-0.000 000 000 742 148 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 9(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
-0.000 000 000 742 148 9(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:

|-0.000 000 000 742 148 9| = 0.000 000 000 742 148 9


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: 0.000 000 000 742 148 9.

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) 0.000 000 000 742 148 9 × 2 = 0 + 0.000 000 001 484 297 8;
  • 2) 0.000 000 001 484 297 8 × 2 = 0 + 0.000 000 002 968 595 6;
  • 3) 0.000 000 002 968 595 6 × 2 = 0 + 0.000 000 005 937 191 2;
  • 4) 0.000 000 005 937 191 2 × 2 = 0 + 0.000 000 011 874 382 4;
  • 5) 0.000 000 011 874 382 4 × 2 = 0 + 0.000 000 023 748 764 8;
  • 6) 0.000 000 023 748 764 8 × 2 = 0 + 0.000 000 047 497 529 6;
  • 7) 0.000 000 047 497 529 6 × 2 = 0 + 0.000 000 094 995 059 2;
  • 8) 0.000 000 094 995 059 2 × 2 = 0 + 0.000 000 189 990 118 4;
  • 9) 0.000 000 189 990 118 4 × 2 = 0 + 0.000 000 379 980 236 8;
  • 10) 0.000 000 379 980 236 8 × 2 = 0 + 0.000 000 759 960 473 6;
  • 11) 0.000 000 759 960 473 6 × 2 = 0 + 0.000 001 519 920 947 2;
  • 12) 0.000 001 519 920 947 2 × 2 = 0 + 0.000 003 039 841 894 4;
  • 13) 0.000 003 039 841 894 4 × 2 = 0 + 0.000 006 079 683 788 8;
  • 14) 0.000 006 079 683 788 8 × 2 = 0 + 0.000 012 159 367 577 6;
  • 15) 0.000 012 159 367 577 6 × 2 = 0 + 0.000 024 318 735 155 2;
  • 16) 0.000 024 318 735 155 2 × 2 = 0 + 0.000 048 637 470 310 4;
  • 17) 0.000 048 637 470 310 4 × 2 = 0 + 0.000 097 274 940 620 8;
  • 18) 0.000 097 274 940 620 8 × 2 = 0 + 0.000 194 549 881 241 6;
  • 19) 0.000 194 549 881 241 6 × 2 = 0 + 0.000 389 099 762 483 2;
  • 20) 0.000 389 099 762 483 2 × 2 = 0 + 0.000 778 199 524 966 4;
  • 21) 0.000 778 199 524 966 4 × 2 = 0 + 0.001 556 399 049 932 8;
  • 22) 0.001 556 399 049 932 8 × 2 = 0 + 0.003 112 798 099 865 6;
  • 23) 0.003 112 798 099 865 6 × 2 = 0 + 0.006 225 596 199 731 2;
  • 24) 0.006 225 596 199 731 2 × 2 = 0 + 0.012 451 192 399 462 4;
  • 25) 0.012 451 192 399 462 4 × 2 = 0 + 0.024 902 384 798 924 8;
  • 26) 0.024 902 384 798 924 8 × 2 = 0 + 0.049 804 769 597 849 6;
  • 27) 0.049 804 769 597 849 6 × 2 = 0 + 0.099 609 539 195 699 2;
  • 28) 0.099 609 539 195 699 2 × 2 = 0 + 0.199 219 078 391 398 4;
  • 29) 0.199 219 078 391 398 4 × 2 = 0 + 0.398 438 156 782 796 8;
  • 30) 0.398 438 156 782 796 8 × 2 = 0 + 0.796 876 313 565 593 6;
  • 31) 0.796 876 313 565 593 6 × 2 = 1 + 0.593 752 627 131 187 2;
  • 32) 0.593 752 627 131 187 2 × 2 = 1 + 0.187 505 254 262 374 4;
  • 33) 0.187 505 254 262 374 4 × 2 = 0 + 0.375 010 508 524 748 8;
  • 34) 0.375 010 508 524 748 8 × 2 = 0 + 0.750 021 017 049 497 6;
  • 35) 0.750 021 017 049 497 6 × 2 = 1 + 0.500 042 034 098 995 2;
  • 36) 0.500 042 034 098 995 2 × 2 = 1 + 0.000 084 068 197 990 4;
  • 37) 0.000 084 068 197 990 4 × 2 = 0 + 0.000 168 136 395 980 8;
  • 38) 0.000 168 136 395 980 8 × 2 = 0 + 0.000 336 272 791 961 6;
  • 39) 0.000 336 272 791 961 6 × 2 = 0 + 0.000 672 545 583 923 2;
  • 40) 0.000 672 545 583 923 2 × 2 = 0 + 0.001 345 091 167 846 4;
  • 41) 0.001 345 091 167 846 4 × 2 = 0 + 0.002 690 182 335 692 8;
  • 42) 0.002 690 182 335 692 8 × 2 = 0 + 0.005 380 364 671 385 6;
  • 43) 0.005 380 364 671 385 6 × 2 = 0 + 0.010 760 729 342 771 2;
  • 44) 0.010 760 729 342 771 2 × 2 = 0 + 0.021 521 458 685 542 4;
  • 45) 0.021 521 458 685 542 4 × 2 = 0 + 0.043 042 917 371 084 8;
  • 46) 0.043 042 917 371 084 8 × 2 = 0 + 0.086 085 834 742 169 6;
  • 47) 0.086 085 834 742 169 6 × 2 = 0 + 0.172 171 669 484 339 2;
  • 48) 0.172 171 669 484 339 2 × 2 = 0 + 0.344 343 338 968 678 4;
  • 49) 0.344 343 338 968 678 4 × 2 = 0 + 0.688 686 677 937 356 8;
  • 50) 0.688 686 677 937 356 8 × 2 = 1 + 0.377 373 355 874 713 6;
  • 51) 0.377 373 355 874 713 6 × 2 = 0 + 0.754 746 711 749 427 2;
  • 52) 0.754 746 711 749 427 2 × 2 = 1 + 0.509 493 423 498 854 4;
  • 53) 0.509 493 423 498 854 4 × 2 = 1 + 0.018 986 846 997 708 8;
  • 54) 0.018 986 846 997 708 8 × 2 = 0 + 0.037 973 693 995 417 6;

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:


0.000 000 000 742 148 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0101 10(2)

6. Positive number before normalization:

0.000 000 000 742 148 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0101 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:


0.000 000 000 742 148 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0101 10(2) × 20 =

1.1001 1000 0000 0000 0010 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 1000 0000 0000 0010 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 1100 0000 0000 0001 0110 =

100 1100 0000 0000 0001 0110


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 0110


Decimal number -0.000 000 000 742 148 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0001 0110

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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