-0.000 000 000 742 147 676 652 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 652 9(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
-0.000 000 000 742 147 676 652 9(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:

|-0.000 000 000 742 147 676 652 9| = 0.000 000 000 742 147 676 652 9


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: 0.000 000 000 742 147 676 652 9.

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) 0.000 000 000 742 147 676 652 9 × 2 = 0 + 0.000 000 001 484 295 353 305 8;
  • 2) 0.000 000 001 484 295 353 305 8 × 2 = 0 + 0.000 000 002 968 590 706 611 6;
  • 3) 0.000 000 002 968 590 706 611 6 × 2 = 0 + 0.000 000 005 937 181 413 223 2;
  • 4) 0.000 000 005 937 181 413 223 2 × 2 = 0 + 0.000 000 011 874 362 826 446 4;
  • 5) 0.000 000 011 874 362 826 446 4 × 2 = 0 + 0.000 000 023 748 725 652 892 8;
  • 6) 0.000 000 023 748 725 652 892 8 × 2 = 0 + 0.000 000 047 497 451 305 785 6;
  • 7) 0.000 000 047 497 451 305 785 6 × 2 = 0 + 0.000 000 094 994 902 611 571 2;
  • 8) 0.000 000 094 994 902 611 571 2 × 2 = 0 + 0.000 000 189 989 805 223 142 4;
  • 9) 0.000 000 189 989 805 223 142 4 × 2 = 0 + 0.000 000 379 979 610 446 284 8;
  • 10) 0.000 000 379 979 610 446 284 8 × 2 = 0 + 0.000 000 759 959 220 892 569 6;
  • 11) 0.000 000 759 959 220 892 569 6 × 2 = 0 + 0.000 001 519 918 441 785 139 2;
  • 12) 0.000 001 519 918 441 785 139 2 × 2 = 0 + 0.000 003 039 836 883 570 278 4;
  • 13) 0.000 003 039 836 883 570 278 4 × 2 = 0 + 0.000 006 079 673 767 140 556 8;
  • 14) 0.000 006 079 673 767 140 556 8 × 2 = 0 + 0.000 012 159 347 534 281 113 6;
  • 15) 0.000 012 159 347 534 281 113 6 × 2 = 0 + 0.000 024 318 695 068 562 227 2;
  • 16) 0.000 024 318 695 068 562 227 2 × 2 = 0 + 0.000 048 637 390 137 124 454 4;
  • 17) 0.000 048 637 390 137 124 454 4 × 2 = 0 + 0.000 097 274 780 274 248 908 8;
  • 18) 0.000 097 274 780 274 248 908 8 × 2 = 0 + 0.000 194 549 560 548 497 817 6;
  • 19) 0.000 194 549 560 548 497 817 6 × 2 = 0 + 0.000 389 099 121 096 995 635 2;
  • 20) 0.000 389 099 121 096 995 635 2 × 2 = 0 + 0.000 778 198 242 193 991 270 4;
  • 21) 0.000 778 198 242 193 991 270 4 × 2 = 0 + 0.001 556 396 484 387 982 540 8;
  • 22) 0.001 556 396 484 387 982 540 8 × 2 = 0 + 0.003 112 792 968 775 965 081 6;
  • 23) 0.003 112 792 968 775 965 081 6 × 2 = 0 + 0.006 225 585 937 551 930 163 2;
  • 24) 0.006 225 585 937 551 930 163 2 × 2 = 0 + 0.012 451 171 875 103 860 326 4;
  • 25) 0.012 451 171 875 103 860 326 4 × 2 = 0 + 0.024 902 343 750 207 720 652 8;
  • 26) 0.024 902 343 750 207 720 652 8 × 2 = 0 + 0.049 804 687 500 415 441 305 6;
  • 27) 0.049 804 687 500 415 441 305 6 × 2 = 0 + 0.099 609 375 000 830 882 611 2;
  • 28) 0.099 609 375 000 830 882 611 2 × 2 = 0 + 0.199 218 750 001 661 765 222 4;
  • 29) 0.199 218 750 001 661 765 222 4 × 2 = 0 + 0.398 437 500 003 323 530 444 8;
  • 30) 0.398 437 500 003 323 530 444 8 × 2 = 0 + 0.796 875 000 006 647 060 889 6;
  • 31) 0.796 875 000 006 647 060 889 6 × 2 = 1 + 0.593 750 000 013 294 121 779 2;
  • 32) 0.593 750 000 013 294 121 779 2 × 2 = 1 + 0.187 500 000 026 588 243 558 4;
  • 33) 0.187 500 000 026 588 243 558 4 × 2 = 0 + 0.375 000 000 053 176 487 116 8;
  • 34) 0.375 000 000 053 176 487 116 8 × 2 = 0 + 0.750 000 000 106 352 974 233 6;
  • 35) 0.750 000 000 106 352 974 233 6 × 2 = 1 + 0.500 000 000 212 705 948 467 2;
  • 36) 0.500 000 000 212 705 948 467 2 × 2 = 1 + 0.000 000 000 425 411 896 934 4;
  • 37) 0.000 000 000 425 411 896 934 4 × 2 = 0 + 0.000 000 000 850 823 793 868 8;
  • 38) 0.000 000 000 850 823 793 868 8 × 2 = 0 + 0.000 000 001 701 647 587 737 6;
  • 39) 0.000 000 001 701 647 587 737 6 × 2 = 0 + 0.000 000 003 403 295 175 475 2;
  • 40) 0.000 000 003 403 295 175 475 2 × 2 = 0 + 0.000 000 006 806 590 350 950 4;
  • 41) 0.000 000 006 806 590 350 950 4 × 2 = 0 + 0.000 000 013 613 180 701 900 8;
  • 42) 0.000 000 013 613 180 701 900 8 × 2 = 0 + 0.000 000 027 226 361 403 801 6;
  • 43) 0.000 000 027 226 361 403 801 6 × 2 = 0 + 0.000 000 054 452 722 807 603 2;
  • 44) 0.000 000 054 452 722 807 603 2 × 2 = 0 + 0.000 000 108 905 445 615 206 4;
  • 45) 0.000 000 108 905 445 615 206 4 × 2 = 0 + 0.000 000 217 810 891 230 412 8;
  • 46) 0.000 000 217 810 891 230 412 8 × 2 = 0 + 0.000 000 435 621 782 460 825 6;
  • 47) 0.000 000 435 621 782 460 825 6 × 2 = 0 + 0.000 000 871 243 564 921 651 2;
  • 48) 0.000 000 871 243 564 921 651 2 × 2 = 0 + 0.000 001 742 487 129 843 302 4;
  • 49) 0.000 001 742 487 129 843 302 4 × 2 = 0 + 0.000 003 484 974 259 686 604 8;
  • 50) 0.000 003 484 974 259 686 604 8 × 2 = 0 + 0.000 006 969 948 519 373 209 6;
  • 51) 0.000 006 969 948 519 373 209 6 × 2 = 0 + 0.000 013 939 897 038 746 419 2;
  • 52) 0.000 013 939 897 038 746 419 2 × 2 = 0 + 0.000 027 879 794 077 492 838 4;
  • 53) 0.000 027 879 794 077 492 838 4 × 2 = 0 + 0.000 055 759 588 154 985 676 8;
  • 54) 0.000 055 759 588 154 985 676 8 × 2 = 0 + 0.000 111 519 176 309 971 353 6;

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:


0.000 000 000 742 147 676 652 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 652 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 652 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 652 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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