-0.000 000 000 742 147 678 84 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 678 84(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 84(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 84| = 0.000 000 000 742 147 678 84


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 84.

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 84 × 2 = 0 + 0.000 000 001 484 295 357 68;
  • 2) 0.000 000 001 484 295 357 68 × 2 = 0 + 0.000 000 002 968 590 715 36;
  • 3) 0.000 000 002 968 590 715 36 × 2 = 0 + 0.000 000 005 937 181 430 72;
  • 4) 0.000 000 005 937 181 430 72 × 2 = 0 + 0.000 000 011 874 362 861 44;
  • 5) 0.000 000 011 874 362 861 44 × 2 = 0 + 0.000 000 023 748 725 722 88;
  • 6) 0.000 000 023 748 725 722 88 × 2 = 0 + 0.000 000 047 497 451 445 76;
  • 7) 0.000 000 047 497 451 445 76 × 2 = 0 + 0.000 000 094 994 902 891 52;
  • 8) 0.000 000 094 994 902 891 52 × 2 = 0 + 0.000 000 189 989 805 783 04;
  • 9) 0.000 000 189 989 805 783 04 × 2 = 0 + 0.000 000 379 979 611 566 08;
  • 10) 0.000 000 379 979 611 566 08 × 2 = 0 + 0.000 000 759 959 223 132 16;
  • 11) 0.000 000 759 959 223 132 16 × 2 = 0 + 0.000 001 519 918 446 264 32;
  • 12) 0.000 001 519 918 446 264 32 × 2 = 0 + 0.000 003 039 836 892 528 64;
  • 13) 0.000 003 039 836 892 528 64 × 2 = 0 + 0.000 006 079 673 785 057 28;
  • 14) 0.000 006 079 673 785 057 28 × 2 = 0 + 0.000 012 159 347 570 114 56;
  • 15) 0.000 012 159 347 570 114 56 × 2 = 0 + 0.000 024 318 695 140 229 12;
  • 16) 0.000 024 318 695 140 229 12 × 2 = 0 + 0.000 048 637 390 280 458 24;
  • 17) 0.000 048 637 390 280 458 24 × 2 = 0 + 0.000 097 274 780 560 916 48;
  • 18) 0.000 097 274 780 560 916 48 × 2 = 0 + 0.000 194 549 561 121 832 96;
  • 19) 0.000 194 549 561 121 832 96 × 2 = 0 + 0.000 389 099 122 243 665 92;
  • 20) 0.000 389 099 122 243 665 92 × 2 = 0 + 0.000 778 198 244 487 331 84;
  • 21) 0.000 778 198 244 487 331 84 × 2 = 0 + 0.001 556 396 488 974 663 68;
  • 22) 0.001 556 396 488 974 663 68 × 2 = 0 + 0.003 112 792 977 949 327 36;
  • 23) 0.003 112 792 977 949 327 36 × 2 = 0 + 0.006 225 585 955 898 654 72;
  • 24) 0.006 225 585 955 898 654 72 × 2 = 0 + 0.012 451 171 911 797 309 44;
  • 25) 0.012 451 171 911 797 309 44 × 2 = 0 + 0.024 902 343 823 594 618 88;
  • 26) 0.024 902 343 823 594 618 88 × 2 = 0 + 0.049 804 687 647 189 237 76;
  • 27) 0.049 804 687 647 189 237 76 × 2 = 0 + 0.099 609 375 294 378 475 52;
  • 28) 0.099 609 375 294 378 475 52 × 2 = 0 + 0.199 218 750 588 756 951 04;
  • 29) 0.199 218 750 588 756 951 04 × 2 = 0 + 0.398 437 501 177 513 902 08;
  • 30) 0.398 437 501 177 513 902 08 × 2 = 0 + 0.796 875 002 355 027 804 16;
  • 31) 0.796 875 002 355 027 804 16 × 2 = 1 + 0.593 750 004 710 055 608 32;
  • 32) 0.593 750 004 710 055 608 32 × 2 = 1 + 0.187 500 009 420 111 216 64;
  • 33) 0.187 500 009 420 111 216 64 × 2 = 0 + 0.375 000 018 840 222 433 28;
  • 34) 0.375 000 018 840 222 433 28 × 2 = 0 + 0.750 000 037 680 444 866 56;
  • 35) 0.750 000 037 680 444 866 56 × 2 = 1 + 0.500 000 075 360 889 733 12;
  • 36) 0.500 000 075 360 889 733 12 × 2 = 1 + 0.000 000 150 721 779 466 24;
  • 37) 0.000 000 150 721 779 466 24 × 2 = 0 + 0.000 000 301 443 558 932 48;
  • 38) 0.000 000 301 443 558 932 48 × 2 = 0 + 0.000 000 602 887 117 864 96;
  • 39) 0.000 000 602 887 117 864 96 × 2 = 0 + 0.000 001 205 774 235 729 92;
  • 40) 0.000 001 205 774 235 729 92 × 2 = 0 + 0.000 002 411 548 471 459 84;
  • 41) 0.000 002 411 548 471 459 84 × 2 = 0 + 0.000 004 823 096 942 919 68;
  • 42) 0.000 004 823 096 942 919 68 × 2 = 0 + 0.000 009 646 193 885 839 36;
  • 43) 0.000 009 646 193 885 839 36 × 2 = 0 + 0.000 019 292 387 771 678 72;
  • 44) 0.000 019 292 387 771 678 72 × 2 = 0 + 0.000 038 584 775 543 357 44;
  • 45) 0.000 038 584 775 543 357 44 × 2 = 0 + 0.000 077 169 551 086 714 88;
  • 46) 0.000 077 169 551 086 714 88 × 2 = 0 + 0.000 154 339 102 173 429 76;
  • 47) 0.000 154 339 102 173 429 76 × 2 = 0 + 0.000 308 678 204 346 859 52;
  • 48) 0.000 308 678 204 346 859 52 × 2 = 0 + 0.000 617 356 408 693 719 04;
  • 49) 0.000 617 356 408 693 719 04 × 2 = 0 + 0.001 234 712 817 387 438 08;
  • 50) 0.001 234 712 817 387 438 08 × 2 = 0 + 0.002 469 425 634 774 876 16;
  • 51) 0.002 469 425 634 774 876 16 × 2 = 0 + 0.004 938 851 269 549 752 32;
  • 52) 0.004 938 851 269 549 752 32 × 2 = 0 + 0.009 877 702 539 099 504 64;
  • 53) 0.009 877 702 539 099 504 64 × 2 = 0 + 0.019 755 405 078 199 009 28;
  • 54) 0.019 755 405 078 199 009 28 × 2 = 0 + 0.039 510 810 156 398 018 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 84(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 84(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 84(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 84 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