  • 29) 0.199 218 749 869 886 799 872 × 2 = 0 + 0.398 437 499 739 773 599 744;
  • 30) 0.398 437 499 739 773 599 744 × 2 = 0 + 0.796 874 999 479 547 199 488;
  • 31) 0.796 874 999 479 547 199 488 × 2 = 1 + 0.593 749 998 959 094 398 976;
  • 32) 0.593 749 998 959 094 398 976 × 2 = 1 + 0.187 499 997 918 188 797 952;
  • 33) 0.187 499 997 918 188 797 952 × 2 = 0 + 0.374 999 995 836 377 595 904;
  • 34) 0.374 999 995 836 377 595 904 × 2 = 0 + 0.749 999 991 672 755 191 808;
  • 35) 0.749 999 991 672 755 191 808 × 2 = 1 + 0.499 999 983 345 510 383 616;
  • 36) 0.499 999 983 345 510 383 616 × 2 = 0 + 0.999 999 966 691 020 767 232;
  • 37) 0.999 999 966 691 020 767 232 × 2 = 1 + 0.999 999 933 382 041 534 464;
  • 38) 0.999 999 933 382 041 534 464 × 2 = 1 + 0.999 999 866 764 083 068 928;
  • 39) 0.999 999 866 764 083 068 928 × 2 = 1 + 0.999 999 733 528 166 137 856;
  • 40) 0.999 999 733 528 166 137 856 × 2 = 1 + 0.999 999 467 056 332 275 712;
  • 41) 0.999 999 467 056 332 275 712 × 2 = 1 + 0.999 998 934 112 664 551 424;
  • 42) 0.999 998 934 112 664 551 424 × 2 = 1 + 0.999 997 868 225 329 102 848;
  • 43) 0.999 997 868 225 329 102 848 × 2 = 1 + 0.999 995 736 450 658 205 696;
  • 44) 0.999 995 736 450 658 205 696 × 2 = 1 + 0.999 991 472 901 316 411 392;
  • 45) 0.999 991 472 901 316 411 392 × 2 = 1 + 0.999 982 945 802 632 822 784;
  • 46) 0.999 982 945 802 632 822 784 × 2 = 1 + 0.999 965 891 605 265 645 568;
  • 47) 0.999 965 891 605 265 645 568 × 2 = 1 + 0.999 931 783 210 531 291 136;
  • 48) 0.999 931 783 210 531 291 136 × 2 = 1 + 0.999 863 566 421 062 582 272;
  • 49) 0.999 863 566 421 062 582 272 × 2 = 1 + 0.999 727 132 842 125 164 544;
  • 50) 0.999 727 132 842 125 164 544 × 2 = 1 + 0.999 454 265 684 250 329 088;
  • 51) 0.999 454 265 684 250 329 088 × 2 = 1 + 0.998 908 531 368 500 658 176;
  • 52) 0.998 908 531 368 500 658 176 × 2 = 1 + 0.997 817 062 737 001 316 352;
  • 53) 0.997 817 062 737 001 316 352 × 2 = 1 + 0.995 634 125 474 002 632 704;
  • 54) 0.995 634 125 474 002 632 704 × 2 = 1 + 0.991 268 250 948 005 265 408;

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:


0.000 000 000 742 147 676 162(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 162(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:


0.000 000 000 742 147 676 162(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 -0.000 000 000 742 147 676 162 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