-0.000 000 000 742 147 676 107 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 107(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
-0.000 000 000 742 147 676 107(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:

|-0.000 000 000 742 147 676 107| = 0.000 000 000 742 147 676 107


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: 0.000 000 000 742 147 676 107.

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) 0.000 000 000 742 147 676 107 × 2 = 0 + 0.000 000 001 484 295 352 214;
  • 2) 0.000 000 001 484 295 352 214 × 2 = 0 + 0.000 000 002 968 590 704 428;
  • 3) 0.000 000 002 968 590 704 428 × 2 = 0 + 0.000 000 005 937 181 408 856;
  • 4) 0.000 000 005 937 181 408 856 × 2 = 0 + 0.000 000 011 874 362 817 712;
  • 5) 0.000 000 011 874 362 817 712 × 2 = 0 + 0.000 000 023 748 725 635 424;
  • 6) 0.000 000 023 748 725 635 424 × 2 = 0 + 0.000 000 047 497 451 270 848;
  • 7) 0.000 000 047 497 451 270 848 × 2 = 0 + 0.000 000 094 994 902 541 696;
  • 8) 0.000 000 094 994 902 541 696 × 2 = 0 + 0.000 000 189 989 805 083 392;
  • 9) 0.000 000 189 989 805 083 392 × 2 = 0 + 0.000 000 379 979 610 166 784;
  • 10) 0.000 000 379 979 610 166 784 × 2 = 0 + 0.000 000 759 959 220 333 568;
  • 11) 0.000 000 759 959 220 333 568 × 2 = 0 + 0.000 001 519 918 440 667 136;
  • 12) 0.000 001 519 918 440 667 136 × 2 = 0 + 0.000 003 039 836 881 334 272;
  • 13) 0.000 003 039 836 881 334 272 × 2 = 0 + 0.000 006 079 673 762 668 544;
  • 14) 0.000 006 079 673 762 668 544 × 2 = 0 + 0.000 012 159 347 525 337 088;
  • 15) 0.000 012 159 347 525 337 088 × 2 = 0 + 0.000 024 318 695 050 674 176;
  • 16) 0.000 024 318 695 050 674 176 × 2 = 0 + 0.000 048 637 390 101 348 352;
  • 17) 0.000 048 637 390 101 348 352 × 2 = 0 + 0.000 097 274 780 202 696 704;
  • 18) 0.000 097 274 780 202 696 704 × 2 = 0 + 0.000 194 549 560 405 393 408;
  • 19) 0.000 194 549 560 405 393 408 × 2 = 0 + 0.000 389 099 120 810 786 816;
  • 20) 0.000 389 099 120 810 786 816 × 2 = 0 + 0.000 778 198 241 621 573 632;
  • 21) 0.000 778 198 241 621 573 632 × 2 = 0 + 0.001 556 396 483 243 147 264;
  • 22) 0.001 556 396 483 243 147 264 × 2 = 0 + 0.003 112 792 966 486 294 528;
  • 23) 0.003 112 792 966 486 294 528 × 2 = 0 + 0.006 225 585 932 972 589 056;
  • 24) 0.006 225 585 932 972 589 056 × 2 = 0 + 0.012 451 171 865 945 178 112;
  • 25) 0.012 451 171 865 945 178 112 × 2 = 0 + 0.024 902 343 731 890 356 224;
  • 26) 0.024 902 343 731 890 356 224 × 2 = 0 + 0.049 804 687 463 780 712 448;
  • 27) 0.049 804 687 463 780 712 448 × 2 = 0 + 0.099 609 374 927 561 424 896;
  • 28) 0.099 609 374 927 561 424 896 × 2 = 0 + 0.199 218 749 855 122 849 792;
  • 29) 0.199 218 749 855 122 849 792 × 2 = 0 + 0.398 437 499 710 245 699 584;
  • 30) 0.398 437 499 710 245 699 584 × 2 = 0 + 0.796 874 999 420 491 399 168;
  • 31) 0.796 874 999 420 491 399 168 × 2 = 1 + 0.593 749 998 840 982 798 336;
  • 32) 0.593 749 998 840 982 798 336 × 2 = 1 + 0.187 499 997 681 965 596 672;
  • 33) 0.187 499 997 681 965 596 672 × 2 = 0 + 0.374 999 995 363 931 193 344;
  • 34) 0.374 999 995 363 931 193 344 × 2 = 0 + 0.749 999 990 727 862 386 688;
  • 35) 0.749 999 990 727 862 386 688 × 2 = 1 + 0.499 999 981 455 724 773 376;
  • 36) 0.499 999 981 455 724 773 376 × 2 = 0 + 0.999 999 962 911 449 546 752;
  • 37) 0.999 999 962 911 449 546 752 × 2 = 1 + 0.999 999 925 822 899 093 504;
  • 38) 0.999 999 925 822 899 093 504 × 2 = 1 + 0.999 999 851 645 798 187 008;
  • 39) 0.999 999 851 645 798 187 008 × 2 = 1 + 0.999 999 703 291 596 374 016;
  • 40) 0.999 999 703 291 596 374 016 × 2 = 1 + 0.999 999 406 583 192 748 032;
  • 41) 0.999 999 406 583 192 748 032 × 2 = 1 + 0.999 998 813 166 385 496 064;
  • 42) 0.999 998 813 166 385 496 064 × 2 = 1 + 0.999 997 626 332 770 992 128;
  • 43) 0.999 997 626 332 770 992 128 × 2 = 1 + 0.999 995 252 665 541 984 256;
  • 44) 0.999 995 252 665 541 984 256 × 2 = 1 + 0.999 990 505 331 083 968 512;
  • 45) 0.999 990 505 331 083 968 512 × 2 = 1 + 0.999 981 010 662 167 937 024;
  • 46) 0.999 981 010 662 167 937 024 × 2 = 1 + 0.999 962 021 324 335 874 048;
  • 47) 0.999 962 021 324 335 874 048 × 2 = 1 + 0.999 924 042 648 671 748 096;
  • 48) 0.999 924 042 648 671 748 096 × 2 = 1 + 0.999 848 085 297 343 496 192;
  • 49) 0.999 848 085 297 343 496 192 × 2 = 1 + 0.999 696 170 594 686 992 384;
  • 50) 0.999 696 170 594 686 992 384 × 2 = 1 + 0.999 392 341 189 373 984 768;
  • 51) 0.999 392 341 189 373 984 768 × 2 = 1 + 0.998 784 682 378 747 969 536;
  • 52) 0.998 784 682 378 747 969 536 × 2 = 1 + 0.997 569 364 757 495 939 072;
  • 53) 0.997 569 364 757 495 939 072 × 2 = 1 + 0.995 138 729 514 991 878 144;
  • 54) 0.995 138 729 514 991 878 144 × 2 = 1 + 0.990 277 459 029 983 756 288;

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:


0.000 000 000 742 147 676 107(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 107(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 107(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 0111 1111 1111 1111 111


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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 107 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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