-0.000 000 000 741 948 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 948(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
-0.000 000 000 741 948(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:

|-0.000 000 000 741 948| = 0.000 000 000 741 948


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: 0.000 000 000 741 948.

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) 0.000 000 000 741 948 × 2 = 0 + 0.000 000 001 483 896;
  • 2) 0.000 000 001 483 896 × 2 = 0 + 0.000 000 002 967 792;
  • 3) 0.000 000 002 967 792 × 2 = 0 + 0.000 000 005 935 584;
  • 4) 0.000 000 005 935 584 × 2 = 0 + 0.000 000 011 871 168;
  • 5) 0.000 000 011 871 168 × 2 = 0 + 0.000 000 023 742 336;
  • 6) 0.000 000 023 742 336 × 2 = 0 + 0.000 000 047 484 672;
  • 7) 0.000 000 047 484 672 × 2 = 0 + 0.000 000 094 969 344;
  • 8) 0.000 000 094 969 344 × 2 = 0 + 0.000 000 189 938 688;
  • 9) 0.000 000 189 938 688 × 2 = 0 + 0.000 000 379 877 376;
  • 10) 0.000 000 379 877 376 × 2 = 0 + 0.000 000 759 754 752;
  • 11) 0.000 000 759 754 752 × 2 = 0 + 0.000 001 519 509 504;
  • 12) 0.000 001 519 509 504 × 2 = 0 + 0.000 003 039 019 008;
  • 13) 0.000 003 039 019 008 × 2 = 0 + 0.000 006 078 038 016;
  • 14) 0.000 006 078 038 016 × 2 = 0 + 0.000 012 156 076 032;
  • 15) 0.000 012 156 076 032 × 2 = 0 + 0.000 024 312 152 064;
  • 16) 0.000 024 312 152 064 × 2 = 0 + 0.000 048 624 304 128;
  • 17) 0.000 048 624 304 128 × 2 = 0 + 0.000 097 248 608 256;
  • 18) 0.000 097 248 608 256 × 2 = 0 + 0.000 194 497 216 512;
  • 19) 0.000 194 497 216 512 × 2 = 0 + 0.000 388 994 433 024;
  • 20) 0.000 388 994 433 024 × 2 = 0 + 0.000 777 988 866 048;
  • 21) 0.000 777 988 866 048 × 2 = 0 + 0.001 555 977 732 096;
  • 22) 0.001 555 977 732 096 × 2 = 0 + 0.003 111 955 464 192;
  • 23) 0.003 111 955 464 192 × 2 = 0 + 0.006 223 910 928 384;
  • 24) 0.006 223 910 928 384 × 2 = 0 + 0.012 447 821 856 768;
  • 25) 0.012 447 821 856 768 × 2 = 0 + 0.024 895 643 713 536;
  • 26) 0.024 895 643 713 536 × 2 = 0 + 0.049 791 287 427 072;
  • 27) 0.049 791 287 427 072 × 2 = 0 + 0.099 582 574 854 144;
  • 28) 0.099 582 574 854 144 × 2 = 0 + 0.199 165 149 708 288;
  • 29) 0.199 165 149 708 288 × 2 = 0 + 0.398 330 299 416 576;
  • 30) 0.398 330 299 416 576 × 2 = 0 + 0.796 660 598 833 152;
  • 31) 0.796 660 598 833 152 × 2 = 1 + 0.593 321 197 666 304;
  • 32) 0.593 321 197 666 304 × 2 = 1 + 0.186 642 395 332 608;
  • 33) 0.186 642 395 332 608 × 2 = 0 + 0.373 284 790 665 216;
  • 34) 0.373 284 790 665 216 × 2 = 0 + 0.746 569 581 330 432;
  • 35) 0.746 569 581 330 432 × 2 = 1 + 0.493 139 162 660 864;
  • 36) 0.493 139 162 660 864 × 2 = 0 + 0.986 278 325 321 728;
  • 37) 0.986 278 325 321 728 × 2 = 1 + 0.972 556 650 643 456;
  • 38) 0.972 556 650 643 456 × 2 = 1 + 0.945 113 301 286 912;
  • 39) 0.945 113 301 286 912 × 2 = 1 + 0.890 226 602 573 824;
  • 40) 0.890 226 602 573 824 × 2 = 1 + 0.780 453 205 147 648;
  • 41) 0.780 453 205 147 648 × 2 = 1 + 0.560 906 410 295 296;
  • 42) 0.560 906 410 295 296 × 2 = 1 + 0.121 812 820 590 592;
  • 43) 0.121 812 820 590 592 × 2 = 0 + 0.243 625 641 181 184;
  • 44) 0.243 625 641 181 184 × 2 = 0 + 0.487 251 282 362 368;
  • 45) 0.487 251 282 362 368 × 2 = 0 + 0.974 502 564 724 736;
  • 46) 0.974 502 564 724 736 × 2 = 1 + 0.949 005 129 449 472;
  • 47) 0.949 005 129 449 472 × 2 = 1 + 0.898 010 258 898 944;
  • 48) 0.898 010 258 898 944 × 2 = 1 + 0.796 020 517 797 888;
  • 49) 0.796 020 517 797 888 × 2 = 1 + 0.592 041 035 595 776;
  • 50) 0.592 041 035 595 776 × 2 = 1 + 0.184 082 071 191 552;
  • 51) 0.184 082 071 191 552 × 2 = 0 + 0.368 164 142 383 104;
  • 52) 0.368 164 142 383 104 × 2 = 0 + 0.736 328 284 766 208;
  • 53) 0.736 328 284 766 208 × 2 = 1 + 0.472 656 569 532 416;
  • 54) 0.472 656 569 532 416 × 2 = 0 + 0.945 313 139 064 832;

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:


0.000 000 000 741 948(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 0111 1100 10(2)

6. Positive number before normalization:

0.000 000 000 741 948(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 0111 1100 10(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:


0.000 000 000 741 948(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 0111 1100 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 0111 1100 10(2) × 20 =

1.1001 0111 1110 0011 1110 010(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 0111 1110 0011 1110 010


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 1011 1111 0001 1111 0010 =

100 1011 1111 0001 1111 0010


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 1011 1111 0001 1111 0010


Decimal number -0.000 000 000 741 948 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 0001 1111 0010

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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