-0.000 000 000 742 147 676 414 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 414(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
-0.000 000 000 742 147 676 414(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:

|-0.000 000 000 742 147 676 414| = 0.000 000 000 742 147 676 414


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: 0.000 000 000 742 147 676 414.

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) 0.000 000 000 742 147 676 414 × 2 = 0 + 0.000 000 001 484 295 352 828;
  • 2) 0.000 000 001 484 295 352 828 × 2 = 0 + 0.000 000 002 968 590 705 656;
  • 3) 0.000 000 002 968 590 705 656 × 2 = 0 + 0.000 000 005 937 181 411 312;
  • 4) 0.000 000 005 937 181 411 312 × 2 = 0 + 0.000 000 011 874 362 822 624;
  • 5) 0.000 000 011 874 362 822 624 × 2 = 0 + 0.000 000 023 748 725 645 248;
  • 6) 0.000 000 023 748 725 645 248 × 2 = 0 + 0.000 000 047 497 451 290 496;
  • 7) 0.000 000 047 497 451 290 496 × 2 = 0 + 0.000 000 094 994 902 580 992;
  • 8) 0.000 000 094 994 902 580 992 × 2 = 0 + 0.000 000 189 989 805 161 984;
  • 9) 0.000 000 189 989 805 161 984 × 2 = 0 + 0.000 000 379 979 610 323 968;
  • 10) 0.000 000 379 979 610 323 968 × 2 = 0 + 0.000 000 759 959 220 647 936;
  • 11) 0.000 000 759 959 220 647 936 × 2 = 0 + 0.000 001 519 918 441 295 872;
  • 12) 0.000 001 519 918 441 295 872 × 2 = 0 + 0.000 003 039 836 882 591 744;
  • 13) 0.000 003 039 836 882 591 744 × 2 = 0 + 0.000 006 079 673 765 183 488;
  • 14) 0.000 006 079 673 765 183 488 × 2 = 0 + 0.000 012 159 347 530 366 976;
  • 15) 0.000 012 159 347 530 366 976 × 2 = 0 + 0.000 024 318 695 060 733 952;
  • 16) 0.000 024 318 695 060 733 952 × 2 = 0 + 0.000 048 637 390 121 467 904;
  • 17) 0.000 048 637 390 121 467 904 × 2 = 0 + 0.000 097 274 780 242 935 808;
  • 18) 0.000 097 274 780 242 935 808 × 2 = 0 + 0.000 194 549 560 485 871 616;
  • 19) 0.000 194 549 560 485 871 616 × 2 = 0 + 0.000 389 099 120 971 743 232;
  • 20) 0.000 389 099 120 971 743 232 × 2 = 0 + 0.000 778 198 241 943 486 464;
  • 21) 0.000 778 198 241 943 486 464 × 2 = 0 + 0.001 556 396 483 886 972 928;
  • 22) 0.001 556 396 483 886 972 928 × 2 = 0 + 0.003 112 792 967 773 945 856;
  • 23) 0.003 112 792 967 773 945 856 × 2 = 0 + 0.006 225 585 935 547 891 712;
  • 24) 0.006 225 585 935 547 891 712 × 2 = 0 + 0.012 451 171 871 095 783 424;
  • 25) 0.012 451 171 871 095 783 424 × 2 = 0 + 0.024 902 343 742 191 566 848;
  • 26) 0.024 902 343 742 191 566 848 × 2 = 0 + 0.049 804 687 484 383 133 696;
  • 27) 0.049 804 687 484 383 133 696 × 2 = 0 + 0.099 609 374 968 766 267 392;
  • 28) 0.099 609 374 968 766 267 392 × 2 = 0 + 0.199 218 749 937 532 534 784;
  • 29) 0.199 218 749 937 532 534 784 × 2 = 0 + 0.398 437 499 875 065 069 568;
  • 30) 0.398 437 499 875 065 069 568 × 2 = 0 + 0.796 874 999 750 130 139 136;
  • 31) 0.796 874 999 750 130 139 136 × 2 = 1 + 0.593 749 999 500 260 278 272;
  • 32) 0.593 749 999 500 260 278 272 × 2 = 1 + 0.187 499 999 000 520 556 544;
  • 33) 0.187 499 999 000 520 556 544 × 2 = 0 + 0.374 999 998 001 041 113 088;
  • 34) 0.374 999 998 001 041 113 088 × 2 = 0 + 0.749 999 996 002 082 226 176;
  • 35) 0.749 999 996 002 082 226 176 × 2 = 1 + 0.499 999 992 004 164 452 352;
  • 36) 0.499 999 992 004 164 452 352 × 2 = 0 + 0.999 999 984 008 328 904 704;
  • 37) 0.999 999 984 008 328 904 704 × 2 = 1 + 0.999 999 968 016 657 809 408;
  • 38) 0.999 999 968 016 657 809 408 × 2 = 1 + 0.999 999 936 033 315 618 816;
  • 39) 0.999 999 936 033 315 618 816 × 2 = 1 + 0.999 999 872 066 631 237 632;
  • 40) 0.999 999 872 066 631 237 632 × 2 = 1 + 0.999 999 744 133 262 475 264;
  • 41) 0.999 999 744 133 262 475 264 × 2 = 1 + 0.999 999 488 266 524 950 528;
  • 42) 0.999 999 488 266 524 950 528 × 2 = 1 + 0.999 998 976 533 049 901 056;
  • 43) 0.999 998 976 533 049 901 056 × 2 = 1 + 0.999 997 953 066 099 802 112;
  • 44) 0.999 997 953 066 099 802 112 × 2 = 1 + 0.999 995 906 132 199 604 224;
  • 45) 0.999 995 906 132 199 604 224 × 2 = 1 + 0.999 991 812 264 399 208 448;
  • 46) 0.999 991 812 264 399 208 448 × 2 = 1 + 0.999 983 624 528 798 416 896;
  • 47) 0.999 983 624 528 798 416 896 × 2 = 1 + 0.999 967 249 057 596 833 792;
  • 48) 0.999 967 249 057 596 833 792 × 2 = 1 + 0.999 934 498 115 193 667 584;
  • 49) 0.999 934 498 115 193 667 584 × 2 = 1 + 0.999 868 996 230 387 335 168;
  • 50) 0.999 868 996 230 387 335 168 × 2 = 1 + 0.999 737 992 460 774 670 336;
  • 51) 0.999 737 992 460 774 670 336 × 2 = 1 + 0.999 475 984 921 549 340 672;
  • 52) 0.999 475 984 921 549 340 672 × 2 = 1 + 0.998 951 969 843 098 681 344;
  • 53) 0.998 951 969 843 098 681 344 × 2 = 1 + 0.997 903 939 686 197 362 688;
  • 54) 0.997 903 939 686 197 362 688 × 2 = 1 + 0.995 807 879 372 394 725 376;

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:


0.000 000 000 742 147 676 414(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 414(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:


0.000 000 000 742 147 676 414(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 -0.000 000 000 742 147 676 414 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