-0.000 000 000 742 147 686 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 686 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
-0.000 000 000 742 147 686 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:

|-0.000 000 000 742 147 686 9| = 0.000 000 000 742 147 686 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: 0.000 000 000 742 147 686 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) 0.000 000 000 742 147 686 9 × 2 = 0 + 0.000 000 001 484 295 373 8;
  • 2) 0.000 000 001 484 295 373 8 × 2 = 0 + 0.000 000 002 968 590 747 6;
  • 3) 0.000 000 002 968 590 747 6 × 2 = 0 + 0.000 000 005 937 181 495 2;
  • 4) 0.000 000 005 937 181 495 2 × 2 = 0 + 0.000 000 011 874 362 990 4;
  • 5) 0.000 000 011 874 362 990 4 × 2 = 0 + 0.000 000 023 748 725 980 8;
  • 6) 0.000 000 023 748 725 980 8 × 2 = 0 + 0.000 000 047 497 451 961 6;
  • 7) 0.000 000 047 497 451 961 6 × 2 = 0 + 0.000 000 094 994 903 923 2;
  • 8) 0.000 000 094 994 903 923 2 × 2 = 0 + 0.000 000 189 989 807 846 4;
  • 9) 0.000 000 189 989 807 846 4 × 2 = 0 + 0.000 000 379 979 615 692 8;
  • 10) 0.000 000 379 979 615 692 8 × 2 = 0 + 0.000 000 759 959 231 385 6;
  • 11) 0.000 000 759 959 231 385 6 × 2 = 0 + 0.000 001 519 918 462 771 2;
  • 12) 0.000 001 519 918 462 771 2 × 2 = 0 + 0.000 003 039 836 925 542 4;
  • 13) 0.000 003 039 836 925 542 4 × 2 = 0 + 0.000 006 079 673 851 084 8;
  • 14) 0.000 006 079 673 851 084 8 × 2 = 0 + 0.000 012 159 347 702 169 6;
  • 15) 0.000 012 159 347 702 169 6 × 2 = 0 + 0.000 024 318 695 404 339 2;
  • 16) 0.000 024 318 695 404 339 2 × 2 = 0 + 0.000 048 637 390 808 678 4;
  • 17) 0.000 048 637 390 808 678 4 × 2 = 0 + 0.000 097 274 781 617 356 8;
  • 18) 0.000 097 274 781 617 356 8 × 2 = 0 + 0.000 194 549 563 234 713 6;
  • 19) 0.000 194 549 563 234 713 6 × 2 = 0 + 0.000 389 099 126 469 427 2;
  • 20) 0.000 389 099 126 469 427 2 × 2 = 0 + 0.000 778 198 252 938 854 4;
  • 21) 0.000 778 198 252 938 854 4 × 2 = 0 + 0.001 556 396 505 877 708 8;
  • 22) 0.001 556 396 505 877 708 8 × 2 = 0 + 0.003 112 793 011 755 417 6;
  • 23) 0.003 112 793 011 755 417 6 × 2 = 0 + 0.006 225 586 023 510 835 2;
  • 24) 0.006 225 586 023 510 835 2 × 2 = 0 + 0.012 451 172 047 021 670 4;
  • 25) 0.012 451 172 047 021 670 4 × 2 = 0 + 0.024 902 344 094 043 340 8;
  • 26) 0.024 902 344 094 043 340 8 × 2 = 0 + 0.049 804 688 188 086 681 6;
  • 27) 0.049 804 688 188 086 681 6 × 2 = 0 + 0.099 609 376 376 173 363 2;
  • 28) 0.099 609 376 376 173 363 2 × 2 = 0 + 0.199 218 752 752 346 726 4;
  • 29) 0.199 218 752 752 346 726 4 × 2 = 0 + 0.398 437 505 504 693 452 8;
  • 30) 0.398 437 505 504 693 452 8 × 2 = 0 + 0.796 875 011 009 386 905 6;
  • 31) 0.796 875 011 009 386 905 6 × 2 = 1 + 0.593 750 022 018 773 811 2;
  • 32) 0.593 750 022 018 773 811 2 × 2 = 1 + 0.187 500 044 037 547 622 4;
  • 33) 0.187 500 044 037 547 622 4 × 2 = 0 + 0.375 000 088 075 095 244 8;
  • 34) 0.375 000 088 075 095 244 8 × 2 = 0 + 0.750 000 176 150 190 489 6;
  • 35) 0.750 000 176 150 190 489 6 × 2 = 1 + 0.500 000 352 300 380 979 2;
  • 36) 0.500 000 352 300 380 979 2 × 2 = 1 + 0.000 000 704 600 761 958 4;
  • 37) 0.000 000 704 600 761 958 4 × 2 = 0 + 0.000 001 409 201 523 916 8;
  • 38) 0.000 001 409 201 523 916 8 × 2 = 0 + 0.000 002 818 403 047 833 6;
  • 39) 0.000 002 818 403 047 833 6 × 2 = 0 + 0.000 005 636 806 095 667 2;
  • 40) 0.000 005 636 806 095 667 2 × 2 = 0 + 0.000 011 273 612 191 334 4;
  • 41) 0.000 011 273 612 191 334 4 × 2 = 0 + 0.000 022 547 224 382 668 8;
  • 42) 0.000 022 547 224 382 668 8 × 2 = 0 + 0.000 045 094 448 765 337 6;
  • 43) 0.000 045 094 448 765 337 6 × 2 = 0 + 0.000 090 188 897 530 675 2;
  • 44) 0.000 090 188 897 530 675 2 × 2 = 0 + 0.000 180 377 795 061 350 4;
  • 45) 0.000 180 377 795 061 350 4 × 2 = 0 + 0.000 360 755 590 122 700 8;
  • 46) 0.000 360 755 590 122 700 8 × 2 = 0 + 0.000 721 511 180 245 401 6;
  • 47) 0.000 721 511 180 245 401 6 × 2 = 0 + 0.001 443 022 360 490 803 2;
  • 48) 0.001 443 022 360 490 803 2 × 2 = 0 + 0.002 886 044 720 981 606 4;
  • 49) 0.002 886 044 720 981 606 4 × 2 = 0 + 0.005 772 089 441 963 212 8;
  • 50) 0.005 772 089 441 963 212 8 × 2 = 0 + 0.011 544 178 883 926 425 6;
  • 51) 0.011 544 178 883 926 425 6 × 2 = 0 + 0.023 088 357 767 852 851 2;
  • 52) 0.023 088 357 767 852 851 2 × 2 = 0 + 0.046 176 715 535 705 702 4;
  • 53) 0.046 176 715 535 705 702 4 × 2 = 0 + 0.092 353 431 071 411 404 8;
  • 54) 0.092 353 431 071 411 404 8 × 2 = 0 + 0.184 706 862 142 822 809 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:


0.000 000 000 742 147 686 9(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 686 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:


0.000 000 000 742 147 686 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 -0.000 000 000 742 147 686 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