-0.000 000 000 742 147 676 646 715 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 715 1(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
-0.000 000 000 742 147 676 646 715 1(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:

|-0.000 000 000 742 147 676 646 715 1| = 0.000 000 000 742 147 676 646 715 1


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: 0.000 000 000 742 147 676 646 715 1.

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) 0.000 000 000 742 147 676 646 715 1 × 2 = 0 + 0.000 000 001 484 295 353 293 430 2;
  • 2) 0.000 000 001 484 295 353 293 430 2 × 2 = 0 + 0.000 000 002 968 590 706 586 860 4;
  • 3) 0.000 000 002 968 590 706 586 860 4 × 2 = 0 + 0.000 000 005 937 181 413 173 720 8;
  • 4) 0.000 000 005 937 181 413 173 720 8 × 2 = 0 + 0.000 000 011 874 362 826 347 441 6;
  • 5) 0.000 000 011 874 362 826 347 441 6 × 2 = 0 + 0.000 000 023 748 725 652 694 883 2;
  • 6) 0.000 000 023 748 725 652 694 883 2 × 2 = 0 + 0.000 000 047 497 451 305 389 766 4;
  • 7) 0.000 000 047 497 451 305 389 766 4 × 2 = 0 + 0.000 000 094 994 902 610 779 532 8;
  • 8) 0.000 000 094 994 902 610 779 532 8 × 2 = 0 + 0.000 000 189 989 805 221 559 065 6;
  • 9) 0.000 000 189 989 805 221 559 065 6 × 2 = 0 + 0.000 000 379 979 610 443 118 131 2;
  • 10) 0.000 000 379 979 610 443 118 131 2 × 2 = 0 + 0.000 000 759 959 220 886 236 262 4;
  • 11) 0.000 000 759 959 220 886 236 262 4 × 2 = 0 + 0.000 001 519 918 441 772 472 524 8;
  • 12) 0.000 001 519 918 441 772 472 524 8 × 2 = 0 + 0.000 003 039 836 883 544 945 049 6;
  • 13) 0.000 003 039 836 883 544 945 049 6 × 2 = 0 + 0.000 006 079 673 767 089 890 099 2;
  • 14) 0.000 006 079 673 767 089 890 099 2 × 2 = 0 + 0.000 012 159 347 534 179 780 198 4;
  • 15) 0.000 012 159 347 534 179 780 198 4 × 2 = 0 + 0.000 024 318 695 068 359 560 396 8;
  • 16) 0.000 024 318 695 068 359 560 396 8 × 2 = 0 + 0.000 048 637 390 136 719 120 793 6;
  • 17) 0.000 048 637 390 136 719 120 793 6 × 2 = 0 + 0.000 097 274 780 273 438 241 587 2;
  • 18) 0.000 097 274 780 273 438 241 587 2 × 2 = 0 + 0.000 194 549 560 546 876 483 174 4;
  • 19) 0.000 194 549 560 546 876 483 174 4 × 2 = 0 + 0.000 389 099 121 093 752 966 348 8;
  • 20) 0.000 389 099 121 093 752 966 348 8 × 2 = 0 + 0.000 778 198 242 187 505 932 697 6;
  • 21) 0.000 778 198 242 187 505 932 697 6 × 2 = 0 + 0.001 556 396 484 375 011 865 395 2;
  • 22) 0.001 556 396 484 375 011 865 395 2 × 2 = 0 + 0.003 112 792 968 750 023 730 790 4;
  • 23) 0.003 112 792 968 750 023 730 790 4 × 2 = 0 + 0.006 225 585 937 500 047 461 580 8;
  • 24) 0.006 225 585 937 500 047 461 580 8 × 2 = 0 + 0.012 451 171 875 000 094 923 161 6;
  • 25) 0.012 451 171 875 000 094 923 161 6 × 2 = 0 + 0.024 902 343 750 000 189 846 323 2;
  • 26) 0.024 902 343 750 000 189 846 323 2 × 2 = 0 + 0.049 804 687 500 000 379 692 646 4;
  • 27) 0.049 804 687 500 000 379 692 646 4 × 2 = 0 + 0.099 609 375 000 000 759 385 292 8;
  • 28) 0.099 609 375 000 000 759 385 292 8 × 2 = 0 + 0.199 218 750 000 001 518 770 585 6;
  • 29) 0.199 218 750 000 001 518 770 585 6 × 2 = 0 + 0.398 437 500 000 003 037 541 171 2;
  • 30) 0.398 437 500 000 003 037 541 171 2 × 2 = 0 + 0.796 875 000 000 006 075 082 342 4;
  • 31) 0.796 875 000 000 006 075 082 342 4 × 2 = 1 + 0.593 750 000 000 012 150 164 684 8;
  • 32) 0.593 750 000 000 012 150 164 684 8 × 2 = 1 + 0.187 500 000 000 024 300 329 369 6;
  • 33) 0.187 500 000 000 024 300 329 369 6 × 2 = 0 + 0.375 000 000 000 048 600 658 739 2;
  • 34) 0.375 000 000 000 048 600 658 739 2 × 2 = 0 + 0.750 000 000 000 097 201 317 478 4;
  • 35) 0.750 000 000 000 097 201 317 478 4 × 2 = 1 + 0.500 000 000 000 194 402 634 956 8;
  • 36) 0.500 000 000 000 194 402 634 956 8 × 2 = 1 + 0.000 000 000 000 388 805 269 913 6;
  • 37) 0.000 000 000 000 388 805 269 913 6 × 2 = 0 + 0.000 000 000 000 777 610 539 827 2;
  • 38) 0.000 000 000 000 777 610 539 827 2 × 2 = 0 + 0.000 000 000 001 555 221 079 654 4;
  • 39) 0.000 000 000 001 555 221 079 654 4 × 2 = 0 + 0.000 000 000 003 110 442 159 308 8;
  • 40) 0.000 000 000 003 110 442 159 308 8 × 2 = 0 + 0.000 000 000 006 220 884 318 617 6;
  • 41) 0.000 000 000 006 220 884 318 617 6 × 2 = 0 + 0.000 000 000 012 441 768 637 235 2;
  • 42) 0.000 000 000 012 441 768 637 235 2 × 2 = 0 + 0.000 000 000 024 883 537 274 470 4;
  • 43) 0.000 000 000 024 883 537 274 470 4 × 2 = 0 + 0.000 000 000 049 767 074 548 940 8;
  • 44) 0.000 000 000 049 767 074 548 940 8 × 2 = 0 + 0.000 000 000 099 534 149 097 881 6;
  • 45) 0.000 000 000 099 534 149 097 881 6 × 2 = 0 + 0.000 000 000 199 068 298 195 763 2;
  • 46) 0.000 000 000 199 068 298 195 763 2 × 2 = 0 + 0.000 000 000 398 136 596 391 526 4;
  • 47) 0.000 000 000 398 136 596 391 526 4 × 2 = 0 + 0.000 000 000 796 273 192 783 052 8;
  • 48) 0.000 000 000 796 273 192 783 052 8 × 2 = 0 + 0.000 000 001 592 546 385 566 105 6;
  • 49) 0.000 000 001 592 546 385 566 105 6 × 2 = 0 + 0.000 000 003 185 092 771 132 211 2;
  • 50) 0.000 000 003 185 092 771 132 211 2 × 2 = 0 + 0.000 000 006 370 185 542 264 422 4;
  • 51) 0.000 000 006 370 185 542 264 422 4 × 2 = 0 + 0.000 000 012 740 371 084 528 844 8;
  • 52) 0.000 000 012 740 371 084 528 844 8 × 2 = 0 + 0.000 000 025 480 742 169 057 689 6;
  • 53) 0.000 000 025 480 742 169 057 689 6 × 2 = 0 + 0.000 000 050 961 484 338 115 379 2;
  • 54) 0.000 000 050 961 484 338 115 379 2 × 2 = 0 + 0.000 000 101 922 968 676 230 758 4;

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:


0.000 000 000 742 147 676 646 715 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 715 1(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:


0.000 000 000 742 147 676 646 715 1(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 -0.000 000 000 742 147 676 646 715 1 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