-0.000 000 000 742 147 676 646 625 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 625(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
-0.000 000 000 742 147 676 646 625(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:

|-0.000 000 000 742 147 676 646 625| = 0.000 000 000 742 147 676 646 625


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: 0.000 000 000 742 147 676 646 625.

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) 0.000 000 000 742 147 676 646 625 × 2 = 0 + 0.000 000 001 484 295 353 293 25;
  • 2) 0.000 000 001 484 295 353 293 25 × 2 = 0 + 0.000 000 002 968 590 706 586 5;
  • 3) 0.000 000 002 968 590 706 586 5 × 2 = 0 + 0.000 000 005 937 181 413 173;
  • 4) 0.000 000 005 937 181 413 173 × 2 = 0 + 0.000 000 011 874 362 826 346;
  • 5) 0.000 000 011 874 362 826 346 × 2 = 0 + 0.000 000 023 748 725 652 692;
  • 6) 0.000 000 023 748 725 652 692 × 2 = 0 + 0.000 000 047 497 451 305 384;
  • 7) 0.000 000 047 497 451 305 384 × 2 = 0 + 0.000 000 094 994 902 610 768;
  • 8) 0.000 000 094 994 902 610 768 × 2 = 0 + 0.000 000 189 989 805 221 536;
  • 9) 0.000 000 189 989 805 221 536 × 2 = 0 + 0.000 000 379 979 610 443 072;
  • 10) 0.000 000 379 979 610 443 072 × 2 = 0 + 0.000 000 759 959 220 886 144;
  • 11) 0.000 000 759 959 220 886 144 × 2 = 0 + 0.000 001 519 918 441 772 288;
  • 12) 0.000 001 519 918 441 772 288 × 2 = 0 + 0.000 003 039 836 883 544 576;
  • 13) 0.000 003 039 836 883 544 576 × 2 = 0 + 0.000 006 079 673 767 089 152;
  • 14) 0.000 006 079 673 767 089 152 × 2 = 0 + 0.000 012 159 347 534 178 304;
  • 15) 0.000 012 159 347 534 178 304 × 2 = 0 + 0.000 024 318 695 068 356 608;
  • 16) 0.000 024 318 695 068 356 608 × 2 = 0 + 0.000 048 637 390 136 713 216;
  • 17) 0.000 048 637 390 136 713 216 × 2 = 0 + 0.000 097 274 780 273 426 432;
  • 18) 0.000 097 274 780 273 426 432 × 2 = 0 + 0.000 194 549 560 546 852 864;
  • 19) 0.000 194 549 560 546 852 864 × 2 = 0 + 0.000 389 099 121 093 705 728;
  • 20) 0.000 389 099 121 093 705 728 × 2 = 0 + 0.000 778 198 242 187 411 456;
  • 21) 0.000 778 198 242 187 411 456 × 2 = 0 + 0.001 556 396 484 374 822 912;
  • 22) 0.001 556 396 484 374 822 912 × 2 = 0 + 0.003 112 792 968 749 645 824;
  • 23) 0.003 112 792 968 749 645 824 × 2 = 0 + 0.006 225 585 937 499 291 648;
  • 24) 0.006 225 585 937 499 291 648 × 2 = 0 + 0.012 451 171 874 998 583 296;
  • 25) 0.012 451 171 874 998 583 296 × 2 = 0 + 0.024 902 343 749 997 166 592;
  • 26) 0.024 902 343 749 997 166 592 × 2 = 0 + 0.049 804 687 499 994 333 184;
  • 27) 0.049 804 687 499 994 333 184 × 2 = 0 + 0.099 609 374 999 988 666 368;
  • 28) 0.099 609 374 999 988 666 368 × 2 = 0 + 0.199 218 749 999 977 332 736;
  • 29) 0.199 218 749 999 977 332 736 × 2 = 0 + 0.398 437 499 999 954 665 472;
  • 30) 0.398 437 499 999 954 665 472 × 2 = 0 + 0.796 874 999 999 909 330 944;
  • 31) 0.796 874 999 999 909 330 944 × 2 = 1 + 0.593 749 999 999 818 661 888;
  • 32) 0.593 749 999 999 818 661 888 × 2 = 1 + 0.187 499 999 999 637 323 776;
  • 33) 0.187 499 999 999 637 323 776 × 2 = 0 + 0.374 999 999 999 274 647 552;
  • 34) 0.374 999 999 999 274 647 552 × 2 = 0 + 0.749 999 999 998 549 295 104;
  • 35) 0.749 999 999 998 549 295 104 × 2 = 1 + 0.499 999 999 997 098 590 208;
  • 36) 0.499 999 999 997 098 590 208 × 2 = 0 + 0.999 999 999 994 197 180 416;
  • 37) 0.999 999 999 994 197 180 416 × 2 = 1 + 0.999 999 999 988 394 360 832;
  • 38) 0.999 999 999 988 394 360 832 × 2 = 1 + 0.999 999 999 976 788 721 664;
  • 39) 0.999 999 999 976 788 721 664 × 2 = 1 + 0.999 999 999 953 577 443 328;
  • 40) 0.999 999 999 953 577 443 328 × 2 = 1 + 0.999 999 999 907 154 886 656;
  • 41) 0.999 999 999 907 154 886 656 × 2 = 1 + 0.999 999 999 814 309 773 312;
  • 42) 0.999 999 999 814 309 773 312 × 2 = 1 + 0.999 999 999 628 619 546 624;
  • 43) 0.999 999 999 628 619 546 624 × 2 = 1 + 0.999 999 999 257 239 093 248;
  • 44) 0.999 999 999 257 239 093 248 × 2 = 1 + 0.999 999 998 514 478 186 496;
  • 45) 0.999 999 998 514 478 186 496 × 2 = 1 + 0.999 999 997 028 956 372 992;
  • 46) 0.999 999 997 028 956 372 992 × 2 = 1 + 0.999 999 994 057 912 745 984;
  • 47) 0.999 999 994 057 912 745 984 × 2 = 1 + 0.999 999 988 115 825 491 968;
  • 48) 0.999 999 988 115 825 491 968 × 2 = 1 + 0.999 999 976 231 650 983 936;
  • 49) 0.999 999 976 231 650 983 936 × 2 = 1 + 0.999 999 952 463 301 967 872;
  • 50) 0.999 999 952 463 301 967 872 × 2 = 1 + 0.999 999 904 926 603 935 744;
  • 51) 0.999 999 904 926 603 935 744 × 2 = 1 + 0.999 999 809 853 207 871 488;
  • 52) 0.999 999 809 853 207 871 488 × 2 = 1 + 0.999 999 619 706 415 742 976;
  • 53) 0.999 999 619 706 415 742 976 × 2 = 1 + 0.999 999 239 412 831 485 952;
  • 54) 0.999 999 239 412 831 485 952 × 2 = 1 + 0.999 998 478 825 662 971 904;

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:


0.000 000 000 742 147 676 646 625(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 625(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:


0.000 000 000 742 147 676 646 625(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 -0.000 000 000 742 147 676 646 625 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