-0.000 000 000 742 147 676 648 82 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 648 82(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
-0.000 000 000 742 147 676 648 82(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:

|-0.000 000 000 742 147 676 648 82| = 0.000 000 000 742 147 676 648 82


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: 0.000 000 000 742 147 676 648 82.

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) 0.000 000 000 742 147 676 648 82 × 2 = 0 + 0.000 000 001 484 295 353 297 64;
  • 2) 0.000 000 001 484 295 353 297 64 × 2 = 0 + 0.000 000 002 968 590 706 595 28;
  • 3) 0.000 000 002 968 590 706 595 28 × 2 = 0 + 0.000 000 005 937 181 413 190 56;
  • 4) 0.000 000 005 937 181 413 190 56 × 2 = 0 + 0.000 000 011 874 362 826 381 12;
  • 5) 0.000 000 011 874 362 826 381 12 × 2 = 0 + 0.000 000 023 748 725 652 762 24;
  • 6) 0.000 000 023 748 725 652 762 24 × 2 = 0 + 0.000 000 047 497 451 305 524 48;
  • 7) 0.000 000 047 497 451 305 524 48 × 2 = 0 + 0.000 000 094 994 902 611 048 96;
  • 8) 0.000 000 094 994 902 611 048 96 × 2 = 0 + 0.000 000 189 989 805 222 097 92;
  • 9) 0.000 000 189 989 805 222 097 92 × 2 = 0 + 0.000 000 379 979 610 444 195 84;
  • 10) 0.000 000 379 979 610 444 195 84 × 2 = 0 + 0.000 000 759 959 220 888 391 68;
  • 11) 0.000 000 759 959 220 888 391 68 × 2 = 0 + 0.000 001 519 918 441 776 783 36;
  • 12) 0.000 001 519 918 441 776 783 36 × 2 = 0 + 0.000 003 039 836 883 553 566 72;
  • 13) 0.000 003 039 836 883 553 566 72 × 2 = 0 + 0.000 006 079 673 767 107 133 44;
  • 14) 0.000 006 079 673 767 107 133 44 × 2 = 0 + 0.000 012 159 347 534 214 266 88;
  • 15) 0.000 012 159 347 534 214 266 88 × 2 = 0 + 0.000 024 318 695 068 428 533 76;
  • 16) 0.000 024 318 695 068 428 533 76 × 2 = 0 + 0.000 048 637 390 136 857 067 52;
  • 17) 0.000 048 637 390 136 857 067 52 × 2 = 0 + 0.000 097 274 780 273 714 135 04;
  • 18) 0.000 097 274 780 273 714 135 04 × 2 = 0 + 0.000 194 549 560 547 428 270 08;
  • 19) 0.000 194 549 560 547 428 270 08 × 2 = 0 + 0.000 389 099 121 094 856 540 16;
  • 20) 0.000 389 099 121 094 856 540 16 × 2 = 0 + 0.000 778 198 242 189 713 080 32;
  • 21) 0.000 778 198 242 189 713 080 32 × 2 = 0 + 0.001 556 396 484 379 426 160 64;
  • 22) 0.001 556 396 484 379 426 160 64 × 2 = 0 + 0.003 112 792 968 758 852 321 28;
  • 23) 0.003 112 792 968 758 852 321 28 × 2 = 0 + 0.006 225 585 937 517 704 642 56;
  • 24) 0.006 225 585 937 517 704 642 56 × 2 = 0 + 0.012 451 171 875 035 409 285 12;
  • 25) 0.012 451 171 875 035 409 285 12 × 2 = 0 + 0.024 902 343 750 070 818 570 24;
  • 26) 0.024 902 343 750 070 818 570 24 × 2 = 0 + 0.049 804 687 500 141 637 140 48;
  • 27) 0.049 804 687 500 141 637 140 48 × 2 = 0 + 0.099 609 375 000 283 274 280 96;
  • 28) 0.099 609 375 000 283 274 280 96 × 2 = 0 + 0.199 218 750 000 566 548 561 92;
  • 29) 0.199 218 750 000 566 548 561 92 × 2 = 0 + 0.398 437 500 001 133 097 123 84;
  • 30) 0.398 437 500 001 133 097 123 84 × 2 = 0 + 0.796 875 000 002 266 194 247 68;
  • 31) 0.796 875 000 002 266 194 247 68 × 2 = 1 + 0.593 750 000 004 532 388 495 36;
  • 32) 0.593 750 000 004 532 388 495 36 × 2 = 1 + 0.187 500 000 009 064 776 990 72;
  • 33) 0.187 500 000 009 064 776 990 72 × 2 = 0 + 0.375 000 000 018 129 553 981 44;
  • 34) 0.375 000 000 018 129 553 981 44 × 2 = 0 + 0.750 000 000 036 259 107 962 88;
  • 35) 0.750 000 000 036 259 107 962 88 × 2 = 1 + 0.500 000 000 072 518 215 925 76;
  • 36) 0.500 000 000 072 518 215 925 76 × 2 = 1 + 0.000 000 000 145 036 431 851 52;
  • 37) 0.000 000 000 145 036 431 851 52 × 2 = 0 + 0.000 000 000 290 072 863 703 04;
  • 38) 0.000 000 000 290 072 863 703 04 × 2 = 0 + 0.000 000 000 580 145 727 406 08;
  • 39) 0.000 000 000 580 145 727 406 08 × 2 = 0 + 0.000 000 001 160 291 454 812 16;
  • 40) 0.000 000 001 160 291 454 812 16 × 2 = 0 + 0.000 000 002 320 582 909 624 32;
  • 41) 0.000 000 002 320 582 909 624 32 × 2 = 0 + 0.000 000 004 641 165 819 248 64;
  • 42) 0.000 000 004 641 165 819 248 64 × 2 = 0 + 0.000 000 009 282 331 638 497 28;
  • 43) 0.000 000 009 282 331 638 497 28 × 2 = 0 + 0.000 000 018 564 663 276 994 56;
  • 44) 0.000 000 018 564 663 276 994 56 × 2 = 0 + 0.000 000 037 129 326 553 989 12;
  • 45) 0.000 000 037 129 326 553 989 12 × 2 = 0 + 0.000 000 074 258 653 107 978 24;
  • 46) 0.000 000 074 258 653 107 978 24 × 2 = 0 + 0.000 000 148 517 306 215 956 48;
  • 47) 0.000 000 148 517 306 215 956 48 × 2 = 0 + 0.000 000 297 034 612 431 912 96;
  • 48) 0.000 000 297 034 612 431 912 96 × 2 = 0 + 0.000 000 594 069 224 863 825 92;
  • 49) 0.000 000 594 069 224 863 825 92 × 2 = 0 + 0.000 001 188 138 449 727 651 84;
  • 50) 0.000 001 188 138 449 727 651 84 × 2 = 0 + 0.000 002 376 276 899 455 303 68;
  • 51) 0.000 002 376 276 899 455 303 68 × 2 = 0 + 0.000 004 752 553 798 910 607 36;
  • 52) 0.000 004 752 553 798 910 607 36 × 2 = 0 + 0.000 009 505 107 597 821 214 72;
  • 53) 0.000 009 505 107 597 821 214 72 × 2 = 0 + 0.000 019 010 215 195 642 429 44;
  • 54) 0.000 019 010 215 195 642 429 44 × 2 = 0 + 0.000 038 020 430 391 284 858 88;

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:


0.000 000 000 742 147 676 648 82(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 648 82(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:


0.000 000 000 742 147 676 648 82(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 -0.000 000 000 742 147 676 648 82 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