-0.000 000 000 742 147 676 478 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 478(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
-0.000 000 000 742 147 676 478(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:

|-0.000 000 000 742 147 676 478| = 0.000 000 000 742 147 676 478


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: 0.000 000 000 742 147 676 478.

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) 0.000 000 000 742 147 676 478 × 2 = 0 + 0.000 000 001 484 295 352 956;
  • 2) 0.000 000 001 484 295 352 956 × 2 = 0 + 0.000 000 002 968 590 705 912;
  • 3) 0.000 000 002 968 590 705 912 × 2 = 0 + 0.000 000 005 937 181 411 824;
  • 4) 0.000 000 005 937 181 411 824 × 2 = 0 + 0.000 000 011 874 362 823 648;
  • 5) 0.000 000 011 874 362 823 648 × 2 = 0 + 0.000 000 023 748 725 647 296;
  • 6) 0.000 000 023 748 725 647 296 × 2 = 0 + 0.000 000 047 497 451 294 592;
  • 7) 0.000 000 047 497 451 294 592 × 2 = 0 + 0.000 000 094 994 902 589 184;
  • 8) 0.000 000 094 994 902 589 184 × 2 = 0 + 0.000 000 189 989 805 178 368;
  • 9) 0.000 000 189 989 805 178 368 × 2 = 0 + 0.000 000 379 979 610 356 736;
  • 10) 0.000 000 379 979 610 356 736 × 2 = 0 + 0.000 000 759 959 220 713 472;
  • 11) 0.000 000 759 959 220 713 472 × 2 = 0 + 0.000 001 519 918 441 426 944;
  • 12) 0.000 001 519 918 441 426 944 × 2 = 0 + 0.000 003 039 836 882 853 888;
  • 13) 0.000 003 039 836 882 853 888 × 2 = 0 + 0.000 006 079 673 765 707 776;
  • 14) 0.000 006 079 673 765 707 776 × 2 = 0 + 0.000 012 159 347 531 415 552;
  • 15) 0.000 012 159 347 531 415 552 × 2 = 0 + 0.000 024 318 695 062 831 104;
  • 16) 0.000 024 318 695 062 831 104 × 2 = 0 + 0.000 048 637 390 125 662 208;
  • 17) 0.000 048 637 390 125 662 208 × 2 = 0 + 0.000 097 274 780 251 324 416;
  • 18) 0.000 097 274 780 251 324 416 × 2 = 0 + 0.000 194 549 560 502 648 832;
  • 19) 0.000 194 549 560 502 648 832 × 2 = 0 + 0.000 389 099 121 005 297 664;
  • 20) 0.000 389 099 121 005 297 664 × 2 = 0 + 0.000 778 198 242 010 595 328;
  • 21) 0.000 778 198 242 010 595 328 × 2 = 0 + 0.001 556 396 484 021 190 656;
  • 22) 0.001 556 396 484 021 190 656 × 2 = 0 + 0.003 112 792 968 042 381 312;
  • 23) 0.003 112 792 968 042 381 312 × 2 = 0 + 0.006 225 585 936 084 762 624;
  • 24) 0.006 225 585 936 084 762 624 × 2 = 0 + 0.012 451 171 872 169 525 248;
  • 25) 0.012 451 171 872 169 525 248 × 2 = 0 + 0.024 902 343 744 339 050 496;
  • 26) 0.024 902 343 744 339 050 496 × 2 = 0 + 0.049 804 687 488 678 100 992;
  • 27) 0.049 804 687 488 678 100 992 × 2 = 0 + 0.099 609 374 977 356 201 984;
  • 28) 0.099 609 374 977 356 201 984 × 2 = 0 + 0.199 218 749 954 712 403 968;
  • 29) 0.199 218 749 954 712 403 968 × 2 = 0 + 0.398 437 499 909 424 807 936;
  • 30) 0.398 437 499 909 424 807 936 × 2 = 0 + 0.796 874 999 818 849 615 872;
  • 31) 0.796 874 999 818 849 615 872 × 2 = 1 + 0.593 749 999 637 699 231 744;
  • 32) 0.593 749 999 637 699 231 744 × 2 = 1 + 0.187 499 999 275 398 463 488;
  • 33) 0.187 499 999 275 398 463 488 × 2 = 0 + 0.374 999 998 550 796 926 976;
  • 34) 0.374 999 998 550 796 926 976 × 2 = 0 + 0.749 999 997 101 593 853 952;
  • 35) 0.749 999 997 101 593 853 952 × 2 = 1 + 0.499 999 994 203 187 707 904;
  • 36) 0.499 999 994 203 187 707 904 × 2 = 0 + 0.999 999 988 406 375 415 808;
  • 37) 0.999 999 988 406 375 415 808 × 2 = 1 + 0.999 999 976 812 750 831 616;
  • 38) 0.999 999 976 812 750 831 616 × 2 = 1 + 0.999 999 953 625 501 663 232;
  • 39) 0.999 999 953 625 501 663 232 × 2 = 1 + 0.999 999 907 251 003 326 464;
  • 40) 0.999 999 907 251 003 326 464 × 2 = 1 + 0.999 999 814 502 006 652 928;
  • 41) 0.999 999 814 502 006 652 928 × 2 = 1 + 0.999 999 629 004 013 305 856;
  • 42) 0.999 999 629 004 013 305 856 × 2 = 1 + 0.999 999 258 008 026 611 712;
  • 43) 0.999 999 258 008 026 611 712 × 2 = 1 + 0.999 998 516 016 053 223 424;
  • 44) 0.999 998 516 016 053 223 424 × 2 = 1 + 0.999 997 032 032 106 446 848;
  • 45) 0.999 997 032 032 106 446 848 × 2 = 1 + 0.999 994 064 064 212 893 696;
  • 46) 0.999 994 064 064 212 893 696 × 2 = 1 + 0.999 988 128 128 425 787 392;
  • 47) 0.999 988 128 128 425 787 392 × 2 = 1 + 0.999 976 256 256 851 574 784;
  • 48) 0.999 976 256 256 851 574 784 × 2 = 1 + 0.999 952 512 513 703 149 568;
  • 49) 0.999 952 512 513 703 149 568 × 2 = 1 + 0.999 905 025 027 406 299 136;
  • 50) 0.999 905 025 027 406 299 136 × 2 = 1 + 0.999 810 050 054 812 598 272;
  • 51) 0.999 810 050 054 812 598 272 × 2 = 1 + 0.999 620 100 109 625 196 544;
  • 52) 0.999 620 100 109 625 196 544 × 2 = 1 + 0.999 240 200 219 250 393 088;
  • 53) 0.999 240 200 219 250 393 088 × 2 = 1 + 0.998 480 400 438 500 786 176;
  • 54) 0.998 480 400 438 500 786 176 × 2 = 1 + 0.996 960 800 877 001 572 352;

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:


0.000 000 000 742 147 676 478(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 478(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:


0.000 000 000 742 147 676 478(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 -0.000 000 000 742 147 676 478 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