-0.000 000 000 742 147 676 646 691 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 691 8(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
-0.000 000 000 742 147 676 646 691 8(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:

|-0.000 000 000 742 147 676 646 691 8| = 0.000 000 000 742 147 676 646 691 8


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: 0.000 000 000 742 147 676 646 691 8.

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) 0.000 000 000 742 147 676 646 691 8 × 2 = 0 + 0.000 000 001 484 295 353 293 383 6;
  • 2) 0.000 000 001 484 295 353 293 383 6 × 2 = 0 + 0.000 000 002 968 590 706 586 767 2;
  • 3) 0.000 000 002 968 590 706 586 767 2 × 2 = 0 + 0.000 000 005 937 181 413 173 534 4;
  • 4) 0.000 000 005 937 181 413 173 534 4 × 2 = 0 + 0.000 000 011 874 362 826 347 068 8;
  • 5) 0.000 000 011 874 362 826 347 068 8 × 2 = 0 + 0.000 000 023 748 725 652 694 137 6;
  • 6) 0.000 000 023 748 725 652 694 137 6 × 2 = 0 + 0.000 000 047 497 451 305 388 275 2;
  • 7) 0.000 000 047 497 451 305 388 275 2 × 2 = 0 + 0.000 000 094 994 902 610 776 550 4;
  • 8) 0.000 000 094 994 902 610 776 550 4 × 2 = 0 + 0.000 000 189 989 805 221 553 100 8;
  • 9) 0.000 000 189 989 805 221 553 100 8 × 2 = 0 + 0.000 000 379 979 610 443 106 201 6;
  • 10) 0.000 000 379 979 610 443 106 201 6 × 2 = 0 + 0.000 000 759 959 220 886 212 403 2;
  • 11) 0.000 000 759 959 220 886 212 403 2 × 2 = 0 + 0.000 001 519 918 441 772 424 806 4;
  • 12) 0.000 001 519 918 441 772 424 806 4 × 2 = 0 + 0.000 003 039 836 883 544 849 612 8;
  • 13) 0.000 003 039 836 883 544 849 612 8 × 2 = 0 + 0.000 006 079 673 767 089 699 225 6;
  • 14) 0.000 006 079 673 767 089 699 225 6 × 2 = 0 + 0.000 012 159 347 534 179 398 451 2;
  • 15) 0.000 012 159 347 534 179 398 451 2 × 2 = 0 + 0.000 024 318 695 068 358 796 902 4;
  • 16) 0.000 024 318 695 068 358 796 902 4 × 2 = 0 + 0.000 048 637 390 136 717 593 804 8;
  • 17) 0.000 048 637 390 136 717 593 804 8 × 2 = 0 + 0.000 097 274 780 273 435 187 609 6;
  • 18) 0.000 097 274 780 273 435 187 609 6 × 2 = 0 + 0.000 194 549 560 546 870 375 219 2;
  • 19) 0.000 194 549 560 546 870 375 219 2 × 2 = 0 + 0.000 389 099 121 093 740 750 438 4;
  • 20) 0.000 389 099 121 093 740 750 438 4 × 2 = 0 + 0.000 778 198 242 187 481 500 876 8;
  • 21) 0.000 778 198 242 187 481 500 876 8 × 2 = 0 + 0.001 556 396 484 374 963 001 753 6;
  • 22) 0.001 556 396 484 374 963 001 753 6 × 2 = 0 + 0.003 112 792 968 749 926 003 507 2;
  • 23) 0.003 112 792 968 749 926 003 507 2 × 2 = 0 + 0.006 225 585 937 499 852 007 014 4;
  • 24) 0.006 225 585 937 499 852 007 014 4 × 2 = 0 + 0.012 451 171 874 999 704 014 028 8;
  • 25) 0.012 451 171 874 999 704 014 028 8 × 2 = 0 + 0.024 902 343 749 999 408 028 057 6;
  • 26) 0.024 902 343 749 999 408 028 057 6 × 2 = 0 + 0.049 804 687 499 998 816 056 115 2;
  • 27) 0.049 804 687 499 998 816 056 115 2 × 2 = 0 + 0.099 609 374 999 997 632 112 230 4;
  • 28) 0.099 609 374 999 997 632 112 230 4 × 2 = 0 + 0.199 218 749 999 995 264 224 460 8;
  • 29) 0.199 218 749 999 995 264 224 460 8 × 2 = 0 + 0.398 437 499 999 990 528 448 921 6;
  • 30) 0.398 437 499 999 990 528 448 921 6 × 2 = 0 + 0.796 874 999 999 981 056 897 843 2;
  • 31) 0.796 874 999 999 981 056 897 843 2 × 2 = 1 + 0.593 749 999 999 962 113 795 686 4;
  • 32) 0.593 749 999 999 962 113 795 686 4 × 2 = 1 + 0.187 499 999 999 924 227 591 372 8;
  • 33) 0.187 499 999 999 924 227 591 372 8 × 2 = 0 + 0.374 999 999 999 848 455 182 745 6;
  • 34) 0.374 999 999 999 848 455 182 745 6 × 2 = 0 + 0.749 999 999 999 696 910 365 491 2;
  • 35) 0.749 999 999 999 696 910 365 491 2 × 2 = 1 + 0.499 999 999 999 393 820 730 982 4;
  • 36) 0.499 999 999 999 393 820 730 982 4 × 2 = 0 + 0.999 999 999 998 787 641 461 964 8;
  • 37) 0.999 999 999 998 787 641 461 964 8 × 2 = 1 + 0.999 999 999 997 575 282 923 929 6;
  • 38) 0.999 999 999 997 575 282 923 929 6 × 2 = 1 + 0.999 999 999 995 150 565 847 859 2;
  • 39) 0.999 999 999 995 150 565 847 859 2 × 2 = 1 + 0.999 999 999 990 301 131 695 718 4;
  • 40) 0.999 999 999 990 301 131 695 718 4 × 2 = 1 + 0.999 999 999 980 602 263 391 436 8;
  • 41) 0.999 999 999 980 602 263 391 436 8 × 2 = 1 + 0.999 999 999 961 204 526 782 873 6;
  • 42) 0.999 999 999 961 204 526 782 873 6 × 2 = 1 + 0.999 999 999 922 409 053 565 747 2;
  • 43) 0.999 999 999 922 409 053 565 747 2 × 2 = 1 + 0.999 999 999 844 818 107 131 494 4;
  • 44) 0.999 999 999 844 818 107 131 494 4 × 2 = 1 + 0.999 999 999 689 636 214 262 988 8;
  • 45) 0.999 999 999 689 636 214 262 988 8 × 2 = 1 + 0.999 999 999 379 272 428 525 977 6;
  • 46) 0.999 999 999 379 272 428 525 977 6 × 2 = 1 + 0.999 999 998 758 544 857 051 955 2;
  • 47) 0.999 999 998 758 544 857 051 955 2 × 2 = 1 + 0.999 999 997 517 089 714 103 910 4;
  • 48) 0.999 999 997 517 089 714 103 910 4 × 2 = 1 + 0.999 999 995 034 179 428 207 820 8;
  • 49) 0.999 999 995 034 179 428 207 820 8 × 2 = 1 + 0.999 999 990 068 358 856 415 641 6;
  • 50) 0.999 999 990 068 358 856 415 641 6 × 2 = 1 + 0.999 999 980 136 717 712 831 283 2;
  • 51) 0.999 999 980 136 717 712 831 283 2 × 2 = 1 + 0.999 999 960 273 435 425 662 566 4;
  • 52) 0.999 999 960 273 435 425 662 566 4 × 2 = 1 + 0.999 999 920 546 870 851 325 132 8;
  • 53) 0.999 999 920 546 870 851 325 132 8 × 2 = 1 + 0.999 999 841 093 741 702 650 265 6;
  • 54) 0.999 999 841 093 741 702 650 265 6 × 2 = 1 + 0.999 999 682 187 483 405 300 531 2;

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:


0.000 000 000 742 147 676 646 691 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 691 8(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:


0.000 000 000 742 147 676 646 691 8(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 -0.000 000 000 742 147 676 646 691 8 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