-0.000 000 000 742 147 684 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 684 1(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
-0.000 000 000 742 147 684 1(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:

|-0.000 000 000 742 147 684 1| = 0.000 000 000 742 147 684 1


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: 0.000 000 000 742 147 684 1.

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) 0.000 000 000 742 147 684 1 × 2 = 0 + 0.000 000 001 484 295 368 2;
  • 2) 0.000 000 001 484 295 368 2 × 2 = 0 + 0.000 000 002 968 590 736 4;
  • 3) 0.000 000 002 968 590 736 4 × 2 = 0 + 0.000 000 005 937 181 472 8;
  • 4) 0.000 000 005 937 181 472 8 × 2 = 0 + 0.000 000 011 874 362 945 6;
  • 5) 0.000 000 011 874 362 945 6 × 2 = 0 + 0.000 000 023 748 725 891 2;
  • 6) 0.000 000 023 748 725 891 2 × 2 = 0 + 0.000 000 047 497 451 782 4;
  • 7) 0.000 000 047 497 451 782 4 × 2 = 0 + 0.000 000 094 994 903 564 8;
  • 8) 0.000 000 094 994 903 564 8 × 2 = 0 + 0.000 000 189 989 807 129 6;
  • 9) 0.000 000 189 989 807 129 6 × 2 = 0 + 0.000 000 379 979 614 259 2;
  • 10) 0.000 000 379 979 614 259 2 × 2 = 0 + 0.000 000 759 959 228 518 4;
  • 11) 0.000 000 759 959 228 518 4 × 2 = 0 + 0.000 001 519 918 457 036 8;
  • 12) 0.000 001 519 918 457 036 8 × 2 = 0 + 0.000 003 039 836 914 073 6;
  • 13) 0.000 003 039 836 914 073 6 × 2 = 0 + 0.000 006 079 673 828 147 2;
  • 14) 0.000 006 079 673 828 147 2 × 2 = 0 + 0.000 012 159 347 656 294 4;
  • 15) 0.000 012 159 347 656 294 4 × 2 = 0 + 0.000 024 318 695 312 588 8;
  • 16) 0.000 024 318 695 312 588 8 × 2 = 0 + 0.000 048 637 390 625 177 6;
  • 17) 0.000 048 637 390 625 177 6 × 2 = 0 + 0.000 097 274 781 250 355 2;
  • 18) 0.000 097 274 781 250 355 2 × 2 = 0 + 0.000 194 549 562 500 710 4;
  • 19) 0.000 194 549 562 500 710 4 × 2 = 0 + 0.000 389 099 125 001 420 8;
  • 20) 0.000 389 099 125 001 420 8 × 2 = 0 + 0.000 778 198 250 002 841 6;
  • 21) 0.000 778 198 250 002 841 6 × 2 = 0 + 0.001 556 396 500 005 683 2;
  • 22) 0.001 556 396 500 005 683 2 × 2 = 0 + 0.003 112 793 000 011 366 4;
  • 23) 0.003 112 793 000 011 366 4 × 2 = 0 + 0.006 225 586 000 022 732 8;
  • 24) 0.006 225 586 000 022 732 8 × 2 = 0 + 0.012 451 172 000 045 465 6;
  • 25) 0.012 451 172 000 045 465 6 × 2 = 0 + 0.024 902 344 000 090 931 2;
  • 26) 0.024 902 344 000 090 931 2 × 2 = 0 + 0.049 804 688 000 181 862 4;
  • 27) 0.049 804 688 000 181 862 4 × 2 = 0 + 0.099 609 376 000 363 724 8;
  • 28) 0.099 609 376 000 363 724 8 × 2 = 0 + 0.199 218 752 000 727 449 6;
  • 29) 0.199 218 752 000 727 449 6 × 2 = 0 + 0.398 437 504 001 454 899 2;
  • 30) 0.398 437 504 001 454 899 2 × 2 = 0 + 0.796 875 008 002 909 798 4;
  • 31) 0.796 875 008 002 909 798 4 × 2 = 1 + 0.593 750 016 005 819 596 8;
  • 32) 0.593 750 016 005 819 596 8 × 2 = 1 + 0.187 500 032 011 639 193 6;
  • 33) 0.187 500 032 011 639 193 6 × 2 = 0 + 0.375 000 064 023 278 387 2;
  • 34) 0.375 000 064 023 278 387 2 × 2 = 0 + 0.750 000 128 046 556 774 4;
  • 35) 0.750 000 128 046 556 774 4 × 2 = 1 + 0.500 000 256 093 113 548 8;
  • 36) 0.500 000 256 093 113 548 8 × 2 = 1 + 0.000 000 512 186 227 097 6;
  • 37) 0.000 000 512 186 227 097 6 × 2 = 0 + 0.000 001 024 372 454 195 2;
  • 38) 0.000 001 024 372 454 195 2 × 2 = 0 + 0.000 002 048 744 908 390 4;
  • 39) 0.000 002 048 744 908 390 4 × 2 = 0 + 0.000 004 097 489 816 780 8;
  • 40) 0.000 004 097 489 816 780 8 × 2 = 0 + 0.000 008 194 979 633 561 6;
  • 41) 0.000 008 194 979 633 561 6 × 2 = 0 + 0.000 016 389 959 267 123 2;
  • 42) 0.000 016 389 959 267 123 2 × 2 = 0 + 0.000 032 779 918 534 246 4;
  • 43) 0.000 032 779 918 534 246 4 × 2 = 0 + 0.000 065 559 837 068 492 8;
  • 44) 0.000 065 559 837 068 492 8 × 2 = 0 + 0.000 131 119 674 136 985 6;
  • 45) 0.000 131 119 674 136 985 6 × 2 = 0 + 0.000 262 239 348 273 971 2;
  • 46) 0.000 262 239 348 273 971 2 × 2 = 0 + 0.000 524 478 696 547 942 4;
  • 47) 0.000 524 478 696 547 942 4 × 2 = 0 + 0.001 048 957 393 095 884 8;
  • 48) 0.001 048 957 393 095 884 8 × 2 = 0 + 0.002 097 914 786 191 769 6;
  • 49) 0.002 097 914 786 191 769 6 × 2 = 0 + 0.004 195 829 572 383 539 2;
  • 50) 0.004 195 829 572 383 539 2 × 2 = 0 + 0.008 391 659 144 767 078 4;
  • 51) 0.008 391 659 144 767 078 4 × 2 = 0 + 0.016 783 318 289 534 156 8;
  • 52) 0.016 783 318 289 534 156 8 × 2 = 0 + 0.033 566 636 579 068 313 6;
  • 53) 0.033 566 636 579 068 313 6 × 2 = 0 + 0.067 133 273 158 136 627 2;
  • 54) 0.067 133 273 158 136 627 2 × 2 = 0 + 0.134 266 546 316 273 254 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:


0.000 000 000 742 147 684 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 684 1(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:


0.000 000 000 742 147 684 1(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 -0.000 000 000 742 147 684 1 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