-0.000 000 000 742 147 676 076 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 076(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
-0.000 000 000 742 147 676 076(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:

|-0.000 000 000 742 147 676 076| = 0.000 000 000 742 147 676 076


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: 0.000 000 000 742 147 676 076.

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) 0.000 000 000 742 147 676 076 × 2 = 0 + 0.000 000 001 484 295 352 152;
  • 2) 0.000 000 001 484 295 352 152 × 2 = 0 + 0.000 000 002 968 590 704 304;
  • 3) 0.000 000 002 968 590 704 304 × 2 = 0 + 0.000 000 005 937 181 408 608;
  • 4) 0.000 000 005 937 181 408 608 × 2 = 0 + 0.000 000 011 874 362 817 216;
  • 5) 0.000 000 011 874 362 817 216 × 2 = 0 + 0.000 000 023 748 725 634 432;
  • 6) 0.000 000 023 748 725 634 432 × 2 = 0 + 0.000 000 047 497 451 268 864;
  • 7) 0.000 000 047 497 451 268 864 × 2 = 0 + 0.000 000 094 994 902 537 728;
  • 8) 0.000 000 094 994 902 537 728 × 2 = 0 + 0.000 000 189 989 805 075 456;
  • 9) 0.000 000 189 989 805 075 456 × 2 = 0 + 0.000 000 379 979 610 150 912;
  • 10) 0.000 000 379 979 610 150 912 × 2 = 0 + 0.000 000 759 959 220 301 824;
  • 11) 0.000 000 759 959 220 301 824 × 2 = 0 + 0.000 001 519 918 440 603 648;
  • 12) 0.000 001 519 918 440 603 648 × 2 = 0 + 0.000 003 039 836 881 207 296;
  • 13) 0.000 003 039 836 881 207 296 × 2 = 0 + 0.000 006 079 673 762 414 592;
  • 14) 0.000 006 079 673 762 414 592 × 2 = 0 + 0.000 012 159 347 524 829 184;
  • 15) 0.000 012 159 347 524 829 184 × 2 = 0 + 0.000 024 318 695 049 658 368;
  • 16) 0.000 024 318 695 049 658 368 × 2 = 0 + 0.000 048 637 390 099 316 736;
  • 17) 0.000 048 637 390 099 316 736 × 2 = 0 + 0.000 097 274 780 198 633 472;
  • 18) 0.000 097 274 780 198 633 472 × 2 = 0 + 0.000 194 549 560 397 266 944;
  • 19) 0.000 194 549 560 397 266 944 × 2 = 0 + 0.000 389 099 120 794 533 888;
  • 20) 0.000 389 099 120 794 533 888 × 2 = 0 + 0.000 778 198 241 589 067 776;
  • 21) 0.000 778 198 241 589 067 776 × 2 = 0 + 0.001 556 396 483 178 135 552;
  • 22) 0.001 556 396 483 178 135 552 × 2 = 0 + 0.003 112 792 966 356 271 104;
  • 23) 0.003 112 792 966 356 271 104 × 2 = 0 + 0.006 225 585 932 712 542 208;
  • 24) 0.006 225 585 932 712 542 208 × 2 = 0 + 0.012 451 171 865 425 084 416;
  • 25) 0.012 451 171 865 425 084 416 × 2 = 0 + 0.024 902 343 730 850 168 832;
  • 26) 0.024 902 343 730 850 168 832 × 2 = 0 + 0.049 804 687 461 700 337 664;
  • 27) 0.049 804 687 461 700 337 664 × 2 = 0 + 0.099 609 374 923 400 675 328;
  • 28) 0.099 609 374 923 400 675 328 × 2 = 0 + 0.199 218 749 846 801 350 656;
  • 29) 0.199 218 749 846 801 350 656 × 2 = 0 + 0.398 437 499 693 602 701 312;
  • 30) 0.398 437 499 693 602 701 312 × 2 = 0 + 0.796 874 999 387 205 402 624;
  • 31) 0.796 874 999 387 205 402 624 × 2 = 1 + 0.593 749 998 774 410 805 248;
  • 32) 0.593 749 998 774 410 805 248 × 2 = 1 + 0.187 499 997 548 821 610 496;
  • 33) 0.187 499 997 548 821 610 496 × 2 = 0 + 0.374 999 995 097 643 220 992;
  • 34) 0.374 999 995 097 643 220 992 × 2 = 0 + 0.749 999 990 195 286 441 984;
  • 35) 0.749 999 990 195 286 441 984 × 2 = 1 + 0.499 999 980 390 572 883 968;
  • 36) 0.499 999 980 390 572 883 968 × 2 = 0 + 0.999 999 960 781 145 767 936;
  • 37) 0.999 999 960 781 145 767 936 × 2 = 1 + 0.999 999 921 562 291 535 872;
  • 38) 0.999 999 921 562 291 535 872 × 2 = 1 + 0.999 999 843 124 583 071 744;
  • 39) 0.999 999 843 124 583 071 744 × 2 = 1 + 0.999 999 686 249 166 143 488;
  • 40) 0.999 999 686 249 166 143 488 × 2 = 1 + 0.999 999 372 498 332 286 976;
  • 41) 0.999 999 372 498 332 286 976 × 2 = 1 + 0.999 998 744 996 664 573 952;
  • 42) 0.999 998 744 996 664 573 952 × 2 = 1 + 0.999 997 489 993 329 147 904;
  • 43) 0.999 997 489 993 329 147 904 × 2 = 1 + 0.999 994 979 986 658 295 808;
  • 44) 0.999 994 979 986 658 295 808 × 2 = 1 + 0.999 989 959 973 316 591 616;
  • 45) 0.999 989 959 973 316 591 616 × 2 = 1 + 0.999 979 919 946 633 183 232;
  • 46) 0.999 979 919 946 633 183 232 × 2 = 1 + 0.999 959 839 893 266 366 464;
  • 47) 0.999 959 839 893 266 366 464 × 2 = 1 + 0.999 919 679 786 532 732 928;
  • 48) 0.999 919 679 786 532 732 928 × 2 = 1 + 0.999 839 359 573 065 465 856;
  • 49) 0.999 839 359 573 065 465 856 × 2 = 1 + 0.999 678 719 146 130 931 712;
  • 50) 0.999 678 719 146 130 931 712 × 2 = 1 + 0.999 357 438 292 261 863 424;
  • 51) 0.999 357 438 292 261 863 424 × 2 = 1 + 0.998 714 876 584 523 726 848;
  • 52) 0.998 714 876 584 523 726 848 × 2 = 1 + 0.997 429 753 169 047 453 696;
  • 53) 0.997 429 753 169 047 453 696 × 2 = 1 + 0.994 859 506 338 094 907 392;
  • 54) 0.994 859 506 338 094 907 392 × 2 = 1 + 0.989 719 012 676 189 814 784;

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:


0.000 000 000 742 147 676 076(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 076(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:


0.000 000 000 742 147 676 076(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 -0.000 000 000 742 147 676 076 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