-0.000 000 000 742 147 676 191 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 191(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
-0.000 000 000 742 147 676 191(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:

|-0.000 000 000 742 147 676 191| = 0.000 000 000 742 147 676 191


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: 0.000 000 000 742 147 676 191.

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) 0.000 000 000 742 147 676 191 × 2 = 0 + 0.000 000 001 484 295 352 382;
  • 2) 0.000 000 001 484 295 352 382 × 2 = 0 + 0.000 000 002 968 590 704 764;
  • 3) 0.000 000 002 968 590 704 764 × 2 = 0 + 0.000 000 005 937 181 409 528;
  • 4) 0.000 000 005 937 181 409 528 × 2 = 0 + 0.000 000 011 874 362 819 056;
  • 5) 0.000 000 011 874 362 819 056 × 2 = 0 + 0.000 000 023 748 725 638 112;
  • 6) 0.000 000 023 748 725 638 112 × 2 = 0 + 0.000 000 047 497 451 276 224;
  • 7) 0.000 000 047 497 451 276 224 × 2 = 0 + 0.000 000 094 994 902 552 448;
  • 8) 0.000 000 094 994 902 552 448 × 2 = 0 + 0.000 000 189 989 805 104 896;
  • 9) 0.000 000 189 989 805 104 896 × 2 = 0 + 0.000 000 379 979 610 209 792;
  • 10) 0.000 000 379 979 610 209 792 × 2 = 0 + 0.000 000 759 959 220 419 584;
  • 11) 0.000 000 759 959 220 419 584 × 2 = 0 + 0.000 001 519 918 440 839 168;
  • 12) 0.000 001 519 918 440 839 168 × 2 = 0 + 0.000 003 039 836 881 678 336;
  • 13) 0.000 003 039 836 881 678 336 × 2 = 0 + 0.000 006 079 673 763 356 672;
  • 14) 0.000 006 079 673 763 356 672 × 2 = 0 + 0.000 012 159 347 526 713 344;
  • 15) 0.000 012 159 347 526 713 344 × 2 = 0 + 0.000 024 318 695 053 426 688;
  • 16) 0.000 024 318 695 053 426 688 × 2 = 0 + 0.000 048 637 390 106 853 376;
  • 17) 0.000 048 637 390 106 853 376 × 2 = 0 + 0.000 097 274 780 213 706 752;
  • 18) 0.000 097 274 780 213 706 752 × 2 = 0 + 0.000 194 549 560 427 413 504;
  • 19) 0.000 194 549 560 427 413 504 × 2 = 0 + 0.000 389 099 120 854 827 008;
  • 20) 0.000 389 099 120 854 827 008 × 2 = 0 + 0.000 778 198 241 709 654 016;
  • 21) 0.000 778 198 241 709 654 016 × 2 = 0 + 0.001 556 396 483 419 308 032;
  • 22) 0.001 556 396 483 419 308 032 × 2 = 0 + 0.003 112 792 966 838 616 064;
  • 23) 0.003 112 792 966 838 616 064 × 2 = 0 + 0.006 225 585 933 677 232 128;
  • 24) 0.006 225 585 933 677 232 128 × 2 = 0 + 0.012 451 171 867 354 464 256;
  • 25) 0.012 451 171 867 354 464 256 × 2 = 0 + 0.024 902 343 734 708 928 512;
  • 26) 0.024 902 343 734 708 928 512 × 2 = 0 + 0.049 804 687 469 417 857 024;
  • 27) 0.049 804 687 469 417 857 024 × 2 = 0 + 0.099 609 374 938 835 714 048;
  • 28) 0.099 609 374 938 835 714 048 × 2 = 0 + 0.199 218 749 877 671 428 096;
  • 29) 0.199 218 749 877 671 428 096 × 2 = 0 + 0.398 437 499 755 342 856 192;
  • 30) 0.398 437 499 755 342 856 192 × 2 = 0 + 0.796 874 999 510 685 712 384;
  • 31) 0.796 874 999 510 685 712 384 × 2 = 1 + 0.593 749 999 021 371 424 768;
  • 32) 0.593 749 999 021 371 424 768 × 2 = 1 + 0.187 499 998 042 742 849 536;
  • 33) 0.187 499 998 042 742 849 536 × 2 = 0 + 0.374 999 996 085 485 699 072;
  • 34) 0.374 999 996 085 485 699 072 × 2 = 0 + 0.749 999 992 170 971 398 144;
  • 35) 0.749 999 992 170 971 398 144 × 2 = 1 + 0.499 999 984 341 942 796 288;
  • 36) 0.499 999 984 341 942 796 288 × 2 = 0 + 0.999 999 968 683 885 592 576;
  • 37) 0.999 999 968 683 885 592 576 × 2 = 1 + 0.999 999 937 367 771 185 152;
  • 38) 0.999 999 937 367 771 185 152 × 2 = 1 + 0.999 999 874 735 542 370 304;
  • 39) 0.999 999 874 735 542 370 304 × 2 = 1 + 0.999 999 749 471 084 740 608;
  • 40) 0.999 999 749 471 084 740 608 × 2 = 1 + 0.999 999 498 942 169 481 216;
  • 41) 0.999 999 498 942 169 481 216 × 2 = 1 + 0.999 998 997 884 338 962 432;
  • 42) 0.999 998 997 884 338 962 432 × 2 = 1 + 0.999 997 995 768 677 924 864;
  • 43) 0.999 997 995 768 677 924 864 × 2 = 1 + 0.999 995 991 537 355 849 728;
  • 44) 0.999 995 991 537 355 849 728 × 2 = 1 + 0.999 991 983 074 711 699 456;
  • 45) 0.999 991 983 074 711 699 456 × 2 = 1 + 0.999 983 966 149 423 398 912;
  • 46) 0.999 983 966 149 423 398 912 × 2 = 1 + 0.999 967 932 298 846 797 824;
  • 47) 0.999 967 932 298 846 797 824 × 2 = 1 + 0.999 935 864 597 693 595 648;
  • 48) 0.999 935 864 597 693 595 648 × 2 = 1 + 0.999 871 729 195 387 191 296;
  • 49) 0.999 871 729 195 387 191 296 × 2 = 1 + 0.999 743 458 390 774 382 592;
  • 50) 0.999 743 458 390 774 382 592 × 2 = 1 + 0.999 486 916 781 548 765 184;
  • 51) 0.999 486 916 781 548 765 184 × 2 = 1 + 0.998 973 833 563 097 530 368;
  • 52) 0.998 973 833 563 097 530 368 × 2 = 1 + 0.997 947 667 126 195 060 736;
  • 53) 0.997 947 667 126 195 060 736 × 2 = 1 + 0.995 895 334 252 390 121 472;
  • 54) 0.995 895 334 252 390 121 472 × 2 = 1 + 0.991 790 668 504 780 242 944;

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:


0.000 000 000 742 147 676 191(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 191(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:


0.000 000 000 742 147 676 191(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 -0.000 000 000 742 147 676 191 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