-0.000 000 000 742 147 678 34 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 678 34(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
-0.000 000 000 742 147 678 34(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:

|-0.000 000 000 742 147 678 34| = 0.000 000 000 742 147 678 34


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: 0.000 000 000 742 147 678 34.

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) 0.000 000 000 742 147 678 34 × 2 = 0 + 0.000 000 001 484 295 356 68;
  • 2) 0.000 000 001 484 295 356 68 × 2 = 0 + 0.000 000 002 968 590 713 36;
  • 3) 0.000 000 002 968 590 713 36 × 2 = 0 + 0.000 000 005 937 181 426 72;
  • 4) 0.000 000 005 937 181 426 72 × 2 = 0 + 0.000 000 011 874 362 853 44;
  • 5) 0.000 000 011 874 362 853 44 × 2 = 0 + 0.000 000 023 748 725 706 88;
  • 6) 0.000 000 023 748 725 706 88 × 2 = 0 + 0.000 000 047 497 451 413 76;
  • 7) 0.000 000 047 497 451 413 76 × 2 = 0 + 0.000 000 094 994 902 827 52;
  • 8) 0.000 000 094 994 902 827 52 × 2 = 0 + 0.000 000 189 989 805 655 04;
  • 9) 0.000 000 189 989 805 655 04 × 2 = 0 + 0.000 000 379 979 611 310 08;
  • 10) 0.000 000 379 979 611 310 08 × 2 = 0 + 0.000 000 759 959 222 620 16;
  • 11) 0.000 000 759 959 222 620 16 × 2 = 0 + 0.000 001 519 918 445 240 32;
  • 12) 0.000 001 519 918 445 240 32 × 2 = 0 + 0.000 003 039 836 890 480 64;
  • 13) 0.000 003 039 836 890 480 64 × 2 = 0 + 0.000 006 079 673 780 961 28;
  • 14) 0.000 006 079 673 780 961 28 × 2 = 0 + 0.000 012 159 347 561 922 56;
  • 15) 0.000 012 159 347 561 922 56 × 2 = 0 + 0.000 024 318 695 123 845 12;
  • 16) 0.000 024 318 695 123 845 12 × 2 = 0 + 0.000 048 637 390 247 690 24;
  • 17) 0.000 048 637 390 247 690 24 × 2 = 0 + 0.000 097 274 780 495 380 48;
  • 18) 0.000 097 274 780 495 380 48 × 2 = 0 + 0.000 194 549 560 990 760 96;
  • 19) 0.000 194 549 560 990 760 96 × 2 = 0 + 0.000 389 099 121 981 521 92;
  • 20) 0.000 389 099 121 981 521 92 × 2 = 0 + 0.000 778 198 243 963 043 84;
  • 21) 0.000 778 198 243 963 043 84 × 2 = 0 + 0.001 556 396 487 926 087 68;
  • 22) 0.001 556 396 487 926 087 68 × 2 = 0 + 0.003 112 792 975 852 175 36;
  • 23) 0.003 112 792 975 852 175 36 × 2 = 0 + 0.006 225 585 951 704 350 72;
  • 24) 0.006 225 585 951 704 350 72 × 2 = 0 + 0.012 451 171 903 408 701 44;
  • 25) 0.012 451 171 903 408 701 44 × 2 = 0 + 0.024 902 343 806 817 402 88;
  • 26) 0.024 902 343 806 817 402 88 × 2 = 0 + 0.049 804 687 613 634 805 76;
  • 27) 0.049 804 687 613 634 805 76 × 2 = 0 + 0.099 609 375 227 269 611 52;
  • 28) 0.099 609 375 227 269 611 52 × 2 = 0 + 0.199 218 750 454 539 223 04;
  • 29) 0.199 218 750 454 539 223 04 × 2 = 0 + 0.398 437 500 909 078 446 08;
  • 30) 0.398 437 500 909 078 446 08 × 2 = 0 + 0.796 875 001 818 156 892 16;
  • 31) 0.796 875 001 818 156 892 16 × 2 = 1 + 0.593 750 003 636 313 784 32;
  • 32) 0.593 750 003 636 313 784 32 × 2 = 1 + 0.187 500 007 272 627 568 64;
  • 33) 0.187 500 007 272 627 568 64 × 2 = 0 + 0.375 000 014 545 255 137 28;
  • 34) 0.375 000 014 545 255 137 28 × 2 = 0 + 0.750 000 029 090 510 274 56;
  • 35) 0.750 000 029 090 510 274 56 × 2 = 1 + 0.500 000 058 181 020 549 12;
  • 36) 0.500 000 058 181 020 549 12 × 2 = 1 + 0.000 000 116 362 041 098 24;
  • 37) 0.000 000 116 362 041 098 24 × 2 = 0 + 0.000 000 232 724 082 196 48;
  • 38) 0.000 000 232 724 082 196 48 × 2 = 0 + 0.000 000 465 448 164 392 96;
  • 39) 0.000 000 465 448 164 392 96 × 2 = 0 + 0.000 000 930 896 328 785 92;
  • 40) 0.000 000 930 896 328 785 92 × 2 = 0 + 0.000 001 861 792 657 571 84;
  • 41) 0.000 001 861 792 657 571 84 × 2 = 0 + 0.000 003 723 585 315 143 68;
  • 42) 0.000 003 723 585 315 143 68 × 2 = 0 + 0.000 007 447 170 630 287 36;
  • 43) 0.000 007 447 170 630 287 36 × 2 = 0 + 0.000 014 894 341 260 574 72;
  • 44) 0.000 014 894 341 260 574 72 × 2 = 0 + 0.000 029 788 682 521 149 44;
  • 45) 0.000 029 788 682 521 149 44 × 2 = 0 + 0.000 059 577 365 042 298 88;
  • 46) 0.000 059 577 365 042 298 88 × 2 = 0 + 0.000 119 154 730 084 597 76;
  • 47) 0.000 119 154 730 084 597 76 × 2 = 0 + 0.000 238 309 460 169 195 52;
  • 48) 0.000 238 309 460 169 195 52 × 2 = 0 + 0.000 476 618 920 338 391 04;
  • 49) 0.000 476 618 920 338 391 04 × 2 = 0 + 0.000 953 237 840 676 782 08;
  • 50) 0.000 953 237 840 676 782 08 × 2 = 0 + 0.001 906 475 681 353 564 16;
  • 51) 0.001 906 475 681 353 564 16 × 2 = 0 + 0.003 812 951 362 707 128 32;
  • 52) 0.003 812 951 362 707 128 32 × 2 = 0 + 0.007 625 902 725 414 256 64;
  • 53) 0.007 625 902 725 414 256 64 × 2 = 0 + 0.015 251 805 450 828 513 28;
  • 54) 0.015 251 805 450 828 513 28 × 2 = 0 + 0.030 503 610 901 657 026 56;

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:


0.000 000 000 742 147 678 34(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 678 34(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:


0.000 000 000 742 147 678 34(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 -0.000 000 000 742 147 678 34 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