-0.000 000 000 742 147 676 447 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 447(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
-0.000 000 000 742 147 676 447(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:

|-0.000 000 000 742 147 676 447| = 0.000 000 000 742 147 676 447


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: 0.000 000 000 742 147 676 447.

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) 0.000 000 000 742 147 676 447 × 2 = 0 + 0.000 000 001 484 295 352 894;
  • 2) 0.000 000 001 484 295 352 894 × 2 = 0 + 0.000 000 002 968 590 705 788;
  • 3) 0.000 000 002 968 590 705 788 × 2 = 0 + 0.000 000 005 937 181 411 576;
  • 4) 0.000 000 005 937 181 411 576 × 2 = 0 + 0.000 000 011 874 362 823 152;
  • 5) 0.000 000 011 874 362 823 152 × 2 = 0 + 0.000 000 023 748 725 646 304;
  • 6) 0.000 000 023 748 725 646 304 × 2 = 0 + 0.000 000 047 497 451 292 608;
  • 7) 0.000 000 047 497 451 292 608 × 2 = 0 + 0.000 000 094 994 902 585 216;
  • 8) 0.000 000 094 994 902 585 216 × 2 = 0 + 0.000 000 189 989 805 170 432;
  • 9) 0.000 000 189 989 805 170 432 × 2 = 0 + 0.000 000 379 979 610 340 864;
  • 10) 0.000 000 379 979 610 340 864 × 2 = 0 + 0.000 000 759 959 220 681 728;
  • 11) 0.000 000 759 959 220 681 728 × 2 = 0 + 0.000 001 519 918 441 363 456;
  • 12) 0.000 001 519 918 441 363 456 × 2 = 0 + 0.000 003 039 836 882 726 912;
  • 13) 0.000 003 039 836 882 726 912 × 2 = 0 + 0.000 006 079 673 765 453 824;
  • 14) 0.000 006 079 673 765 453 824 × 2 = 0 + 0.000 012 159 347 530 907 648;
  • 15) 0.000 012 159 347 530 907 648 × 2 = 0 + 0.000 024 318 695 061 815 296;
  • 16) 0.000 024 318 695 061 815 296 × 2 = 0 + 0.000 048 637 390 123 630 592;
  • 17) 0.000 048 637 390 123 630 592 × 2 = 0 + 0.000 097 274 780 247 261 184;
  • 18) 0.000 097 274 780 247 261 184 × 2 = 0 + 0.000 194 549 560 494 522 368;
  • 19) 0.000 194 549 560 494 522 368 × 2 = 0 + 0.000 389 099 120 989 044 736;
  • 20) 0.000 389 099 120 989 044 736 × 2 = 0 + 0.000 778 198 241 978 089 472;
  • 21) 0.000 778 198 241 978 089 472 × 2 = 0 + 0.001 556 396 483 956 178 944;
  • 22) 0.001 556 396 483 956 178 944 × 2 = 0 + 0.003 112 792 967 912 357 888;
  • 23) 0.003 112 792 967 912 357 888 × 2 = 0 + 0.006 225 585 935 824 715 776;
  • 24) 0.006 225 585 935 824 715 776 × 2 = 0 + 0.012 451 171 871 649 431 552;
  • 25) 0.012 451 171 871 649 431 552 × 2 = 0 + 0.024 902 343 743 298 863 104;
  • 26) 0.024 902 343 743 298 863 104 × 2 = 0 + 0.049 804 687 486 597 726 208;
  • 27) 0.049 804 687 486 597 726 208 × 2 = 0 + 0.099 609 374 973 195 452 416;
  • 28) 0.099 609 374 973 195 452 416 × 2 = 0 + 0.199 218 749 946 390 904 832;
  • 29) 0.199 218 749 946 390 904 832 × 2 = 0 + 0.398 437 499 892 781 809 664;
  • 30) 0.398 437 499 892 781 809 664 × 2 = 0 + 0.796 874 999 785 563 619 328;
  • 31) 0.796 874 999 785 563 619 328 × 2 = 1 + 0.593 749 999 571 127 238 656;
  • 32) 0.593 749 999 571 127 238 656 × 2 = 1 + 0.187 499 999 142 254 477 312;
  • 33) 0.187 499 999 142 254 477 312 × 2 = 0 + 0.374 999 998 284 508 954 624;
  • 34) 0.374 999 998 284 508 954 624 × 2 = 0 + 0.749 999 996 569 017 909 248;
  • 35) 0.749 999 996 569 017 909 248 × 2 = 1 + 0.499 999 993 138 035 818 496;
  • 36) 0.499 999 993 138 035 818 496 × 2 = 0 + 0.999 999 986 276 071 636 992;
  • 37) 0.999 999 986 276 071 636 992 × 2 = 1 + 0.999 999 972 552 143 273 984;
  • 38) 0.999 999 972 552 143 273 984 × 2 = 1 + 0.999 999 945 104 286 547 968;
  • 39) 0.999 999 945 104 286 547 968 × 2 = 1 + 0.999 999 890 208 573 095 936;
  • 40) 0.999 999 890 208 573 095 936 × 2 = 1 + 0.999 999 780 417 146 191 872;
  • 41) 0.999 999 780 417 146 191 872 × 2 = 1 + 0.999 999 560 834 292 383 744;
  • 42) 0.999 999 560 834 292 383 744 × 2 = 1 + 0.999 999 121 668 584 767 488;
  • 43) 0.999 999 121 668 584 767 488 × 2 = 1 + 0.999 998 243 337 169 534 976;
  • 44) 0.999 998 243 337 169 534 976 × 2 = 1 + 0.999 996 486 674 339 069 952;
  • 45) 0.999 996 486 674 339 069 952 × 2 = 1 + 0.999 992 973 348 678 139 904;
  • 46) 0.999 992 973 348 678 139 904 × 2 = 1 + 0.999 985 946 697 356 279 808;
  • 47) 0.999 985 946 697 356 279 808 × 2 = 1 + 0.999 971 893 394 712 559 616;
  • 48) 0.999 971 893 394 712 559 616 × 2 = 1 + 0.999 943 786 789 425 119 232;
  • 49) 0.999 943 786 789 425 119 232 × 2 = 1 + 0.999 887 573 578 850 238 464;
  • 50) 0.999 887 573 578 850 238 464 × 2 = 1 + 0.999 775 147 157 700 476 928;
  • 51) 0.999 775 147 157 700 476 928 × 2 = 1 + 0.999 550 294 315 400 953 856;
  • 52) 0.999 550 294 315 400 953 856 × 2 = 1 + 0.999 100 588 630 801 907 712;
  • 53) 0.999 100 588 630 801 907 712 × 2 = 1 + 0.998 201 177 261 603 815 424;
  • 54) 0.998 201 177 261 603 815 424 × 2 = 1 + 0.996 402 354 523 207 630 848;

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:


0.000 000 000 742 147 676 447(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 447(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:


0.000 000 000 742 147 676 447(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 -0.000 000 000 742 147 676 447 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