-0.000 000 000 742 149 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 149 2(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
-0.000 000 000 742 149 2(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:

|-0.000 000 000 742 149 2| = 0.000 000 000 742 149 2


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: 0.000 000 000 742 149 2.

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) 0.000 000 000 742 149 2 × 2 = 0 + 0.000 000 001 484 298 4;
  • 2) 0.000 000 001 484 298 4 × 2 = 0 + 0.000 000 002 968 596 8;
  • 3) 0.000 000 002 968 596 8 × 2 = 0 + 0.000 000 005 937 193 6;
  • 4) 0.000 000 005 937 193 6 × 2 = 0 + 0.000 000 011 874 387 2;
  • 5) 0.000 000 011 874 387 2 × 2 = 0 + 0.000 000 023 748 774 4;
  • 6) 0.000 000 023 748 774 4 × 2 = 0 + 0.000 000 047 497 548 8;
  • 7) 0.000 000 047 497 548 8 × 2 = 0 + 0.000 000 094 995 097 6;
  • 8) 0.000 000 094 995 097 6 × 2 = 0 + 0.000 000 189 990 195 2;
  • 9) 0.000 000 189 990 195 2 × 2 = 0 + 0.000 000 379 980 390 4;
  • 10) 0.000 000 379 980 390 4 × 2 = 0 + 0.000 000 759 960 780 8;
  • 11) 0.000 000 759 960 780 8 × 2 = 0 + 0.000 001 519 921 561 6;
  • 12) 0.000 001 519 921 561 6 × 2 = 0 + 0.000 003 039 843 123 2;
  • 13) 0.000 003 039 843 123 2 × 2 = 0 + 0.000 006 079 686 246 4;
  • 14) 0.000 006 079 686 246 4 × 2 = 0 + 0.000 012 159 372 492 8;
  • 15) 0.000 012 159 372 492 8 × 2 = 0 + 0.000 024 318 744 985 6;
  • 16) 0.000 024 318 744 985 6 × 2 = 0 + 0.000 048 637 489 971 2;
  • 17) 0.000 048 637 489 971 2 × 2 = 0 + 0.000 097 274 979 942 4;
  • 18) 0.000 097 274 979 942 4 × 2 = 0 + 0.000 194 549 959 884 8;
  • 19) 0.000 194 549 959 884 8 × 2 = 0 + 0.000 389 099 919 769 6;
  • 20) 0.000 389 099 919 769 6 × 2 = 0 + 0.000 778 199 839 539 2;
  • 21) 0.000 778 199 839 539 2 × 2 = 0 + 0.001 556 399 679 078 4;
  • 22) 0.001 556 399 679 078 4 × 2 = 0 + 0.003 112 799 358 156 8;
  • 23) 0.003 112 799 358 156 8 × 2 = 0 + 0.006 225 598 716 313 6;
  • 24) 0.006 225 598 716 313 6 × 2 = 0 + 0.012 451 197 432 627 2;
  • 25) 0.012 451 197 432 627 2 × 2 = 0 + 0.024 902 394 865 254 4;
  • 26) 0.024 902 394 865 254 4 × 2 = 0 + 0.049 804 789 730 508 8;
  • 27) 0.049 804 789 730 508 8 × 2 = 0 + 0.099 609 579 461 017 6;
  • 28) 0.099 609 579 461 017 6 × 2 = 0 + 0.199 219 158 922 035 2;
  • 29) 0.199 219 158 922 035 2 × 2 = 0 + 0.398 438 317 844 070 4;
  • 30) 0.398 438 317 844 070 4 × 2 = 0 + 0.796 876 635 688 140 8;
  • 31) 0.796 876 635 688 140 8 × 2 = 1 + 0.593 753 271 376 281 6;
  • 32) 0.593 753 271 376 281 6 × 2 = 1 + 0.187 506 542 752 563 2;
  • 33) 0.187 506 542 752 563 2 × 2 = 0 + 0.375 013 085 505 126 4;
  • 34) 0.375 013 085 505 126 4 × 2 = 0 + 0.750 026 171 010 252 8;
  • 35) 0.750 026 171 010 252 8 × 2 = 1 + 0.500 052 342 020 505 6;
  • 36) 0.500 052 342 020 505 6 × 2 = 1 + 0.000 104 684 041 011 2;
  • 37) 0.000 104 684 041 011 2 × 2 = 0 + 0.000 209 368 082 022 4;
  • 38) 0.000 209 368 082 022 4 × 2 = 0 + 0.000 418 736 164 044 8;
  • 39) 0.000 418 736 164 044 8 × 2 = 0 + 0.000 837 472 328 089 6;
  • 40) 0.000 837 472 328 089 6 × 2 = 0 + 0.001 674 944 656 179 2;
  • 41) 0.001 674 944 656 179 2 × 2 = 0 + 0.003 349 889 312 358 4;
  • 42) 0.003 349 889 312 358 4 × 2 = 0 + 0.006 699 778 624 716 8;
  • 43) 0.006 699 778 624 716 8 × 2 = 0 + 0.013 399 557 249 433 6;
  • 44) 0.013 399 557 249 433 6 × 2 = 0 + 0.026 799 114 498 867 2;
  • 45) 0.026 799 114 498 867 2 × 2 = 0 + 0.053 598 228 997 734 4;
  • 46) 0.053 598 228 997 734 4 × 2 = 0 + 0.107 196 457 995 468 8;
  • 47) 0.107 196 457 995 468 8 × 2 = 0 + 0.214 392 915 990 937 6;
  • 48) 0.214 392 915 990 937 6 × 2 = 0 + 0.428 785 831 981 875 2;
  • 49) 0.428 785 831 981 875 2 × 2 = 0 + 0.857 571 663 963 750 4;
  • 50) 0.857 571 663 963 750 4 × 2 = 1 + 0.715 143 327 927 500 8;
  • 51) 0.715 143 327 927 500 8 × 2 = 1 + 0.430 286 655 855 001 6;
  • 52) 0.430 286 655 855 001 6 × 2 = 0 + 0.860 573 311 710 003 2;
  • 53) 0.860 573 311 710 003 2 × 2 = 1 + 0.721 146 623 420 006 4;
  • 54) 0.721 146 623 420 006 4 × 2 = 1 + 0.442 293 246 840 012 8;

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:


0.000 000 000 742 149 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 11(2)

6. Positive number before normalization:

0.000 000 000 742 149 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 11(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:


0.000 000 000 742 149 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 11(2) × 20 =

1.1001 1000 0000 0000 0011 011(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 0011 011


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 0001 1011 =

100 1100 0000 0000 0001 1011


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 0001 1011


Decimal number -0.000 000 000 742 149 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0001 1011

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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