-0.000 000 000 781 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 781(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
-0.000 000 000 781(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:

|-0.000 000 000 781| = 0.000 000 000 781


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: 0.000 000 000 781.

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) 0.000 000 000 781 × 2 = 0 + 0.000 000 001 562;
  • 2) 0.000 000 001 562 × 2 = 0 + 0.000 000 003 124;
  • 3) 0.000 000 003 124 × 2 = 0 + 0.000 000 006 248;
  • 4) 0.000 000 006 248 × 2 = 0 + 0.000 000 012 496;
  • 5) 0.000 000 012 496 × 2 = 0 + 0.000 000 024 992;
  • 6) 0.000 000 024 992 × 2 = 0 + 0.000 000 049 984;
  • 7) 0.000 000 049 984 × 2 = 0 + 0.000 000 099 968;
  • 8) 0.000 000 099 968 × 2 = 0 + 0.000 000 199 936;
  • 9) 0.000 000 199 936 × 2 = 0 + 0.000 000 399 872;
  • 10) 0.000 000 399 872 × 2 = 0 + 0.000 000 799 744;
  • 11) 0.000 000 799 744 × 2 = 0 + 0.000 001 599 488;
  • 12) 0.000 001 599 488 × 2 = 0 + 0.000 003 198 976;
  • 13) 0.000 003 198 976 × 2 = 0 + 0.000 006 397 952;
  • 14) 0.000 006 397 952 × 2 = 0 + 0.000 012 795 904;
  • 15) 0.000 012 795 904 × 2 = 0 + 0.000 025 591 808;
  • 16) 0.000 025 591 808 × 2 = 0 + 0.000 051 183 616;
  • 17) 0.000 051 183 616 × 2 = 0 + 0.000 102 367 232;
  • 18) 0.000 102 367 232 × 2 = 0 + 0.000 204 734 464;
  • 19) 0.000 204 734 464 × 2 = 0 + 0.000 409 468 928;
  • 20) 0.000 409 468 928 × 2 = 0 + 0.000 818 937 856;
  • 21) 0.000 818 937 856 × 2 = 0 + 0.001 637 875 712;
  • 22) 0.001 637 875 712 × 2 = 0 + 0.003 275 751 424;
  • 23) 0.003 275 751 424 × 2 = 0 + 0.006 551 502 848;
  • 24) 0.006 551 502 848 × 2 = 0 + 0.013 103 005 696;
  • 25) 0.013 103 005 696 × 2 = 0 + 0.026 206 011 392;
  • 26) 0.026 206 011 392 × 2 = 0 + 0.052 412 022 784;
  • 27) 0.052 412 022 784 × 2 = 0 + 0.104 824 045 568;
  • 28) 0.104 824 045 568 × 2 = 0 + 0.209 648 091 136;
  • 29) 0.209 648 091 136 × 2 = 0 + 0.419 296 182 272;
  • 30) 0.419 296 182 272 × 2 = 0 + 0.838 592 364 544;
  • 31) 0.838 592 364 544 × 2 = 1 + 0.677 184 729 088;
  • 32) 0.677 184 729 088 × 2 = 1 + 0.354 369 458 176;
  • 33) 0.354 369 458 176 × 2 = 0 + 0.708 738 916 352;
  • 34) 0.708 738 916 352 × 2 = 1 + 0.417 477 832 704;
  • 35) 0.417 477 832 704 × 2 = 0 + 0.834 955 665 408;
  • 36) 0.834 955 665 408 × 2 = 1 + 0.669 911 330 816;
  • 37) 0.669 911 330 816 × 2 = 1 + 0.339 822 661 632;
  • 38) 0.339 822 661 632 × 2 = 0 + 0.679 645 323 264;
  • 39) 0.679 645 323 264 × 2 = 1 + 0.359 290 646 528;
  • 40) 0.359 290 646 528 × 2 = 0 + 0.718 581 293 056;
  • 41) 0.718 581 293 056 × 2 = 1 + 0.437 162 586 112;
  • 42) 0.437 162 586 112 × 2 = 0 + 0.874 325 172 224;
  • 43) 0.874 325 172 224 × 2 = 1 + 0.748 650 344 448;
  • 44) 0.748 650 344 448 × 2 = 1 + 0.497 300 688 896;
  • 45) 0.497 300 688 896 × 2 = 0 + 0.994 601 377 792;
  • 46) 0.994 601 377 792 × 2 = 1 + 0.989 202 755 584;
  • 47) 0.989 202 755 584 × 2 = 1 + 0.978 405 511 168;
  • 48) 0.978 405 511 168 × 2 = 1 + 0.956 811 022 336;
  • 49) 0.956 811 022 336 × 2 = 1 + 0.913 622 044 672;
  • 50) 0.913 622 044 672 × 2 = 1 + 0.827 244 089 344;
  • 51) 0.827 244 089 344 × 2 = 1 + 0.654 488 178 688;
  • 52) 0.654 488 178 688 × 2 = 1 + 0.308 976 357 376;
  • 53) 0.308 976 357 376 × 2 = 0 + 0.617 952 714 752;
  • 54) 0.617 952 714 752 × 2 = 1 + 0.235 905 429 504;

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:


0.000 000 000 781(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0101 1010 1011 0111 1111 01(2)

6. Positive number before normalization:

0.000 000 000 781(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0101 1010 1011 0111 1111 01(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:


0.000 000 000 781(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0101 1010 1011 0111 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0101 1010 1011 0111 1111 01(2) × 20 =

1.1010 1101 0101 1011 1111 101(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):
1.1010 1101 0101 1011 1111 101


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. 101 0110 1010 1101 1111 1101 =

101 0110 1010 1101 1111 1101


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) =
101 0110 1010 1101 1111 1101


Decimal number -0.000 000 000 781 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 0110 1010 1101 1111 1101

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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