-0.000 000 000 742 147 411 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 411(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
-0.000 000 000 742 147 411(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:

|-0.000 000 000 742 147 411| = 0.000 000 000 742 147 411


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: 0.000 000 000 742 147 411.

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) 0.000 000 000 742 147 411 × 2 = 0 + 0.000 000 001 484 294 822;
  • 2) 0.000 000 001 484 294 822 × 2 = 0 + 0.000 000 002 968 589 644;
  • 3) 0.000 000 002 968 589 644 × 2 = 0 + 0.000 000 005 937 179 288;
  • 4) 0.000 000 005 937 179 288 × 2 = 0 + 0.000 000 011 874 358 576;
  • 5) 0.000 000 011 874 358 576 × 2 = 0 + 0.000 000 023 748 717 152;
  • 6) 0.000 000 023 748 717 152 × 2 = 0 + 0.000 000 047 497 434 304;
  • 7) 0.000 000 047 497 434 304 × 2 = 0 + 0.000 000 094 994 868 608;
  • 8) 0.000 000 094 994 868 608 × 2 = 0 + 0.000 000 189 989 737 216;
  • 9) 0.000 000 189 989 737 216 × 2 = 0 + 0.000 000 379 979 474 432;
  • 10) 0.000 000 379 979 474 432 × 2 = 0 + 0.000 000 759 958 948 864;
  • 11) 0.000 000 759 958 948 864 × 2 = 0 + 0.000 001 519 917 897 728;
  • 12) 0.000 001 519 917 897 728 × 2 = 0 + 0.000 003 039 835 795 456;
  • 13) 0.000 003 039 835 795 456 × 2 = 0 + 0.000 006 079 671 590 912;
  • 14) 0.000 006 079 671 590 912 × 2 = 0 + 0.000 012 159 343 181 824;
  • 15) 0.000 012 159 343 181 824 × 2 = 0 + 0.000 024 318 686 363 648;
  • 16) 0.000 024 318 686 363 648 × 2 = 0 + 0.000 048 637 372 727 296;
  • 17) 0.000 048 637 372 727 296 × 2 = 0 + 0.000 097 274 745 454 592;
  • 18) 0.000 097 274 745 454 592 × 2 = 0 + 0.000 194 549 490 909 184;
  • 19) 0.000 194 549 490 909 184 × 2 = 0 + 0.000 389 098 981 818 368;
  • 20) 0.000 389 098 981 818 368 × 2 = 0 + 0.000 778 197 963 636 736;
  • 21) 0.000 778 197 963 636 736 × 2 = 0 + 0.001 556 395 927 273 472;
  • 22) 0.001 556 395 927 273 472 × 2 = 0 + 0.003 112 791 854 546 944;
  • 23) 0.003 112 791 854 546 944 × 2 = 0 + 0.006 225 583 709 093 888;
  • 24) 0.006 225 583 709 093 888 × 2 = 0 + 0.012 451 167 418 187 776;
  • 25) 0.012 451 167 418 187 776 × 2 = 0 + 0.024 902 334 836 375 552;
  • 26) 0.024 902 334 836 375 552 × 2 = 0 + 0.049 804 669 672 751 104;
  • 27) 0.049 804 669 672 751 104 × 2 = 0 + 0.099 609 339 345 502 208;
  • 28) 0.099 609 339 345 502 208 × 2 = 0 + 0.199 218 678 691 004 416;
  • 29) 0.199 218 678 691 004 416 × 2 = 0 + 0.398 437 357 382 008 832;
  • 30) 0.398 437 357 382 008 832 × 2 = 0 + 0.796 874 714 764 017 664;
  • 31) 0.796 874 714 764 017 664 × 2 = 1 + 0.593 749 429 528 035 328;
  • 32) 0.593 749 429 528 035 328 × 2 = 1 + 0.187 498 859 056 070 656;
  • 33) 0.187 498 859 056 070 656 × 2 = 0 + 0.374 997 718 112 141 312;
  • 34) 0.374 997 718 112 141 312 × 2 = 0 + 0.749 995 436 224 282 624;
  • 35) 0.749 995 436 224 282 624 × 2 = 1 + 0.499 990 872 448 565 248;
  • 36) 0.499 990 872 448 565 248 × 2 = 0 + 0.999 981 744 897 130 496;
  • 37) 0.999 981 744 897 130 496 × 2 = 1 + 0.999 963 489 794 260 992;
  • 38) 0.999 963 489 794 260 992 × 2 = 1 + 0.999 926 979 588 521 984;
  • 39) 0.999 926 979 588 521 984 × 2 = 1 + 0.999 853 959 177 043 968;
  • 40) 0.999 853 959 177 043 968 × 2 = 1 + 0.999 707 918 354 087 936;
  • 41) 0.999 707 918 354 087 936 × 2 = 1 + 0.999 415 836 708 175 872;
  • 42) 0.999 415 836 708 175 872 × 2 = 1 + 0.998 831 673 416 351 744;
  • 43) 0.998 831 673 416 351 744 × 2 = 1 + 0.997 663 346 832 703 488;
  • 44) 0.997 663 346 832 703 488 × 2 = 1 + 0.995 326 693 665 406 976;
  • 45) 0.995 326 693 665 406 976 × 2 = 1 + 0.990 653 387 330 813 952;
  • 46) 0.990 653 387 330 813 952 × 2 = 1 + 0.981 306 774 661 627 904;
  • 47) 0.981 306 774 661 627 904 × 2 = 1 + 0.962 613 549 323 255 808;
  • 48) 0.962 613 549 323 255 808 × 2 = 1 + 0.925 227 098 646 511 616;
  • 49) 0.925 227 098 646 511 616 × 2 = 1 + 0.850 454 197 293 023 232;
  • 50) 0.850 454 197 293 023 232 × 2 = 1 + 0.700 908 394 586 046 464;
  • 51) 0.700 908 394 586 046 464 × 2 = 1 + 0.401 816 789 172 092 928;
  • 52) 0.401 816 789 172 092 928 × 2 = 0 + 0.803 633 578 344 185 856;
  • 53) 0.803 633 578 344 185 856 × 2 = 1 + 0.607 267 156 688 371 712;
  • 54) 0.607 267 156 688 371 712 × 2 = 1 + 0.214 534 313 376 743 424;

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:


0.000 000 000 742 147 411(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 411(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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:


0.000 000 000 742 147 411(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) × 20 =

1.1001 0111 1111 1111 1111 011(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 011


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 1011 =

100 1011 1111 1111 1111 1011


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 1011


Decimal number -0.000 000 000 742 147 411 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1011

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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