-0.000 000 000 742 147 676 646 710 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 710 2(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
-0.000 000 000 742 147 676 646 710 2(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:

|-0.000 000 000 742 147 676 646 710 2| = 0.000 000 000 742 147 676 646 710 2


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: 0.000 000 000 742 147 676 646 710 2.

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) 0.000 000 000 742 147 676 646 710 2 × 2 = 0 + 0.000 000 001 484 295 353 293 420 4;
  • 2) 0.000 000 001 484 295 353 293 420 4 × 2 = 0 + 0.000 000 002 968 590 706 586 840 8;
  • 3) 0.000 000 002 968 590 706 586 840 8 × 2 = 0 + 0.000 000 005 937 181 413 173 681 6;
  • 4) 0.000 000 005 937 181 413 173 681 6 × 2 = 0 + 0.000 000 011 874 362 826 347 363 2;
  • 5) 0.000 000 011 874 362 826 347 363 2 × 2 = 0 + 0.000 000 023 748 725 652 694 726 4;
  • 6) 0.000 000 023 748 725 652 694 726 4 × 2 = 0 + 0.000 000 047 497 451 305 389 452 8;
  • 7) 0.000 000 047 497 451 305 389 452 8 × 2 = 0 + 0.000 000 094 994 902 610 778 905 6;
  • 8) 0.000 000 094 994 902 610 778 905 6 × 2 = 0 + 0.000 000 189 989 805 221 557 811 2;
  • 9) 0.000 000 189 989 805 221 557 811 2 × 2 = 0 + 0.000 000 379 979 610 443 115 622 4;
  • 10) 0.000 000 379 979 610 443 115 622 4 × 2 = 0 + 0.000 000 759 959 220 886 231 244 8;
  • 11) 0.000 000 759 959 220 886 231 244 8 × 2 = 0 + 0.000 001 519 918 441 772 462 489 6;
  • 12) 0.000 001 519 918 441 772 462 489 6 × 2 = 0 + 0.000 003 039 836 883 544 924 979 2;
  • 13) 0.000 003 039 836 883 544 924 979 2 × 2 = 0 + 0.000 006 079 673 767 089 849 958 4;
  • 14) 0.000 006 079 673 767 089 849 958 4 × 2 = 0 + 0.000 012 159 347 534 179 699 916 8;
  • 15) 0.000 012 159 347 534 179 699 916 8 × 2 = 0 + 0.000 024 318 695 068 359 399 833 6;
  • 16) 0.000 024 318 695 068 359 399 833 6 × 2 = 0 + 0.000 048 637 390 136 718 799 667 2;
  • 17) 0.000 048 637 390 136 718 799 667 2 × 2 = 0 + 0.000 097 274 780 273 437 599 334 4;
  • 18) 0.000 097 274 780 273 437 599 334 4 × 2 = 0 + 0.000 194 549 560 546 875 198 668 8;
  • 19) 0.000 194 549 560 546 875 198 668 8 × 2 = 0 + 0.000 389 099 121 093 750 397 337 6;
  • 20) 0.000 389 099 121 093 750 397 337 6 × 2 = 0 + 0.000 778 198 242 187 500 794 675 2;
  • 21) 0.000 778 198 242 187 500 794 675 2 × 2 = 0 + 0.001 556 396 484 375 001 589 350 4;
  • 22) 0.001 556 396 484 375 001 589 350 4 × 2 = 0 + 0.003 112 792 968 750 003 178 700 8;
  • 23) 0.003 112 792 968 750 003 178 700 8 × 2 = 0 + 0.006 225 585 937 500 006 357 401 6;
  • 24) 0.006 225 585 937 500 006 357 401 6 × 2 = 0 + 0.012 451 171 875 000 012 714 803 2;
  • 25) 0.012 451 171 875 000 012 714 803 2 × 2 = 0 + 0.024 902 343 750 000 025 429 606 4;
  • 26) 0.024 902 343 750 000 025 429 606 4 × 2 = 0 + 0.049 804 687 500 000 050 859 212 8;
  • 27) 0.049 804 687 500 000 050 859 212 8 × 2 = 0 + 0.099 609 375 000 000 101 718 425 6;
  • 28) 0.099 609 375 000 000 101 718 425 6 × 2 = 0 + 0.199 218 750 000 000 203 436 851 2;
  • 29) 0.199 218 750 000 000 203 436 851 2 × 2 = 0 + 0.398 437 500 000 000 406 873 702 4;
  • 30) 0.398 437 500 000 000 406 873 702 4 × 2 = 0 + 0.796 875 000 000 000 813 747 404 8;
  • 31) 0.796 875 000 000 000 813 747 404 8 × 2 = 1 + 0.593 750 000 000 001 627 494 809 6;
  • 32) 0.593 750 000 000 001 627 494 809 6 × 2 = 1 + 0.187 500 000 000 003 254 989 619 2;
  • 33) 0.187 500 000 000 003 254 989 619 2 × 2 = 0 + 0.375 000 000 000 006 509 979 238 4;
  • 34) 0.375 000 000 000 006 509 979 238 4 × 2 = 0 + 0.750 000 000 000 013 019 958 476 8;
  • 35) 0.750 000 000 000 013 019 958 476 8 × 2 = 1 + 0.500 000 000 000 026 039 916 953 6;
  • 36) 0.500 000 000 000 026 039 916 953 6 × 2 = 1 + 0.000 000 000 000 052 079 833 907 2;
  • 37) 0.000 000 000 000 052 079 833 907 2 × 2 = 0 + 0.000 000 000 000 104 159 667 814 4;
  • 38) 0.000 000 000 000 104 159 667 814 4 × 2 = 0 + 0.000 000 000 000 208 319 335 628 8;
  • 39) 0.000 000 000 000 208 319 335 628 8 × 2 = 0 + 0.000 000 000 000 416 638 671 257 6;
  • 40) 0.000 000 000 000 416 638 671 257 6 × 2 = 0 + 0.000 000 000 000 833 277 342 515 2;
  • 41) 0.000 000 000 000 833 277 342 515 2 × 2 = 0 + 0.000 000 000 001 666 554 685 030 4;
  • 42) 0.000 000 000 001 666 554 685 030 4 × 2 = 0 + 0.000 000 000 003 333 109 370 060 8;
  • 43) 0.000 000 000 003 333 109 370 060 8 × 2 = 0 + 0.000 000 000 006 666 218 740 121 6;
  • 44) 0.000 000 000 006 666 218 740 121 6 × 2 = 0 + 0.000 000 000 013 332 437 480 243 2;
  • 45) 0.000 000 000 013 332 437 480 243 2 × 2 = 0 + 0.000 000 000 026 664 874 960 486 4;
  • 46) 0.000 000 000 026 664 874 960 486 4 × 2 = 0 + 0.000 000 000 053 329 749 920 972 8;
  • 47) 0.000 000 000 053 329 749 920 972 8 × 2 = 0 + 0.000 000 000 106 659 499 841 945 6;
  • 48) 0.000 000 000 106 659 499 841 945 6 × 2 = 0 + 0.000 000 000 213 318 999 683 891 2;
  • 49) 0.000 000 000 213 318 999 683 891 2 × 2 = 0 + 0.000 000 000 426 637 999 367 782 4;
  • 50) 0.000 000 000 426 637 999 367 782 4 × 2 = 0 + 0.000 000 000 853 275 998 735 564 8;
  • 51) 0.000 000 000 853 275 998 735 564 8 × 2 = 0 + 0.000 000 001 706 551 997 471 129 6;
  • 52) 0.000 000 001 706 551 997 471 129 6 × 2 = 0 + 0.000 000 003 413 103 994 942 259 2;
  • 53) 0.000 000 003 413 103 994 942 259 2 × 2 = 0 + 0.000 000 006 826 207 989 884 518 4;
  • 54) 0.000 000 006 826 207 989 884 518 4 × 2 = 0 + 0.000 000 013 652 415 979 769 036 8;

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:


0.000 000 000 742 147 676 646 710 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 710 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 646 710 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 710 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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