-0.000 000 000 742 147 676 26 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 26(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 26(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 26| = 0.000 000 000 742 147 676 26


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 26.

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 26 × 2 = 0 + 0.000 000 001 484 295 352 52;
  • 2) 0.000 000 001 484 295 352 52 × 2 = 0 + 0.000 000 002 968 590 705 04;
  • 3) 0.000 000 002 968 590 705 04 × 2 = 0 + 0.000 000 005 937 181 410 08;
  • 4) 0.000 000 005 937 181 410 08 × 2 = 0 + 0.000 000 011 874 362 820 16;
  • 5) 0.000 000 011 874 362 820 16 × 2 = 0 + 0.000 000 023 748 725 640 32;
  • 6) 0.000 000 023 748 725 640 32 × 2 = 0 + 0.000 000 047 497 451 280 64;
  • 7) 0.000 000 047 497 451 280 64 × 2 = 0 + 0.000 000 094 994 902 561 28;
  • 8) 0.000 000 094 994 902 561 28 × 2 = 0 + 0.000 000 189 989 805 122 56;
  • 9) 0.000 000 189 989 805 122 56 × 2 = 0 + 0.000 000 379 979 610 245 12;
  • 10) 0.000 000 379 979 610 245 12 × 2 = 0 + 0.000 000 759 959 220 490 24;
  • 11) 0.000 000 759 959 220 490 24 × 2 = 0 + 0.000 001 519 918 440 980 48;
  • 12) 0.000 001 519 918 440 980 48 × 2 = 0 + 0.000 003 039 836 881 960 96;
  • 13) 0.000 003 039 836 881 960 96 × 2 = 0 + 0.000 006 079 673 763 921 92;
  • 14) 0.000 006 079 673 763 921 92 × 2 = 0 + 0.000 012 159 347 527 843 84;
  • 15) 0.000 012 159 347 527 843 84 × 2 = 0 + 0.000 024 318 695 055 687 68;
  • 16) 0.000 024 318 695 055 687 68 × 2 = 0 + 0.000 048 637 390 111 375 36;
  • 17) 0.000 048 637 390 111 375 36 × 2 = 0 + 0.000 097 274 780 222 750 72;
  • 18) 0.000 097 274 780 222 750 72 × 2 = 0 + 0.000 194 549 560 445 501 44;
  • 19) 0.000 194 549 560 445 501 44 × 2 = 0 + 0.000 389 099 120 891 002 88;
  • 20) 0.000 389 099 120 891 002 88 × 2 = 0 + 0.000 778 198 241 782 005 76;
  • 21) 0.000 778 198 241 782 005 76 × 2 = 0 + 0.001 556 396 483 564 011 52;
  • 22) 0.001 556 396 483 564 011 52 × 2 = 0 + 0.003 112 792 967 128 023 04;
  • 23) 0.003 112 792 967 128 023 04 × 2 = 0 + 0.006 225 585 934 256 046 08;
  • 24) 0.006 225 585 934 256 046 08 × 2 = 0 + 0.012 451 171 868 512 092 16;
  • 25) 0.012 451 171 868 512 092 16 × 2 = 0 + 0.024 902 343 737 024 184 32;
  • 26) 0.024 902 343 737 024 184 32 × 2 = 0 + 0.049 804 687 474 048 368 64;
  • 27) 0.049 804 687 474 048 368 64 × 2 = 0 + 0.099 609 374 948 096 737 28;
  • 28) 0.099 609 374 948 096 737 28 × 2 = 0 + 0.199 218 749 896 193 474 56;
  • 29) 0.199 218 749 896 193 474 56 × 2 = 0 + 0.398 437 499 792 386 949 12;
  • 30) 0.398 437 499 792 386 949 12 × 2 = 0 + 0.796 874 999 584 773 898 24;
  • 31) 0.796 874 999 584 773 898 24 × 2 = 1 + 0.593 749 999 169 547 796 48;
  • 32) 0.593 749 999 169 547 796 48 × 2 = 1 + 0.187 499 998 339 095 592 96;
  • 33) 0.187 499 998 339 095 592 96 × 2 = 0 + 0.374 999 996 678 191 185 92;
  • 34) 0.374 999 996 678 191 185 92 × 2 = 0 + 0.749 999 993 356 382 371 84;
  • 35) 0.749 999 993 356 382 371 84 × 2 = 1 + 0.499 999 986 712 764 743 68;
  • 36) 0.499 999 986 712 764 743 68 × 2 = 0 + 0.999 999 973 425 529 487 36;
  • 37) 0.999 999 973 425 529 487 36 × 2 = 1 + 0.999 999 946 851 058 974 72;
  • 38) 0.999 999 946 851 058 974 72 × 2 = 1 + 0.999 999 893 702 117 949 44;
  • 39) 0.999 999 893 702 117 949 44 × 2 = 1 + 0.999 999 787 404 235 898 88;
  • 40) 0.999 999 787 404 235 898 88 × 2 = 1 + 0.999 999 574 808 471 797 76;
  • 41) 0.999 999 574 808 471 797 76 × 2 = 1 + 0.999 999 149 616 943 595 52;
  • 42) 0.999 999 149 616 943 595 52 × 2 = 1 + 0.999 998 299 233 887 191 04;
  • 43) 0.999 998 299 233 887 191 04 × 2 = 1 + 0.999 996 598 467 774 382 08;
  • 44) 0.999 996 598 467 774 382 08 × 2 = 1 + 0.999 993 196 935 548 764 16;
  • 45) 0.999 993 196 935 548 764 16 × 2 = 1 + 0.999 986 393 871 097 528 32;
  • 46) 0.999 986 393 871 097 528 32 × 2 = 1 + 0.999 972 787 742 195 056 64;
  • 47) 0.999 972 787 742 195 056 64 × 2 = 1 + 0.999 945 575 484 390 113 28;
  • 48) 0.999 945 575 484 390 113 28 × 2 = 1 + 0.999 891 150 968 780 226 56;
  • 49) 0.999 891 150 968 780 226 56 × 2 = 1 + 0.999 782 301 937 560 453 12;
  • 50) 0.999 782 301 937 560 453 12 × 2 = 1 + 0.999 564 603 875 120 906 24;
  • 51) 0.999 564 603 875 120 906 24 × 2 = 1 + 0.999 129 207 750 241 812 48;
  • 52) 0.999 129 207 750 241 812 48 × 2 = 1 + 0.998 258 415 500 483 624 96;
  • 53) 0.998 258 415 500 483 624 96 × 2 = 1 + 0.996 516 831 000 967 249 92;
  • 54) 0.996 516 831 000 967 249 92 × 2 = 1 + 0.993 033 662 001 934 499 84;

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 26(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 26(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 26(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 26 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