-0.000 000 000 742 147 676 631 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 631 4(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
-0.000 000 000 742 147 676 631 4(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:

|-0.000 000 000 742 147 676 631 4| = 0.000 000 000 742 147 676 631 4


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: 0.000 000 000 742 147 676 631 4.

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) 0.000 000 000 742 147 676 631 4 × 2 = 0 + 0.000 000 001 484 295 353 262 8;
  • 2) 0.000 000 001 484 295 353 262 8 × 2 = 0 + 0.000 000 002 968 590 706 525 6;
  • 3) 0.000 000 002 968 590 706 525 6 × 2 = 0 + 0.000 000 005 937 181 413 051 2;
  • 4) 0.000 000 005 937 181 413 051 2 × 2 = 0 + 0.000 000 011 874 362 826 102 4;
  • 5) 0.000 000 011 874 362 826 102 4 × 2 = 0 + 0.000 000 023 748 725 652 204 8;
  • 6) 0.000 000 023 748 725 652 204 8 × 2 = 0 + 0.000 000 047 497 451 304 409 6;
  • 7) 0.000 000 047 497 451 304 409 6 × 2 = 0 + 0.000 000 094 994 902 608 819 2;
  • 8) 0.000 000 094 994 902 608 819 2 × 2 = 0 + 0.000 000 189 989 805 217 638 4;
  • 9) 0.000 000 189 989 805 217 638 4 × 2 = 0 + 0.000 000 379 979 610 435 276 8;
  • 10) 0.000 000 379 979 610 435 276 8 × 2 = 0 + 0.000 000 759 959 220 870 553 6;
  • 11) 0.000 000 759 959 220 870 553 6 × 2 = 0 + 0.000 001 519 918 441 741 107 2;
  • 12) 0.000 001 519 918 441 741 107 2 × 2 = 0 + 0.000 003 039 836 883 482 214 4;
  • 13) 0.000 003 039 836 883 482 214 4 × 2 = 0 + 0.000 006 079 673 766 964 428 8;
  • 14) 0.000 006 079 673 766 964 428 8 × 2 = 0 + 0.000 012 159 347 533 928 857 6;
  • 15) 0.000 012 159 347 533 928 857 6 × 2 = 0 + 0.000 024 318 695 067 857 715 2;
  • 16) 0.000 024 318 695 067 857 715 2 × 2 = 0 + 0.000 048 637 390 135 715 430 4;
  • 17) 0.000 048 637 390 135 715 430 4 × 2 = 0 + 0.000 097 274 780 271 430 860 8;
  • 18) 0.000 097 274 780 271 430 860 8 × 2 = 0 + 0.000 194 549 560 542 861 721 6;
  • 19) 0.000 194 549 560 542 861 721 6 × 2 = 0 + 0.000 389 099 121 085 723 443 2;
  • 20) 0.000 389 099 121 085 723 443 2 × 2 = 0 + 0.000 778 198 242 171 446 886 4;
  • 21) 0.000 778 198 242 171 446 886 4 × 2 = 0 + 0.001 556 396 484 342 893 772 8;
  • 22) 0.001 556 396 484 342 893 772 8 × 2 = 0 + 0.003 112 792 968 685 787 545 6;
  • 23) 0.003 112 792 968 685 787 545 6 × 2 = 0 + 0.006 225 585 937 371 575 091 2;
  • 24) 0.006 225 585 937 371 575 091 2 × 2 = 0 + 0.012 451 171 874 743 150 182 4;
  • 25) 0.012 451 171 874 743 150 182 4 × 2 = 0 + 0.024 902 343 749 486 300 364 8;
  • 26) 0.024 902 343 749 486 300 364 8 × 2 = 0 + 0.049 804 687 498 972 600 729 6;
  • 27) 0.049 804 687 498 972 600 729 6 × 2 = 0 + 0.099 609 374 997 945 201 459 2;
  • 28) 0.099 609 374 997 945 201 459 2 × 2 = 0 + 0.199 218 749 995 890 402 918 4;
  • 29) 0.199 218 749 995 890 402 918 4 × 2 = 0 + 0.398 437 499 991 780 805 836 8;
  • 30) 0.398 437 499 991 780 805 836 8 × 2 = 0 + 0.796 874 999 983 561 611 673 6;
  • 31) 0.796 874 999 983 561 611 673 6 × 2 = 1 + 0.593 749 999 967 123 223 347 2;
  • 32) 0.593 749 999 967 123 223 347 2 × 2 = 1 + 0.187 499 999 934 246 446 694 4;
  • 33) 0.187 499 999 934 246 446 694 4 × 2 = 0 + 0.374 999 999 868 492 893 388 8;
  • 34) 0.374 999 999 868 492 893 388 8 × 2 = 0 + 0.749 999 999 736 985 786 777 6;
  • 35) 0.749 999 999 736 985 786 777 6 × 2 = 1 + 0.499 999 999 473 971 573 555 2;
  • 36) 0.499 999 999 473 971 573 555 2 × 2 = 0 + 0.999 999 998 947 943 147 110 4;
  • 37) 0.999 999 998 947 943 147 110 4 × 2 = 1 + 0.999 999 997 895 886 294 220 8;
  • 38) 0.999 999 997 895 886 294 220 8 × 2 = 1 + 0.999 999 995 791 772 588 441 6;
  • 39) 0.999 999 995 791 772 588 441 6 × 2 = 1 + 0.999 999 991 583 545 176 883 2;
  • 40) 0.999 999 991 583 545 176 883 2 × 2 = 1 + 0.999 999 983 167 090 353 766 4;
  • 41) 0.999 999 983 167 090 353 766 4 × 2 = 1 + 0.999 999 966 334 180 707 532 8;
  • 42) 0.999 999 966 334 180 707 532 8 × 2 = 1 + 0.999 999 932 668 361 415 065 6;
  • 43) 0.999 999 932 668 361 415 065 6 × 2 = 1 + 0.999 999 865 336 722 830 131 2;
  • 44) 0.999 999 865 336 722 830 131 2 × 2 = 1 + 0.999 999 730 673 445 660 262 4;
  • 45) 0.999 999 730 673 445 660 262 4 × 2 = 1 + 0.999 999 461 346 891 320 524 8;
  • 46) 0.999 999 461 346 891 320 524 8 × 2 = 1 + 0.999 998 922 693 782 641 049 6;
  • 47) 0.999 998 922 693 782 641 049 6 × 2 = 1 + 0.999 997 845 387 565 282 099 2;
  • 48) 0.999 997 845 387 565 282 099 2 × 2 = 1 + 0.999 995 690 775 130 564 198 4;
  • 49) 0.999 995 690 775 130 564 198 4 × 2 = 1 + 0.999 991 381 550 261 128 396 8;
  • 50) 0.999 991 381 550 261 128 396 8 × 2 = 1 + 0.999 982 763 100 522 256 793 6;
  • 51) 0.999 982 763 100 522 256 793 6 × 2 = 1 + 0.999 965 526 201 044 513 587 2;
  • 52) 0.999 965 526 201 044 513 587 2 × 2 = 1 + 0.999 931 052 402 089 027 174 4;
  • 53) 0.999 931 052 402 089 027 174 4 × 2 = 1 + 0.999 862 104 804 178 054 348 8;
  • 54) 0.999 862 104 804 178 054 348 8 × 2 = 1 + 0.999 724 209 608 356 108 697 6;

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:


0.000 000 000 742 147 676 631 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 631 4(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:


0.000 000 000 742 147 676 631 4(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 -0.000 000 000 742 147 676 631 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 624 9 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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