-0.000 000 000 742 147 633 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 633(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
-0.000 000 000 742 147 633(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:

|-0.000 000 000 742 147 633| = 0.000 000 000 742 147 633


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: 0.000 000 000 742 147 633.

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) 0.000 000 000 742 147 633 × 2 = 0 + 0.000 000 001 484 295 266;
  • 2) 0.000 000 001 484 295 266 × 2 = 0 + 0.000 000 002 968 590 532;
  • 3) 0.000 000 002 968 590 532 × 2 = 0 + 0.000 000 005 937 181 064;
  • 4) 0.000 000 005 937 181 064 × 2 = 0 + 0.000 000 011 874 362 128;
  • 5) 0.000 000 011 874 362 128 × 2 = 0 + 0.000 000 023 748 724 256;
  • 6) 0.000 000 023 748 724 256 × 2 = 0 + 0.000 000 047 497 448 512;
  • 7) 0.000 000 047 497 448 512 × 2 = 0 + 0.000 000 094 994 897 024;
  • 8) 0.000 000 094 994 897 024 × 2 = 0 + 0.000 000 189 989 794 048;
  • 9) 0.000 000 189 989 794 048 × 2 = 0 + 0.000 000 379 979 588 096;
  • 10) 0.000 000 379 979 588 096 × 2 = 0 + 0.000 000 759 959 176 192;
  • 11) 0.000 000 759 959 176 192 × 2 = 0 + 0.000 001 519 918 352 384;
  • 12) 0.000 001 519 918 352 384 × 2 = 0 + 0.000 003 039 836 704 768;
  • 13) 0.000 003 039 836 704 768 × 2 = 0 + 0.000 006 079 673 409 536;
  • 14) 0.000 006 079 673 409 536 × 2 = 0 + 0.000 012 159 346 819 072;
  • 15) 0.000 012 159 346 819 072 × 2 = 0 + 0.000 024 318 693 638 144;
  • 16) 0.000 024 318 693 638 144 × 2 = 0 + 0.000 048 637 387 276 288;
  • 17) 0.000 048 637 387 276 288 × 2 = 0 + 0.000 097 274 774 552 576;
  • 18) 0.000 097 274 774 552 576 × 2 = 0 + 0.000 194 549 549 105 152;
  • 19) 0.000 194 549 549 105 152 × 2 = 0 + 0.000 389 099 098 210 304;
  • 20) 0.000 389 099 098 210 304 × 2 = 0 + 0.000 778 198 196 420 608;
  • 21) 0.000 778 198 196 420 608 × 2 = 0 + 0.001 556 396 392 841 216;
  • 22) 0.001 556 396 392 841 216 × 2 = 0 + 0.003 112 792 785 682 432;
  • 23) 0.003 112 792 785 682 432 × 2 = 0 + 0.006 225 585 571 364 864;
  • 24) 0.006 225 585 571 364 864 × 2 = 0 + 0.012 451 171 142 729 728;
  • 25) 0.012 451 171 142 729 728 × 2 = 0 + 0.024 902 342 285 459 456;
  • 26) 0.024 902 342 285 459 456 × 2 = 0 + 0.049 804 684 570 918 912;
  • 27) 0.049 804 684 570 918 912 × 2 = 0 + 0.099 609 369 141 837 824;
  • 28) 0.099 609 369 141 837 824 × 2 = 0 + 0.199 218 738 283 675 648;
  • 29) 0.199 218 738 283 675 648 × 2 = 0 + 0.398 437 476 567 351 296;
  • 30) 0.398 437 476 567 351 296 × 2 = 0 + 0.796 874 953 134 702 592;
  • 31) 0.796 874 953 134 702 592 × 2 = 1 + 0.593 749 906 269 405 184;
  • 32) 0.593 749 906 269 405 184 × 2 = 1 + 0.187 499 812 538 810 368;
  • 33) 0.187 499 812 538 810 368 × 2 = 0 + 0.374 999 625 077 620 736;
  • 34) 0.374 999 625 077 620 736 × 2 = 0 + 0.749 999 250 155 241 472;
  • 35) 0.749 999 250 155 241 472 × 2 = 1 + 0.499 998 500 310 482 944;
  • 36) 0.499 998 500 310 482 944 × 2 = 0 + 0.999 997 000 620 965 888;
  • 37) 0.999 997 000 620 965 888 × 2 = 1 + 0.999 994 001 241 931 776;
  • 38) 0.999 994 001 241 931 776 × 2 = 1 + 0.999 988 002 483 863 552;
  • 39) 0.999 988 002 483 863 552 × 2 = 1 + 0.999 976 004 967 727 104;
  • 40) 0.999 976 004 967 727 104 × 2 = 1 + 0.999 952 009 935 454 208;
  • 41) 0.999 952 009 935 454 208 × 2 = 1 + 0.999 904 019 870 908 416;
  • 42) 0.999 904 019 870 908 416 × 2 = 1 + 0.999 808 039 741 816 832;
  • 43) 0.999 808 039 741 816 832 × 2 = 1 + 0.999 616 079 483 633 664;
  • 44) 0.999 616 079 483 633 664 × 2 = 1 + 0.999 232 158 967 267 328;
  • 45) 0.999 232 158 967 267 328 × 2 = 1 + 0.998 464 317 934 534 656;
  • 46) 0.998 464 317 934 534 656 × 2 = 1 + 0.996 928 635 869 069 312;
  • 47) 0.996 928 635 869 069 312 × 2 = 1 + 0.993 857 271 738 138 624;
  • 48) 0.993 857 271 738 138 624 × 2 = 1 + 0.987 714 543 476 277 248;
  • 49) 0.987 714 543 476 277 248 × 2 = 1 + 0.975 429 086 952 554 496;
  • 50) 0.975 429 086 952 554 496 × 2 = 1 + 0.950 858 173 905 108 992;
  • 51) 0.950 858 173 905 108 992 × 2 = 1 + 0.901 716 347 810 217 984;
  • 52) 0.901 716 347 810 217 984 × 2 = 1 + 0.803 432 695 620 435 968;
  • 53) 0.803 432 695 620 435 968 × 2 = 1 + 0.606 865 391 240 871 936;
  • 54) 0.606 865 391 240 871 936 × 2 = 1 + 0.213 730 782 481 743 872;

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:


0.000 000 000 742 147 633(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 633(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:


0.000 000 000 742 147 633(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 -0.000 000 000 742 147 633 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