-0.000 000 000 742 147 676 646 710 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 710 7(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
-0.000 000 000 742 147 676 646 710 7(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:

|-0.000 000 000 742 147 676 646 710 7| = 0.000 000 000 742 147 676 646 710 7


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: 0.000 000 000 742 147 676 646 710 7.

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) 0.000 000 000 742 147 676 646 710 7 × 2 = 0 + 0.000 000 001 484 295 353 293 421 4;
  • 2) 0.000 000 001 484 295 353 293 421 4 × 2 = 0 + 0.000 000 002 968 590 706 586 842 8;
  • 3) 0.000 000 002 968 590 706 586 842 8 × 2 = 0 + 0.000 000 005 937 181 413 173 685 6;
  • 4) 0.000 000 005 937 181 413 173 685 6 × 2 = 0 + 0.000 000 011 874 362 826 347 371 2;
  • 5) 0.000 000 011 874 362 826 347 371 2 × 2 = 0 + 0.000 000 023 748 725 652 694 742 4;
  • 6) 0.000 000 023 748 725 652 694 742 4 × 2 = 0 + 0.000 000 047 497 451 305 389 484 8;
  • 7) 0.000 000 047 497 451 305 389 484 8 × 2 = 0 + 0.000 000 094 994 902 610 778 969 6;
  • 8) 0.000 000 094 994 902 610 778 969 6 × 2 = 0 + 0.000 000 189 989 805 221 557 939 2;
  • 9) 0.000 000 189 989 805 221 557 939 2 × 2 = 0 + 0.000 000 379 979 610 443 115 878 4;
  • 10) 0.000 000 379 979 610 443 115 878 4 × 2 = 0 + 0.000 000 759 959 220 886 231 756 8;
  • 11) 0.000 000 759 959 220 886 231 756 8 × 2 = 0 + 0.000 001 519 918 441 772 463 513 6;
  • 12) 0.000 001 519 918 441 772 463 513 6 × 2 = 0 + 0.000 003 039 836 883 544 927 027 2;
  • 13) 0.000 003 039 836 883 544 927 027 2 × 2 = 0 + 0.000 006 079 673 767 089 854 054 4;
  • 14) 0.000 006 079 673 767 089 854 054 4 × 2 = 0 + 0.000 012 159 347 534 179 708 108 8;
  • 15) 0.000 012 159 347 534 179 708 108 8 × 2 = 0 + 0.000 024 318 695 068 359 416 217 6;
  • 16) 0.000 024 318 695 068 359 416 217 6 × 2 = 0 + 0.000 048 637 390 136 718 832 435 2;
  • 17) 0.000 048 637 390 136 718 832 435 2 × 2 = 0 + 0.000 097 274 780 273 437 664 870 4;
  • 18) 0.000 097 274 780 273 437 664 870 4 × 2 = 0 + 0.000 194 549 560 546 875 329 740 8;
  • 19) 0.000 194 549 560 546 875 329 740 8 × 2 = 0 + 0.000 389 099 121 093 750 659 481 6;
  • 20) 0.000 389 099 121 093 750 659 481 6 × 2 = 0 + 0.000 778 198 242 187 501 318 963 2;
  • 21) 0.000 778 198 242 187 501 318 963 2 × 2 = 0 + 0.001 556 396 484 375 002 637 926 4;
  • 22) 0.001 556 396 484 375 002 637 926 4 × 2 = 0 + 0.003 112 792 968 750 005 275 852 8;
  • 23) 0.003 112 792 968 750 005 275 852 8 × 2 = 0 + 0.006 225 585 937 500 010 551 705 6;
  • 24) 0.006 225 585 937 500 010 551 705 6 × 2 = 0 + 0.012 451 171 875 000 021 103 411 2;
  • 25) 0.012 451 171 875 000 021 103 411 2 × 2 = 0 + 0.024 902 343 750 000 042 206 822 4;
  • 26) 0.024 902 343 750 000 042 206 822 4 × 2 = 0 + 0.049 804 687 500 000 084 413 644 8;
  • 27) 0.049 804 687 500 000 084 413 644 8 × 2 = 0 + 0.099 609 375 000 000 168 827 289 6;
  • 28) 0.099 609 375 000 000 168 827 289 6 × 2 = 0 + 0.199 218 750 000 000 337 654 579 2;
  • 29) 0.199 218 750 000 000 337 654 579 2 × 2 = 0 + 0.398 437 500 000 000 675 309 158 4;
  • 30) 0.398 437 500 000 000 675 309 158 4 × 2 = 0 + 0.796 875 000 000 001 350 618 316 8;
  • 31) 0.796 875 000 000 001 350 618 316 8 × 2 = 1 + 0.593 750 000 000 002 701 236 633 6;
  • 32) 0.593 750 000 000 002 701 236 633 6 × 2 = 1 + 0.187 500 000 000 005 402 473 267 2;
  • 33) 0.187 500 000 000 005 402 473 267 2 × 2 = 0 + 0.375 000 000 000 010 804 946 534 4;
  • 34) 0.375 000 000 000 010 804 946 534 4 × 2 = 0 + 0.750 000 000 000 021 609 893 068 8;
  • 35) 0.750 000 000 000 021 609 893 068 8 × 2 = 1 + 0.500 000 000 000 043 219 786 137 6;
  • 36) 0.500 000 000 000 043 219 786 137 6 × 2 = 1 + 0.000 000 000 000 086 439 572 275 2;
  • 37) 0.000 000 000 000 086 439 572 275 2 × 2 = 0 + 0.000 000 000 000 172 879 144 550 4;
  • 38) 0.000 000 000 000 172 879 144 550 4 × 2 = 0 + 0.000 000 000 000 345 758 289 100 8;
  • 39) 0.000 000 000 000 345 758 289 100 8 × 2 = 0 + 0.000 000 000 000 691 516 578 201 6;
  • 40) 0.000 000 000 000 691 516 578 201 6 × 2 = 0 + 0.000 000 000 001 383 033 156 403 2;
  • 41) 0.000 000 000 001 383 033 156 403 2 × 2 = 0 + 0.000 000 000 002 766 066 312 806 4;
  • 42) 0.000 000 000 002 766 066 312 806 4 × 2 = 0 + 0.000 000 000 005 532 132 625 612 8;
  • 43) 0.000 000 000 005 532 132 625 612 8 × 2 = 0 + 0.000 000 000 011 064 265 251 225 6;
  • 44) 0.000 000 000 011 064 265 251 225 6 × 2 = 0 + 0.000 000 000 022 128 530 502 451 2;
  • 45) 0.000 000 000 022 128 530 502 451 2 × 2 = 0 + 0.000 000 000 044 257 061 004 902 4;
  • 46) 0.000 000 000 044 257 061 004 902 4 × 2 = 0 + 0.000 000 000 088 514 122 009 804 8;
  • 47) 0.000 000 000 088 514 122 009 804 8 × 2 = 0 + 0.000 000 000 177 028 244 019 609 6;
  • 48) 0.000 000 000 177 028 244 019 609 6 × 2 = 0 + 0.000 000 000 354 056 488 039 219 2;
  • 49) 0.000 000 000 354 056 488 039 219 2 × 2 = 0 + 0.000 000 000 708 112 976 078 438 4;
  • 50) 0.000 000 000 708 112 976 078 438 4 × 2 = 0 + 0.000 000 001 416 225 952 156 876 8;
  • 51) 0.000 000 001 416 225 952 156 876 8 × 2 = 0 + 0.000 000 002 832 451 904 313 753 6;
  • 52) 0.000 000 002 832 451 904 313 753 6 × 2 = 0 + 0.000 000 005 664 903 808 627 507 2;
  • 53) 0.000 000 005 664 903 808 627 507 2 × 2 = 0 + 0.000 000 011 329 807 617 255 014 4;
  • 54) 0.000 000 011 329 807 617 255 014 4 × 2 = 0 + 0.000 000 022 659 615 234 510 028 8;

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:


0.000 000 000 742 147 676 646 710 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 710 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 646 710 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 710 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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