-0.000 000 000 742 147 795 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 795(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
-0.000 000 000 742 147 795(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:

|-0.000 000 000 742 147 795| = 0.000 000 000 742 147 795


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: 0.000 000 000 742 147 795.

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) 0.000 000 000 742 147 795 × 2 = 0 + 0.000 000 001 484 295 59;
  • 2) 0.000 000 001 484 295 59 × 2 = 0 + 0.000 000 002 968 591 18;
  • 3) 0.000 000 002 968 591 18 × 2 = 0 + 0.000 000 005 937 182 36;
  • 4) 0.000 000 005 937 182 36 × 2 = 0 + 0.000 000 011 874 364 72;
  • 5) 0.000 000 011 874 364 72 × 2 = 0 + 0.000 000 023 748 729 44;
  • 6) 0.000 000 023 748 729 44 × 2 = 0 + 0.000 000 047 497 458 88;
  • 7) 0.000 000 047 497 458 88 × 2 = 0 + 0.000 000 094 994 917 76;
  • 8) 0.000 000 094 994 917 76 × 2 = 0 + 0.000 000 189 989 835 52;
  • 9) 0.000 000 189 989 835 52 × 2 = 0 + 0.000 000 379 979 671 04;
  • 10) 0.000 000 379 979 671 04 × 2 = 0 + 0.000 000 759 959 342 08;
  • 11) 0.000 000 759 959 342 08 × 2 = 0 + 0.000 001 519 918 684 16;
  • 12) 0.000 001 519 918 684 16 × 2 = 0 + 0.000 003 039 837 368 32;
  • 13) 0.000 003 039 837 368 32 × 2 = 0 + 0.000 006 079 674 736 64;
  • 14) 0.000 006 079 674 736 64 × 2 = 0 + 0.000 012 159 349 473 28;
  • 15) 0.000 012 159 349 473 28 × 2 = 0 + 0.000 024 318 698 946 56;
  • 16) 0.000 024 318 698 946 56 × 2 = 0 + 0.000 048 637 397 893 12;
  • 17) 0.000 048 637 397 893 12 × 2 = 0 + 0.000 097 274 795 786 24;
  • 18) 0.000 097 274 795 786 24 × 2 = 0 + 0.000 194 549 591 572 48;
  • 19) 0.000 194 549 591 572 48 × 2 = 0 + 0.000 389 099 183 144 96;
  • 20) 0.000 389 099 183 144 96 × 2 = 0 + 0.000 778 198 366 289 92;
  • 21) 0.000 778 198 366 289 92 × 2 = 0 + 0.001 556 396 732 579 84;
  • 22) 0.001 556 396 732 579 84 × 2 = 0 + 0.003 112 793 465 159 68;
  • 23) 0.003 112 793 465 159 68 × 2 = 0 + 0.006 225 586 930 319 36;
  • 24) 0.006 225 586 930 319 36 × 2 = 0 + 0.012 451 173 860 638 72;
  • 25) 0.012 451 173 860 638 72 × 2 = 0 + 0.024 902 347 721 277 44;
  • 26) 0.024 902 347 721 277 44 × 2 = 0 + 0.049 804 695 442 554 88;
  • 27) 0.049 804 695 442 554 88 × 2 = 0 + 0.099 609 390 885 109 76;
  • 28) 0.099 609 390 885 109 76 × 2 = 0 + 0.199 218 781 770 219 52;
  • 29) 0.199 218 781 770 219 52 × 2 = 0 + 0.398 437 563 540 439 04;
  • 30) 0.398 437 563 540 439 04 × 2 = 0 + 0.796 875 127 080 878 08;
  • 31) 0.796 875 127 080 878 08 × 2 = 1 + 0.593 750 254 161 756 16;
  • 32) 0.593 750 254 161 756 16 × 2 = 1 + 0.187 500 508 323 512 32;
  • 33) 0.187 500 508 323 512 32 × 2 = 0 + 0.375 001 016 647 024 64;
  • 34) 0.375 001 016 647 024 64 × 2 = 0 + 0.750 002 033 294 049 28;
  • 35) 0.750 002 033 294 049 28 × 2 = 1 + 0.500 004 066 588 098 56;
  • 36) 0.500 004 066 588 098 56 × 2 = 1 + 0.000 008 133 176 197 12;
  • 37) 0.000 008 133 176 197 12 × 2 = 0 + 0.000 016 266 352 394 24;
  • 38) 0.000 016 266 352 394 24 × 2 = 0 + 0.000 032 532 704 788 48;
  • 39) 0.000 032 532 704 788 48 × 2 = 0 + 0.000 065 065 409 576 96;
  • 40) 0.000 065 065 409 576 96 × 2 = 0 + 0.000 130 130 819 153 92;
  • 41) 0.000 130 130 819 153 92 × 2 = 0 + 0.000 260 261 638 307 84;
  • 42) 0.000 260 261 638 307 84 × 2 = 0 + 0.000 520 523 276 615 68;
  • 43) 0.000 520 523 276 615 68 × 2 = 0 + 0.001 041 046 553 231 36;
  • 44) 0.001 041 046 553 231 36 × 2 = 0 + 0.002 082 093 106 462 72;
  • 45) 0.002 082 093 106 462 72 × 2 = 0 + 0.004 164 186 212 925 44;
  • 46) 0.004 164 186 212 925 44 × 2 = 0 + 0.008 328 372 425 850 88;
  • 47) 0.008 328 372 425 850 88 × 2 = 0 + 0.016 656 744 851 701 76;
  • 48) 0.016 656 744 851 701 76 × 2 = 0 + 0.033 313 489 703 403 52;
  • 49) 0.033 313 489 703 403 52 × 2 = 0 + 0.066 626 979 406 807 04;
  • 50) 0.066 626 979 406 807 04 × 2 = 0 + 0.133 253 958 813 614 08;
  • 51) 0.133 253 958 813 614 08 × 2 = 0 + 0.266 507 917 627 228 16;
  • 52) 0.266 507 917 627 228 16 × 2 = 0 + 0.533 015 835 254 456 32;
  • 53) 0.533 015 835 254 456 32 × 2 = 1 + 0.066 031 670 508 912 64;
  • 54) 0.066 031 670 508 912 64 × 2 = 0 + 0.132 063 341 017 825 28;

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:


0.000 000 000 742 147 795(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 795(10) =

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


0.000 000 000 742 147 795(10) =

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

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

1.1001 1000 0000 0000 0000 010(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 010


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 0010 =

100 1100 0000 0000 0000 0010


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 0010


Decimal number -0.000 000 000 742 147 795 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0010

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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