-0.000 000 000 742 147 678 36 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 678 36(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 36(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 36| = 0.000 000 000 742 147 678 36


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 36.

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 36 × 2 = 0 + 0.000 000 001 484 295 356 72;
  • 2) 0.000 000 001 484 295 356 72 × 2 = 0 + 0.000 000 002 968 590 713 44;
  • 3) 0.000 000 002 968 590 713 44 × 2 = 0 + 0.000 000 005 937 181 426 88;
  • 4) 0.000 000 005 937 181 426 88 × 2 = 0 + 0.000 000 011 874 362 853 76;
  • 5) 0.000 000 011 874 362 853 76 × 2 = 0 + 0.000 000 023 748 725 707 52;
  • 6) 0.000 000 023 748 725 707 52 × 2 = 0 + 0.000 000 047 497 451 415 04;
  • 7) 0.000 000 047 497 451 415 04 × 2 = 0 + 0.000 000 094 994 902 830 08;
  • 8) 0.000 000 094 994 902 830 08 × 2 = 0 + 0.000 000 189 989 805 660 16;
  • 9) 0.000 000 189 989 805 660 16 × 2 = 0 + 0.000 000 379 979 611 320 32;
  • 10) 0.000 000 379 979 611 320 32 × 2 = 0 + 0.000 000 759 959 222 640 64;
  • 11) 0.000 000 759 959 222 640 64 × 2 = 0 + 0.000 001 519 918 445 281 28;
  • 12) 0.000 001 519 918 445 281 28 × 2 = 0 + 0.000 003 039 836 890 562 56;
  • 13) 0.000 003 039 836 890 562 56 × 2 = 0 + 0.000 006 079 673 781 125 12;
  • 14) 0.000 006 079 673 781 125 12 × 2 = 0 + 0.000 012 159 347 562 250 24;
  • 15) 0.000 012 159 347 562 250 24 × 2 = 0 + 0.000 024 318 695 124 500 48;
  • 16) 0.000 024 318 695 124 500 48 × 2 = 0 + 0.000 048 637 390 249 000 96;
  • 17) 0.000 048 637 390 249 000 96 × 2 = 0 + 0.000 097 274 780 498 001 92;
  • 18) 0.000 097 274 780 498 001 92 × 2 = 0 + 0.000 194 549 560 996 003 84;
  • 19) 0.000 194 549 560 996 003 84 × 2 = 0 + 0.000 389 099 121 992 007 68;
  • 20) 0.000 389 099 121 992 007 68 × 2 = 0 + 0.000 778 198 243 984 015 36;
  • 21) 0.000 778 198 243 984 015 36 × 2 = 0 + 0.001 556 396 487 968 030 72;
  • 22) 0.001 556 396 487 968 030 72 × 2 = 0 + 0.003 112 792 975 936 061 44;
  • 23) 0.003 112 792 975 936 061 44 × 2 = 0 + 0.006 225 585 951 872 122 88;
  • 24) 0.006 225 585 951 872 122 88 × 2 = 0 + 0.012 451 171 903 744 245 76;
  • 25) 0.012 451 171 903 744 245 76 × 2 = 0 + 0.024 902 343 807 488 491 52;
  • 26) 0.024 902 343 807 488 491 52 × 2 = 0 + 0.049 804 687 614 976 983 04;
  • 27) 0.049 804 687 614 976 983 04 × 2 = 0 + 0.099 609 375 229 953 966 08;
  • 28) 0.099 609 375 229 953 966 08 × 2 = 0 + 0.199 218 750 459 907 932 16;
  • 29) 0.199 218 750 459 907 932 16 × 2 = 0 + 0.398 437 500 919 815 864 32;
  • 30) 0.398 437 500 919 815 864 32 × 2 = 0 + 0.796 875 001 839 631 728 64;
  • 31) 0.796 875 001 839 631 728 64 × 2 = 1 + 0.593 750 003 679 263 457 28;
  • 32) 0.593 750 003 679 263 457 28 × 2 = 1 + 0.187 500 007 358 526 914 56;
  • 33) 0.187 500 007 358 526 914 56 × 2 = 0 + 0.375 000 014 717 053 829 12;
  • 34) 0.375 000 014 717 053 829 12 × 2 = 0 + 0.750 000 029 434 107 658 24;
  • 35) 0.750 000 029 434 107 658 24 × 2 = 1 + 0.500 000 058 868 215 316 48;
  • 36) 0.500 000 058 868 215 316 48 × 2 = 1 + 0.000 000 117 736 430 632 96;
  • 37) 0.000 000 117 736 430 632 96 × 2 = 0 + 0.000 000 235 472 861 265 92;
  • 38) 0.000 000 235 472 861 265 92 × 2 = 0 + 0.000 000 470 945 722 531 84;
  • 39) 0.000 000 470 945 722 531 84 × 2 = 0 + 0.000 000 941 891 445 063 68;
  • 40) 0.000 000 941 891 445 063 68 × 2 = 0 + 0.000 001 883 782 890 127 36;
  • 41) 0.000 001 883 782 890 127 36 × 2 = 0 + 0.000 003 767 565 780 254 72;
  • 42) 0.000 003 767 565 780 254 72 × 2 = 0 + 0.000 007 535 131 560 509 44;
  • 43) 0.000 007 535 131 560 509 44 × 2 = 0 + 0.000 015 070 263 121 018 88;
  • 44) 0.000 015 070 263 121 018 88 × 2 = 0 + 0.000 030 140 526 242 037 76;
  • 45) 0.000 030 140 526 242 037 76 × 2 = 0 + 0.000 060 281 052 484 075 52;
  • 46) 0.000 060 281 052 484 075 52 × 2 = 0 + 0.000 120 562 104 968 151 04;
  • 47) 0.000 120 562 104 968 151 04 × 2 = 0 + 0.000 241 124 209 936 302 08;
  • 48) 0.000 241 124 209 936 302 08 × 2 = 0 + 0.000 482 248 419 872 604 16;
  • 49) 0.000 482 248 419 872 604 16 × 2 = 0 + 0.000 964 496 839 745 208 32;
  • 50) 0.000 964 496 839 745 208 32 × 2 = 0 + 0.001 928 993 679 490 416 64;
  • 51) 0.001 928 993 679 490 416 64 × 2 = 0 + 0.003 857 987 358 980 833 28;
  • 52) 0.003 857 987 358 980 833 28 × 2 = 0 + 0.007 715 974 717 961 666 56;
  • 53) 0.007 715 974 717 961 666 56 × 2 = 0 + 0.015 431 949 435 923 333 12;
  • 54) 0.015 431 949 435 923 333 12 × 2 = 0 + 0.030 863 898 871 846 666 24;

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 36(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 36(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 36(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 36 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