-0.000 000 000 742 147 678 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 678 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
-0.000 000 000 742 147 678 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:

|-0.000 000 000 742 147 678 9| = 0.000 000 000 742 147 678 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: 0.000 000 000 742 147 678 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) 0.000 000 000 742 147 678 9 × 2 = 0 + 0.000 000 001 484 295 357 8;
  • 2) 0.000 000 001 484 295 357 8 × 2 = 0 + 0.000 000 002 968 590 715 6;
  • 3) 0.000 000 002 968 590 715 6 × 2 = 0 + 0.000 000 005 937 181 431 2;
  • 4) 0.000 000 005 937 181 431 2 × 2 = 0 + 0.000 000 011 874 362 862 4;
  • 5) 0.000 000 011 874 362 862 4 × 2 = 0 + 0.000 000 023 748 725 724 8;
  • 6) 0.000 000 023 748 725 724 8 × 2 = 0 + 0.000 000 047 497 451 449 6;
  • 7) 0.000 000 047 497 451 449 6 × 2 = 0 + 0.000 000 094 994 902 899 2;
  • 8) 0.000 000 094 994 902 899 2 × 2 = 0 + 0.000 000 189 989 805 798 4;
  • 9) 0.000 000 189 989 805 798 4 × 2 = 0 + 0.000 000 379 979 611 596 8;
  • 10) 0.000 000 379 979 611 596 8 × 2 = 0 + 0.000 000 759 959 223 193 6;
  • 11) 0.000 000 759 959 223 193 6 × 2 = 0 + 0.000 001 519 918 446 387 2;
  • 12) 0.000 001 519 918 446 387 2 × 2 = 0 + 0.000 003 039 836 892 774 4;
  • 13) 0.000 003 039 836 892 774 4 × 2 = 0 + 0.000 006 079 673 785 548 8;
  • 14) 0.000 006 079 673 785 548 8 × 2 = 0 + 0.000 012 159 347 571 097 6;
  • 15) 0.000 012 159 347 571 097 6 × 2 = 0 + 0.000 024 318 695 142 195 2;
  • 16) 0.000 024 318 695 142 195 2 × 2 = 0 + 0.000 048 637 390 284 390 4;
  • 17) 0.000 048 637 390 284 390 4 × 2 = 0 + 0.000 097 274 780 568 780 8;
  • 18) 0.000 097 274 780 568 780 8 × 2 = 0 + 0.000 194 549 561 137 561 6;
  • 19) 0.000 194 549 561 137 561 6 × 2 = 0 + 0.000 389 099 122 275 123 2;
  • 20) 0.000 389 099 122 275 123 2 × 2 = 0 + 0.000 778 198 244 550 246 4;
  • 21) 0.000 778 198 244 550 246 4 × 2 = 0 + 0.001 556 396 489 100 492 8;
  • 22) 0.001 556 396 489 100 492 8 × 2 = 0 + 0.003 112 792 978 200 985 6;
  • 23) 0.003 112 792 978 200 985 6 × 2 = 0 + 0.006 225 585 956 401 971 2;
  • 24) 0.006 225 585 956 401 971 2 × 2 = 0 + 0.012 451 171 912 803 942 4;
  • 25) 0.012 451 171 912 803 942 4 × 2 = 0 + 0.024 902 343 825 607 884 8;
  • 26) 0.024 902 343 825 607 884 8 × 2 = 0 + 0.049 804 687 651 215 769 6;
  • 27) 0.049 804 687 651 215 769 6 × 2 = 0 + 0.099 609 375 302 431 539 2;
  • 28) 0.099 609 375 302 431 539 2 × 2 = 0 + 0.199 218 750 604 863 078 4;
  • 29) 0.199 218 750 604 863 078 4 × 2 = 0 + 0.398 437 501 209 726 156 8;
  • 30) 0.398 437 501 209 726 156 8 × 2 = 0 + 0.796 875 002 419 452 313 6;
  • 31) 0.796 875 002 419 452 313 6 × 2 = 1 + 0.593 750 004 838 904 627 2;
  • 32) 0.593 750 004 838 904 627 2 × 2 = 1 + 0.187 500 009 677 809 254 4;
  • 33) 0.187 500 009 677 809 254 4 × 2 = 0 + 0.375 000 019 355 618 508 8;
  • 34) 0.375 000 019 355 618 508 8 × 2 = 0 + 0.750 000 038 711 237 017 6;
  • 35) 0.750 000 038 711 237 017 6 × 2 = 1 + 0.500 000 077 422 474 035 2;
  • 36) 0.500 000 077 422 474 035 2 × 2 = 1 + 0.000 000 154 844 948 070 4;
  • 37) 0.000 000 154 844 948 070 4 × 2 = 0 + 0.000 000 309 689 896 140 8;
  • 38) 0.000 000 309 689 896 140 8 × 2 = 0 + 0.000 000 619 379 792 281 6;
  • 39) 0.000 000 619 379 792 281 6 × 2 = 0 + 0.000 001 238 759 584 563 2;
  • 40) 0.000 001 238 759 584 563 2 × 2 = 0 + 0.000 002 477 519 169 126 4;
  • 41) 0.000 002 477 519 169 126 4 × 2 = 0 + 0.000 004 955 038 338 252 8;
  • 42) 0.000 004 955 038 338 252 8 × 2 = 0 + 0.000 009 910 076 676 505 6;
  • 43) 0.000 009 910 076 676 505 6 × 2 = 0 + 0.000 019 820 153 353 011 2;
  • 44) 0.000 019 820 153 353 011 2 × 2 = 0 + 0.000 039 640 306 706 022 4;
  • 45) 0.000 039 640 306 706 022 4 × 2 = 0 + 0.000 079 280 613 412 044 8;
  • 46) 0.000 079 280 613 412 044 8 × 2 = 0 + 0.000 158 561 226 824 089 6;
  • 47) 0.000 158 561 226 824 089 6 × 2 = 0 + 0.000 317 122 453 648 179 2;
  • 48) 0.000 317 122 453 648 179 2 × 2 = 0 + 0.000 634 244 907 296 358 4;
  • 49) 0.000 634 244 907 296 358 4 × 2 = 0 + 0.001 268 489 814 592 716 8;
  • 50) 0.001 268 489 814 592 716 8 × 2 = 0 + 0.002 536 979 629 185 433 6;
  • 51) 0.002 536 979 629 185 433 6 × 2 = 0 + 0.005 073 959 258 370 867 2;
  • 52) 0.005 073 959 258 370 867 2 × 2 = 0 + 0.010 147 918 516 741 734 4;
  • 53) 0.010 147 918 516 741 734 4 × 2 = 0 + 0.020 295 837 033 483 468 8;
  • 54) 0.020 295 837 033 483 468 8 × 2 = 0 + 0.040 591 674 066 966 937 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:


0.000 000 000 742 147 678 9(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 678 9(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:


0.000 000 000 742 147 678 9(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 -0.000 000 000 742 147 678 9 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