-0.000 000 000 742 147 676 103 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 103(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
-0.000 000 000 742 147 676 103(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:

|-0.000 000 000 742 147 676 103| = 0.000 000 000 742 147 676 103


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: 0.000 000 000 742 147 676 103.

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) 0.000 000 000 742 147 676 103 × 2 = 0 + 0.000 000 001 484 295 352 206;
  • 2) 0.000 000 001 484 295 352 206 × 2 = 0 + 0.000 000 002 968 590 704 412;
  • 3) 0.000 000 002 968 590 704 412 × 2 = 0 + 0.000 000 005 937 181 408 824;
  • 4) 0.000 000 005 937 181 408 824 × 2 = 0 + 0.000 000 011 874 362 817 648;
  • 5) 0.000 000 011 874 362 817 648 × 2 = 0 + 0.000 000 023 748 725 635 296;
  • 6) 0.000 000 023 748 725 635 296 × 2 = 0 + 0.000 000 047 497 451 270 592;
  • 7) 0.000 000 047 497 451 270 592 × 2 = 0 + 0.000 000 094 994 902 541 184;
  • 8) 0.000 000 094 994 902 541 184 × 2 = 0 + 0.000 000 189 989 805 082 368;
  • 9) 0.000 000 189 989 805 082 368 × 2 = 0 + 0.000 000 379 979 610 164 736;
  • 10) 0.000 000 379 979 610 164 736 × 2 = 0 + 0.000 000 759 959 220 329 472;
  • 11) 0.000 000 759 959 220 329 472 × 2 = 0 + 0.000 001 519 918 440 658 944;
  • 12) 0.000 001 519 918 440 658 944 × 2 = 0 + 0.000 003 039 836 881 317 888;
  • 13) 0.000 003 039 836 881 317 888 × 2 = 0 + 0.000 006 079 673 762 635 776;
  • 14) 0.000 006 079 673 762 635 776 × 2 = 0 + 0.000 012 159 347 525 271 552;
  • 15) 0.000 012 159 347 525 271 552 × 2 = 0 + 0.000 024 318 695 050 543 104;
  • 16) 0.000 024 318 695 050 543 104 × 2 = 0 + 0.000 048 637 390 101 086 208;
  • 17) 0.000 048 637 390 101 086 208 × 2 = 0 + 0.000 097 274 780 202 172 416;
  • 18) 0.000 097 274 780 202 172 416 × 2 = 0 + 0.000 194 549 560 404 344 832;
  • 19) 0.000 194 549 560 404 344 832 × 2 = 0 + 0.000 389 099 120 808 689 664;
  • 20) 0.000 389 099 120 808 689 664 × 2 = 0 + 0.000 778 198 241 617 379 328;
  • 21) 0.000 778 198 241 617 379 328 × 2 = 0 + 0.001 556 396 483 234 758 656;
  • 22) 0.001 556 396 483 234 758 656 × 2 = 0 + 0.003 112 792 966 469 517 312;
  • 23) 0.003 112 792 966 469 517 312 × 2 = 0 + 0.006 225 585 932 939 034 624;
  • 24) 0.006 225 585 932 939 034 624 × 2 = 0 + 0.012 451 171 865 878 069 248;
  • 25) 0.012 451 171 865 878 069 248 × 2 = 0 + 0.024 902 343 731 756 138 496;
  • 26) 0.024 902 343 731 756 138 496 × 2 = 0 + 0.049 804 687 463 512 276 992;
  • 27) 0.049 804 687 463 512 276 992 × 2 = 0 + 0.099 609 374 927 024 553 984;
  • 28) 0.099 609 374 927 024 553 984 × 2 = 0 + 0.199 218 749 854 049 107 968;
  • 29) 0.199 218 749 854 049 107 968 × 2 = 0 + 0.398 437 499 708 098 215 936;
  • 30) 0.398 437 499 708 098 215 936 × 2 = 0 + 0.796 874 999 416 196 431 872;
  • 31) 0.796 874 999 416 196 431 872 × 2 = 1 + 0.593 749 998 832 392 863 744;
  • 32) 0.593 749 998 832 392 863 744 × 2 = 1 + 0.187 499 997 664 785 727 488;
  • 33) 0.187 499 997 664 785 727 488 × 2 = 0 + 0.374 999 995 329 571 454 976;
  • 34) 0.374 999 995 329 571 454 976 × 2 = 0 + 0.749 999 990 659 142 909 952;
  • 35) 0.749 999 990 659 142 909 952 × 2 = 1 + 0.499 999 981 318 285 819 904;
  • 36) 0.499 999 981 318 285 819 904 × 2 = 0 + 0.999 999 962 636 571 639 808;
  • 37) 0.999 999 962 636 571 639 808 × 2 = 1 + 0.999 999 925 273 143 279 616;
  • 38) 0.999 999 925 273 143 279 616 × 2 = 1 + 0.999 999 850 546 286 559 232;
  • 39) 0.999 999 850 546 286 559 232 × 2 = 1 + 0.999 999 701 092 573 118 464;
  • 40) 0.999 999 701 092 573 118 464 × 2 = 1 + 0.999 999 402 185 146 236 928;
  • 41) 0.999 999 402 185 146 236 928 × 2 = 1 + 0.999 998 804 370 292 473 856;
  • 42) 0.999 998 804 370 292 473 856 × 2 = 1 + 0.999 997 608 740 584 947 712;
  • 43) 0.999 997 608 740 584 947 712 × 2 = 1 + 0.999 995 217 481 169 895 424;
  • 44) 0.999 995 217 481 169 895 424 × 2 = 1 + 0.999 990 434 962 339 790 848;
  • 45) 0.999 990 434 962 339 790 848 × 2 = 1 + 0.999 980 869 924 679 581 696;
  • 46) 0.999 980 869 924 679 581 696 × 2 = 1 + 0.999 961 739 849 359 163 392;
  • 47) 0.999 961 739 849 359 163 392 × 2 = 1 + 0.999 923 479 698 718 326 784;
  • 48) 0.999 923 479 698 718 326 784 × 2 = 1 + 0.999 846 959 397 436 653 568;
  • 49) 0.999 846 959 397 436 653 568 × 2 = 1 + 0.999 693 918 794 873 307 136;
  • 50) 0.999 693 918 794 873 307 136 × 2 = 1 + 0.999 387 837 589 746 614 272;
  • 51) 0.999 387 837 589 746 614 272 × 2 = 1 + 0.998 775 675 179 493 228 544;
  • 52) 0.998 775 675 179 493 228 544 × 2 = 1 + 0.997 551 350 358 986 457 088;
  • 53) 0.997 551 350 358 986 457 088 × 2 = 1 + 0.995 102 700 717 972 914 176;
  • 54) 0.995 102 700 717 972 914 176 × 2 = 1 + 0.990 205 401 435 945 828 352;

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:


0.000 000 000 742 147 676 103(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 103(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:


0.000 000 000 742 147 676 103(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 -0.000 000 000 742 147 676 103 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 13 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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