-0.000 000 000 742 041 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 041(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
-0.000 000 000 742 041(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:

|-0.000 000 000 742 041| = 0.000 000 000 742 041


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: 0.000 000 000 742 041.

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) 0.000 000 000 742 041 × 2 = 0 + 0.000 000 001 484 082;
  • 2) 0.000 000 001 484 082 × 2 = 0 + 0.000 000 002 968 164;
  • 3) 0.000 000 002 968 164 × 2 = 0 + 0.000 000 005 936 328;
  • 4) 0.000 000 005 936 328 × 2 = 0 + 0.000 000 011 872 656;
  • 5) 0.000 000 011 872 656 × 2 = 0 + 0.000 000 023 745 312;
  • 6) 0.000 000 023 745 312 × 2 = 0 + 0.000 000 047 490 624;
  • 7) 0.000 000 047 490 624 × 2 = 0 + 0.000 000 094 981 248;
  • 8) 0.000 000 094 981 248 × 2 = 0 + 0.000 000 189 962 496;
  • 9) 0.000 000 189 962 496 × 2 = 0 + 0.000 000 379 924 992;
  • 10) 0.000 000 379 924 992 × 2 = 0 + 0.000 000 759 849 984;
  • 11) 0.000 000 759 849 984 × 2 = 0 + 0.000 001 519 699 968;
  • 12) 0.000 001 519 699 968 × 2 = 0 + 0.000 003 039 399 936;
  • 13) 0.000 003 039 399 936 × 2 = 0 + 0.000 006 078 799 872;
  • 14) 0.000 006 078 799 872 × 2 = 0 + 0.000 012 157 599 744;
  • 15) 0.000 012 157 599 744 × 2 = 0 + 0.000 024 315 199 488;
  • 16) 0.000 024 315 199 488 × 2 = 0 + 0.000 048 630 398 976;
  • 17) 0.000 048 630 398 976 × 2 = 0 + 0.000 097 260 797 952;
  • 18) 0.000 097 260 797 952 × 2 = 0 + 0.000 194 521 595 904;
  • 19) 0.000 194 521 595 904 × 2 = 0 + 0.000 389 043 191 808;
  • 20) 0.000 389 043 191 808 × 2 = 0 + 0.000 778 086 383 616;
  • 21) 0.000 778 086 383 616 × 2 = 0 + 0.001 556 172 767 232;
  • 22) 0.001 556 172 767 232 × 2 = 0 + 0.003 112 345 534 464;
  • 23) 0.003 112 345 534 464 × 2 = 0 + 0.006 224 691 068 928;
  • 24) 0.006 224 691 068 928 × 2 = 0 + 0.012 449 382 137 856;
  • 25) 0.012 449 382 137 856 × 2 = 0 + 0.024 898 764 275 712;
  • 26) 0.024 898 764 275 712 × 2 = 0 + 0.049 797 528 551 424;
  • 27) 0.049 797 528 551 424 × 2 = 0 + 0.099 595 057 102 848;
  • 28) 0.099 595 057 102 848 × 2 = 0 + 0.199 190 114 205 696;
  • 29) 0.199 190 114 205 696 × 2 = 0 + 0.398 380 228 411 392;
  • 30) 0.398 380 228 411 392 × 2 = 0 + 0.796 760 456 822 784;
  • 31) 0.796 760 456 822 784 × 2 = 1 + 0.593 520 913 645 568;
  • 32) 0.593 520 913 645 568 × 2 = 1 + 0.187 041 827 291 136;
  • 33) 0.187 041 827 291 136 × 2 = 0 + 0.374 083 654 582 272;
  • 34) 0.374 083 654 582 272 × 2 = 0 + 0.748 167 309 164 544;
  • 35) 0.748 167 309 164 544 × 2 = 1 + 0.496 334 618 329 088;
  • 36) 0.496 334 618 329 088 × 2 = 0 + 0.992 669 236 658 176;
  • 37) 0.992 669 236 658 176 × 2 = 1 + 0.985 338 473 316 352;
  • 38) 0.985 338 473 316 352 × 2 = 1 + 0.970 676 946 632 704;
  • 39) 0.970 676 946 632 704 × 2 = 1 + 0.941 353 893 265 408;
  • 40) 0.941 353 893 265 408 × 2 = 1 + 0.882 707 786 530 816;
  • 41) 0.882 707 786 530 816 × 2 = 1 + 0.765 415 573 061 632;
  • 42) 0.765 415 573 061 632 × 2 = 1 + 0.530 831 146 123 264;
  • 43) 0.530 831 146 123 264 × 2 = 1 + 0.061 662 292 246 528;
  • 44) 0.061 662 292 246 528 × 2 = 0 + 0.123 324 584 493 056;
  • 45) 0.123 324 584 493 056 × 2 = 0 + 0.246 649 168 986 112;
  • 46) 0.246 649 168 986 112 × 2 = 0 + 0.493 298 337 972 224;
  • 47) 0.493 298 337 972 224 × 2 = 0 + 0.986 596 675 944 448;
  • 48) 0.986 596 675 944 448 × 2 = 1 + 0.973 193 351 888 896;
  • 49) 0.973 193 351 888 896 × 2 = 1 + 0.946 386 703 777 792;
  • 50) 0.946 386 703 777 792 × 2 = 1 + 0.892 773 407 555 584;
  • 51) 0.892 773 407 555 584 × 2 = 1 + 0.785 546 815 111 168;
  • 52) 0.785 546 815 111 168 × 2 = 1 + 0.571 093 630 222 336;
  • 53) 0.571 093 630 222 336 × 2 = 1 + 0.142 187 260 444 672;
  • 54) 0.142 187 260 444 672 × 2 = 0 + 0.284 374 520 889 344;

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:


0.000 000 000 742 041(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 0001 1111 10(2)

6. Positive number before normalization:

0.000 000 000 742 041(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 0001 1111 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:


0.000 000 000 742 041(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 0001 1111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 0001 1111 10(2) × 20 =

1.1001 0111 1111 0000 1111 110(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 1111 0000 1111 110


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 1000 0111 1110 =

100 1011 1111 1000 0111 1110


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 1000 0111 1110


Decimal number -0.000 000 000 742 041 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1000 0111 1110

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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