-0.000 000 000 742 147 725 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 725(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
-0.000 000 000 742 147 725(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:

|-0.000 000 000 742 147 725| = 0.000 000 000 742 147 725


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: 0.000 000 000 742 147 725.

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) 0.000 000 000 742 147 725 × 2 = 0 + 0.000 000 001 484 295 45;
  • 2) 0.000 000 001 484 295 45 × 2 = 0 + 0.000 000 002 968 590 9;
  • 3) 0.000 000 002 968 590 9 × 2 = 0 + 0.000 000 005 937 181 8;
  • 4) 0.000 000 005 937 181 8 × 2 = 0 + 0.000 000 011 874 363 6;
  • 5) 0.000 000 011 874 363 6 × 2 = 0 + 0.000 000 023 748 727 2;
  • 6) 0.000 000 023 748 727 2 × 2 = 0 + 0.000 000 047 497 454 4;
  • 7) 0.000 000 047 497 454 4 × 2 = 0 + 0.000 000 094 994 908 8;
  • 8) 0.000 000 094 994 908 8 × 2 = 0 + 0.000 000 189 989 817 6;
  • 9) 0.000 000 189 989 817 6 × 2 = 0 + 0.000 000 379 979 635 2;
  • 10) 0.000 000 379 979 635 2 × 2 = 0 + 0.000 000 759 959 270 4;
  • 11) 0.000 000 759 959 270 4 × 2 = 0 + 0.000 001 519 918 540 8;
  • 12) 0.000 001 519 918 540 8 × 2 = 0 + 0.000 003 039 837 081 6;
  • 13) 0.000 003 039 837 081 6 × 2 = 0 + 0.000 006 079 674 163 2;
  • 14) 0.000 006 079 674 163 2 × 2 = 0 + 0.000 012 159 348 326 4;
  • 15) 0.000 012 159 348 326 4 × 2 = 0 + 0.000 024 318 696 652 8;
  • 16) 0.000 024 318 696 652 8 × 2 = 0 + 0.000 048 637 393 305 6;
  • 17) 0.000 048 637 393 305 6 × 2 = 0 + 0.000 097 274 786 611 2;
  • 18) 0.000 097 274 786 611 2 × 2 = 0 + 0.000 194 549 573 222 4;
  • 19) 0.000 194 549 573 222 4 × 2 = 0 + 0.000 389 099 146 444 8;
  • 20) 0.000 389 099 146 444 8 × 2 = 0 + 0.000 778 198 292 889 6;
  • 21) 0.000 778 198 292 889 6 × 2 = 0 + 0.001 556 396 585 779 2;
  • 22) 0.001 556 396 585 779 2 × 2 = 0 + 0.003 112 793 171 558 4;
  • 23) 0.003 112 793 171 558 4 × 2 = 0 + 0.006 225 586 343 116 8;
  • 24) 0.006 225 586 343 116 8 × 2 = 0 + 0.012 451 172 686 233 6;
  • 25) 0.012 451 172 686 233 6 × 2 = 0 + 0.024 902 345 372 467 2;
  • 26) 0.024 902 345 372 467 2 × 2 = 0 + 0.049 804 690 744 934 4;
  • 27) 0.049 804 690 744 934 4 × 2 = 0 + 0.099 609 381 489 868 8;
  • 28) 0.099 609 381 489 868 8 × 2 = 0 + 0.199 218 762 979 737 6;
  • 29) 0.199 218 762 979 737 6 × 2 = 0 + 0.398 437 525 959 475 2;
  • 30) 0.398 437 525 959 475 2 × 2 = 0 + 0.796 875 051 918 950 4;
  • 31) 0.796 875 051 918 950 4 × 2 = 1 + 0.593 750 103 837 900 8;
  • 32) 0.593 750 103 837 900 8 × 2 = 1 + 0.187 500 207 675 801 6;
  • 33) 0.187 500 207 675 801 6 × 2 = 0 + 0.375 000 415 351 603 2;
  • 34) 0.375 000 415 351 603 2 × 2 = 0 + 0.750 000 830 703 206 4;
  • 35) 0.750 000 830 703 206 4 × 2 = 1 + 0.500 001 661 406 412 8;
  • 36) 0.500 001 661 406 412 8 × 2 = 1 + 0.000 003 322 812 825 6;
  • 37) 0.000 003 322 812 825 6 × 2 = 0 + 0.000 006 645 625 651 2;
  • 38) 0.000 006 645 625 651 2 × 2 = 0 + 0.000 013 291 251 302 4;
  • 39) 0.000 013 291 251 302 4 × 2 = 0 + 0.000 026 582 502 604 8;
  • 40) 0.000 026 582 502 604 8 × 2 = 0 + 0.000 053 165 005 209 6;
  • 41) 0.000 053 165 005 209 6 × 2 = 0 + 0.000 106 330 010 419 2;
  • 42) 0.000 106 330 010 419 2 × 2 = 0 + 0.000 212 660 020 838 4;
  • 43) 0.000 212 660 020 838 4 × 2 = 0 + 0.000 425 320 041 676 8;
  • 44) 0.000 425 320 041 676 8 × 2 = 0 + 0.000 850 640 083 353 6;
  • 45) 0.000 850 640 083 353 6 × 2 = 0 + 0.001 701 280 166 707 2;
  • 46) 0.001 701 280 166 707 2 × 2 = 0 + 0.003 402 560 333 414 4;
  • 47) 0.003 402 560 333 414 4 × 2 = 0 + 0.006 805 120 666 828 8;
  • 48) 0.006 805 120 666 828 8 × 2 = 0 + 0.013 610 241 333 657 6;
  • 49) 0.013 610 241 333 657 6 × 2 = 0 + 0.027 220 482 667 315 2;
  • 50) 0.027 220 482 667 315 2 × 2 = 0 + 0.054 440 965 334 630 4;
  • 51) 0.054 440 965 334 630 4 × 2 = 0 + 0.108 881 930 669 260 8;
  • 52) 0.108 881 930 669 260 8 × 2 = 0 + 0.217 763 861 338 521 6;
  • 53) 0.217 763 861 338 521 6 × 2 = 0 + 0.435 527 722 677 043 2;
  • 54) 0.435 527 722 677 043 2 × 2 = 0 + 0.871 055 445 354 086 4;

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:


0.000 000 000 742 147 725(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 725(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:


0.000 000 000 742 147 725(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 -0.000 000 000 742 147 725 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