-0.000 000 000 742 147 676 633 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 633 1(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
-0.000 000 000 742 147 676 633 1(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:

|-0.000 000 000 742 147 676 633 1| = 0.000 000 000 742 147 676 633 1


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: 0.000 000 000 742 147 676 633 1.

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) 0.000 000 000 742 147 676 633 1 × 2 = 0 + 0.000 000 001 484 295 353 266 2;
  • 2) 0.000 000 001 484 295 353 266 2 × 2 = 0 + 0.000 000 002 968 590 706 532 4;
  • 3) 0.000 000 002 968 590 706 532 4 × 2 = 0 + 0.000 000 005 937 181 413 064 8;
  • 4) 0.000 000 005 937 181 413 064 8 × 2 = 0 + 0.000 000 011 874 362 826 129 6;
  • 5) 0.000 000 011 874 362 826 129 6 × 2 = 0 + 0.000 000 023 748 725 652 259 2;
  • 6) 0.000 000 023 748 725 652 259 2 × 2 = 0 + 0.000 000 047 497 451 304 518 4;
  • 7) 0.000 000 047 497 451 304 518 4 × 2 = 0 + 0.000 000 094 994 902 609 036 8;
  • 8) 0.000 000 094 994 902 609 036 8 × 2 = 0 + 0.000 000 189 989 805 218 073 6;
  • 9) 0.000 000 189 989 805 218 073 6 × 2 = 0 + 0.000 000 379 979 610 436 147 2;
  • 10) 0.000 000 379 979 610 436 147 2 × 2 = 0 + 0.000 000 759 959 220 872 294 4;
  • 11) 0.000 000 759 959 220 872 294 4 × 2 = 0 + 0.000 001 519 918 441 744 588 8;
  • 12) 0.000 001 519 918 441 744 588 8 × 2 = 0 + 0.000 003 039 836 883 489 177 6;
  • 13) 0.000 003 039 836 883 489 177 6 × 2 = 0 + 0.000 006 079 673 766 978 355 2;
  • 14) 0.000 006 079 673 766 978 355 2 × 2 = 0 + 0.000 012 159 347 533 956 710 4;
  • 15) 0.000 012 159 347 533 956 710 4 × 2 = 0 + 0.000 024 318 695 067 913 420 8;
  • 16) 0.000 024 318 695 067 913 420 8 × 2 = 0 + 0.000 048 637 390 135 826 841 6;
  • 17) 0.000 048 637 390 135 826 841 6 × 2 = 0 + 0.000 097 274 780 271 653 683 2;
  • 18) 0.000 097 274 780 271 653 683 2 × 2 = 0 + 0.000 194 549 560 543 307 366 4;
  • 19) 0.000 194 549 560 543 307 366 4 × 2 = 0 + 0.000 389 099 121 086 614 732 8;
  • 20) 0.000 389 099 121 086 614 732 8 × 2 = 0 + 0.000 778 198 242 173 229 465 6;
  • 21) 0.000 778 198 242 173 229 465 6 × 2 = 0 + 0.001 556 396 484 346 458 931 2;
  • 22) 0.001 556 396 484 346 458 931 2 × 2 = 0 + 0.003 112 792 968 692 917 862 4;
  • 23) 0.003 112 792 968 692 917 862 4 × 2 = 0 + 0.006 225 585 937 385 835 724 8;
  • 24) 0.006 225 585 937 385 835 724 8 × 2 = 0 + 0.012 451 171 874 771 671 449 6;
  • 25) 0.012 451 171 874 771 671 449 6 × 2 = 0 + 0.024 902 343 749 543 342 899 2;
  • 26) 0.024 902 343 749 543 342 899 2 × 2 = 0 + 0.049 804 687 499 086 685 798 4;
  • 27) 0.049 804 687 499 086 685 798 4 × 2 = 0 + 0.099 609 374 998 173 371 596 8;
  • 28) 0.099 609 374 998 173 371 596 8 × 2 = 0 + 0.199 218 749 996 346 743 193 6;
  • 29) 0.199 218 749 996 346 743 193 6 × 2 = 0 + 0.398 437 499 992 693 486 387 2;
  • 30) 0.398 437 499 992 693 486 387 2 × 2 = 0 + 0.796 874 999 985 386 972 774 4;
  • 31) 0.796 874 999 985 386 972 774 4 × 2 = 1 + 0.593 749 999 970 773 945 548 8;
  • 32) 0.593 749 999 970 773 945 548 8 × 2 = 1 + 0.187 499 999 941 547 891 097 6;
  • 33) 0.187 499 999 941 547 891 097 6 × 2 = 0 + 0.374 999 999 883 095 782 195 2;
  • 34) 0.374 999 999 883 095 782 195 2 × 2 = 0 + 0.749 999 999 766 191 564 390 4;
  • 35) 0.749 999 999 766 191 564 390 4 × 2 = 1 + 0.499 999 999 532 383 128 780 8;
  • 36) 0.499 999 999 532 383 128 780 8 × 2 = 0 + 0.999 999 999 064 766 257 561 6;
  • 37) 0.999 999 999 064 766 257 561 6 × 2 = 1 + 0.999 999 998 129 532 515 123 2;
  • 38) 0.999 999 998 129 532 515 123 2 × 2 = 1 + 0.999 999 996 259 065 030 246 4;
  • 39) 0.999 999 996 259 065 030 246 4 × 2 = 1 + 0.999 999 992 518 130 060 492 8;
  • 40) 0.999 999 992 518 130 060 492 8 × 2 = 1 + 0.999 999 985 036 260 120 985 6;
  • 41) 0.999 999 985 036 260 120 985 6 × 2 = 1 + 0.999 999 970 072 520 241 971 2;
  • 42) 0.999 999 970 072 520 241 971 2 × 2 = 1 + 0.999 999 940 145 040 483 942 4;
  • 43) 0.999 999 940 145 040 483 942 4 × 2 = 1 + 0.999 999 880 290 080 967 884 8;
  • 44) 0.999 999 880 290 080 967 884 8 × 2 = 1 + 0.999 999 760 580 161 935 769 6;
  • 45) 0.999 999 760 580 161 935 769 6 × 2 = 1 + 0.999 999 521 160 323 871 539 2;
  • 46) 0.999 999 521 160 323 871 539 2 × 2 = 1 + 0.999 999 042 320 647 743 078 4;
  • 47) 0.999 999 042 320 647 743 078 4 × 2 = 1 + 0.999 998 084 641 295 486 156 8;
  • 48) 0.999 998 084 641 295 486 156 8 × 2 = 1 + 0.999 996 169 282 590 972 313 6;
  • 49) 0.999 996 169 282 590 972 313 6 × 2 = 1 + 0.999 992 338 565 181 944 627 2;
  • 50) 0.999 992 338 565 181 944 627 2 × 2 = 1 + 0.999 984 677 130 363 889 254 4;
  • 51) 0.999 984 677 130 363 889 254 4 × 2 = 1 + 0.999 969 354 260 727 778 508 8;
  • 52) 0.999 969 354 260 727 778 508 8 × 2 = 1 + 0.999 938 708 521 455 557 017 6;
  • 53) 0.999 938 708 521 455 557 017 6 × 2 = 1 + 0.999 877 417 042 911 114 035 2;
  • 54) 0.999 877 417 042 911 114 035 2 × 2 = 1 + 0.999 754 834 085 822 228 070 4;

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:


0.000 000 000 742 147 676 633 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 633 1(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:


0.000 000 000 742 147 676 633 1(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 -0.000 000 000 742 147 676 633 1 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