-0.000 000 000 742 147 676 026 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 026(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
-0.000 000 000 742 147 676 026(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:

|-0.000 000 000 742 147 676 026| = 0.000 000 000 742 147 676 026


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: 0.000 000 000 742 147 676 026.

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) 0.000 000 000 742 147 676 026 × 2 = 0 + 0.000 000 001 484 295 352 052;
  • 2) 0.000 000 001 484 295 352 052 × 2 = 0 + 0.000 000 002 968 590 704 104;
  • 3) 0.000 000 002 968 590 704 104 × 2 = 0 + 0.000 000 005 937 181 408 208;
  • 4) 0.000 000 005 937 181 408 208 × 2 = 0 + 0.000 000 011 874 362 816 416;
  • 5) 0.000 000 011 874 362 816 416 × 2 = 0 + 0.000 000 023 748 725 632 832;
  • 6) 0.000 000 023 748 725 632 832 × 2 = 0 + 0.000 000 047 497 451 265 664;
  • 7) 0.000 000 047 497 451 265 664 × 2 = 0 + 0.000 000 094 994 902 531 328;
  • 8) 0.000 000 094 994 902 531 328 × 2 = 0 + 0.000 000 189 989 805 062 656;
  • 9) 0.000 000 189 989 805 062 656 × 2 = 0 + 0.000 000 379 979 610 125 312;
  • 10) 0.000 000 379 979 610 125 312 × 2 = 0 + 0.000 000 759 959 220 250 624;
  • 11) 0.000 000 759 959 220 250 624 × 2 = 0 + 0.000 001 519 918 440 501 248;
  • 12) 0.000 001 519 918 440 501 248 × 2 = 0 + 0.000 003 039 836 881 002 496;
  • 13) 0.000 003 039 836 881 002 496 × 2 = 0 + 0.000 006 079 673 762 004 992;
  • 14) 0.000 006 079 673 762 004 992 × 2 = 0 + 0.000 012 159 347 524 009 984;
  • 15) 0.000 012 159 347 524 009 984 × 2 = 0 + 0.000 024 318 695 048 019 968;
  • 16) 0.000 024 318 695 048 019 968 × 2 = 0 + 0.000 048 637 390 096 039 936;
  • 17) 0.000 048 637 390 096 039 936 × 2 = 0 + 0.000 097 274 780 192 079 872;
  • 18) 0.000 097 274 780 192 079 872 × 2 = 0 + 0.000 194 549 560 384 159 744;
  • 19) 0.000 194 549 560 384 159 744 × 2 = 0 + 0.000 389 099 120 768 319 488;
  • 20) 0.000 389 099 120 768 319 488 × 2 = 0 + 0.000 778 198 241 536 638 976;
  • 21) 0.000 778 198 241 536 638 976 × 2 = 0 + 0.001 556 396 483 073 277 952;
  • 22) 0.001 556 396 483 073 277 952 × 2 = 0 + 0.003 112 792 966 146 555 904;
  • 23) 0.003 112 792 966 146 555 904 × 2 = 0 + 0.006 225 585 932 293 111 808;
  • 24) 0.006 225 585 932 293 111 808 × 2 = 0 + 0.012 451 171 864 586 223 616;
  • 25) 0.012 451 171 864 586 223 616 × 2 = 0 + 0.024 902 343 729 172 447 232;
  • 26) 0.024 902 343 729 172 447 232 × 2 = 0 + 0.049 804 687 458 344 894 464;
  • 27) 0.049 804 687 458 344 894 464 × 2 = 0 + 0.099 609 374 916 689 788 928;
  • 28) 0.099 609 374 916 689 788 928 × 2 = 0 + 0.199 218 749 833 379 577 856;
  • 29) 0.199 218 749 833 379 577 856 × 2 = 0 + 0.398 437 499 666 759 155 712;
  • 30) 0.398 437 499 666 759 155 712 × 2 = 0 + 0.796 874 999 333 518 311 424;
  • 31) 0.796 874 999 333 518 311 424 × 2 = 1 + 0.593 749 998 667 036 622 848;
  • 32) 0.593 749 998 667 036 622 848 × 2 = 1 + 0.187 499 997 334 073 245 696;
  • 33) 0.187 499 997 334 073 245 696 × 2 = 0 + 0.374 999 994 668 146 491 392;
  • 34) 0.374 999 994 668 146 491 392 × 2 = 0 + 0.749 999 989 336 292 982 784;
  • 35) 0.749 999 989 336 292 982 784 × 2 = 1 + 0.499 999 978 672 585 965 568;
  • 36) 0.499 999 978 672 585 965 568 × 2 = 0 + 0.999 999 957 345 171 931 136;
  • 37) 0.999 999 957 345 171 931 136 × 2 = 1 + 0.999 999 914 690 343 862 272;
  • 38) 0.999 999 914 690 343 862 272 × 2 = 1 + 0.999 999 829 380 687 724 544;
  • 39) 0.999 999 829 380 687 724 544 × 2 = 1 + 0.999 999 658 761 375 449 088;
  • 40) 0.999 999 658 761 375 449 088 × 2 = 1 + 0.999 999 317 522 750 898 176;
  • 41) 0.999 999 317 522 750 898 176 × 2 = 1 + 0.999 998 635 045 501 796 352;
  • 42) 0.999 998 635 045 501 796 352 × 2 = 1 + 0.999 997 270 091 003 592 704;
  • 43) 0.999 997 270 091 003 592 704 × 2 = 1 + 0.999 994 540 182 007 185 408;
  • 44) 0.999 994 540 182 007 185 408 × 2 = 1 + 0.999 989 080 364 014 370 816;
  • 45) 0.999 989 080 364 014 370 816 × 2 = 1 + 0.999 978 160 728 028 741 632;
  • 46) 0.999 978 160 728 028 741 632 × 2 = 1 + 0.999 956 321 456 057 483 264;
  • 47) 0.999 956 321 456 057 483 264 × 2 = 1 + 0.999 912 642 912 114 966 528;
  • 48) 0.999 912 642 912 114 966 528 × 2 = 1 + 0.999 825 285 824 229 933 056;
  • 49) 0.999 825 285 824 229 933 056 × 2 = 1 + 0.999 650 571 648 459 866 112;
  • 50) 0.999 650 571 648 459 866 112 × 2 = 1 + 0.999 301 143 296 919 732 224;
  • 51) 0.999 301 143 296 919 732 224 × 2 = 1 + 0.998 602 286 593 839 464 448;
  • 52) 0.998 602 286 593 839 464 448 × 2 = 1 + 0.997 204 573 187 678 928 896;
  • 53) 0.997 204 573 187 678 928 896 × 2 = 1 + 0.994 409 146 375 357 857 792;
  • 54) 0.994 409 146 375 357 857 792 × 2 = 1 + 0.988 818 292 750 715 715 584;

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:


0.000 000 000 742 147 676 026(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 026(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:


0.000 000 000 742 147 676 026(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 -0.000 000 000 742 147 676 026 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