-0.000 000 000 742 147 676 708 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 708(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
-0.000 000 000 742 147 676 708(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:

|-0.000 000 000 742 147 676 708| = 0.000 000 000 742 147 676 708


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: 0.000 000 000 742 147 676 708.

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) 0.000 000 000 742 147 676 708 × 2 = 0 + 0.000 000 001 484 295 353 416;
  • 2) 0.000 000 001 484 295 353 416 × 2 = 0 + 0.000 000 002 968 590 706 832;
  • 3) 0.000 000 002 968 590 706 832 × 2 = 0 + 0.000 000 005 937 181 413 664;
  • 4) 0.000 000 005 937 181 413 664 × 2 = 0 + 0.000 000 011 874 362 827 328;
  • 5) 0.000 000 011 874 362 827 328 × 2 = 0 + 0.000 000 023 748 725 654 656;
  • 6) 0.000 000 023 748 725 654 656 × 2 = 0 + 0.000 000 047 497 451 309 312;
  • 7) 0.000 000 047 497 451 309 312 × 2 = 0 + 0.000 000 094 994 902 618 624;
  • 8) 0.000 000 094 994 902 618 624 × 2 = 0 + 0.000 000 189 989 805 237 248;
  • 9) 0.000 000 189 989 805 237 248 × 2 = 0 + 0.000 000 379 979 610 474 496;
  • 10) 0.000 000 379 979 610 474 496 × 2 = 0 + 0.000 000 759 959 220 948 992;
  • 11) 0.000 000 759 959 220 948 992 × 2 = 0 + 0.000 001 519 918 441 897 984;
  • 12) 0.000 001 519 918 441 897 984 × 2 = 0 + 0.000 003 039 836 883 795 968;
  • 13) 0.000 003 039 836 883 795 968 × 2 = 0 + 0.000 006 079 673 767 591 936;
  • 14) 0.000 006 079 673 767 591 936 × 2 = 0 + 0.000 012 159 347 535 183 872;
  • 15) 0.000 012 159 347 535 183 872 × 2 = 0 + 0.000 024 318 695 070 367 744;
  • 16) 0.000 024 318 695 070 367 744 × 2 = 0 + 0.000 048 637 390 140 735 488;
  • 17) 0.000 048 637 390 140 735 488 × 2 = 0 + 0.000 097 274 780 281 470 976;
  • 18) 0.000 097 274 780 281 470 976 × 2 = 0 + 0.000 194 549 560 562 941 952;
  • 19) 0.000 194 549 560 562 941 952 × 2 = 0 + 0.000 389 099 121 125 883 904;
  • 20) 0.000 389 099 121 125 883 904 × 2 = 0 + 0.000 778 198 242 251 767 808;
  • 21) 0.000 778 198 242 251 767 808 × 2 = 0 + 0.001 556 396 484 503 535 616;
  • 22) 0.001 556 396 484 503 535 616 × 2 = 0 + 0.003 112 792 969 007 071 232;
  • 23) 0.003 112 792 969 007 071 232 × 2 = 0 + 0.006 225 585 938 014 142 464;
  • 24) 0.006 225 585 938 014 142 464 × 2 = 0 + 0.012 451 171 876 028 284 928;
  • 25) 0.012 451 171 876 028 284 928 × 2 = 0 + 0.024 902 343 752 056 569 856;
  • 26) 0.024 902 343 752 056 569 856 × 2 = 0 + 0.049 804 687 504 113 139 712;
  • 27) 0.049 804 687 504 113 139 712 × 2 = 0 + 0.099 609 375 008 226 279 424;
  • 28) 0.099 609 375 008 226 279 424 × 2 = 0 + 0.199 218 750 016 452 558 848;
  • 29) 0.199 218 750 016 452 558 848 × 2 = 0 + 0.398 437 500 032 905 117 696;
  • 30) 0.398 437 500 032 905 117 696 × 2 = 0 + 0.796 875 000 065 810 235 392;
  • 31) 0.796 875 000 065 810 235 392 × 2 = 1 + 0.593 750 000 131 620 470 784;
  • 32) 0.593 750 000 131 620 470 784 × 2 = 1 + 0.187 500 000 263 240 941 568;
  • 33) 0.187 500 000 263 240 941 568 × 2 = 0 + 0.375 000 000 526 481 883 136;
  • 34) 0.375 000 000 526 481 883 136 × 2 = 0 + 0.750 000 001 052 963 766 272;
  • 35) 0.750 000 001 052 963 766 272 × 2 = 1 + 0.500 000 002 105 927 532 544;
  • 36) 0.500 000 002 105 927 532 544 × 2 = 1 + 0.000 000 004 211 855 065 088;
  • 37) 0.000 000 004 211 855 065 088 × 2 = 0 + 0.000 000 008 423 710 130 176;
  • 38) 0.000 000 008 423 710 130 176 × 2 = 0 + 0.000 000 016 847 420 260 352;
  • 39) 0.000 000 016 847 420 260 352 × 2 = 0 + 0.000 000 033 694 840 520 704;
  • 40) 0.000 000 033 694 840 520 704 × 2 = 0 + 0.000 000 067 389 681 041 408;
  • 41) 0.000 000 067 389 681 041 408 × 2 = 0 + 0.000 000 134 779 362 082 816;
  • 42) 0.000 000 134 779 362 082 816 × 2 = 0 + 0.000 000 269 558 724 165 632;
  • 43) 0.000 000 269 558 724 165 632 × 2 = 0 + 0.000 000 539 117 448 331 264;
  • 44) 0.000 000 539 117 448 331 264 × 2 = 0 + 0.000 001 078 234 896 662 528;
  • 45) 0.000 001 078 234 896 662 528 × 2 = 0 + 0.000 002 156 469 793 325 056;
  • 46) 0.000 002 156 469 793 325 056 × 2 = 0 + 0.000 004 312 939 586 650 112;
  • 47) 0.000 004 312 939 586 650 112 × 2 = 0 + 0.000 008 625 879 173 300 224;
  • 48) 0.000 008 625 879 173 300 224 × 2 = 0 + 0.000 017 251 758 346 600 448;
  • 49) 0.000 017 251 758 346 600 448 × 2 = 0 + 0.000 034 503 516 693 200 896;
  • 50) 0.000 034 503 516 693 200 896 × 2 = 0 + 0.000 069 007 033 386 401 792;
  • 51) 0.000 069 007 033 386 401 792 × 2 = 0 + 0.000 138 014 066 772 803 584;
  • 52) 0.000 138 014 066 772 803 584 × 2 = 0 + 0.000 276 028 133 545 607 168;
  • 53) 0.000 276 028 133 545 607 168 × 2 = 0 + 0.000 552 056 267 091 214 336;
  • 54) 0.000 552 056 267 091 214 336 × 2 = 0 + 0.001 104 112 534 182 428 672;

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:


0.000 000 000 742 147 676 708(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 708(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:


0.000 000 000 742 147 676 708(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 -0.000 000 000 742 147 676 708 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