1.414 213 562 373 095 048 801 688 724 209 698 078 569 671 859 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.414 213 562 373 095 048 801 688 724 209 698 078 569 671 859 8(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
1.414 213 562 373 095 048 801 688 724 209 698 078 569 671 859 8(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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

2. 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.414 213 562 373 095 048 801 688 724 209 698 078 569 671 859 8.

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) 0.414 213 562 373 095 048 801 688 724 209 698 078 569 671 859 8 × 2 = 0 + 0.828 427 124 746 190 097 603 377 448 419 396 157 139 343 719 6;
  • 2) 0.828 427 124 746 190 097 603 377 448 419 396 157 139 343 719 6 × 2 = 1 + 0.656 854 249 492 380 195 206 754 896 838 792 314 278 687 439 2;
  • 3) 0.656 854 249 492 380 195 206 754 896 838 792 314 278 687 439 2 × 2 = 1 + 0.313 708 498 984 760 390 413 509 793 677 584 628 557 374 878 4;
  • 4) 0.313 708 498 984 760 390 413 509 793 677 584 628 557 374 878 4 × 2 = 0 + 0.627 416 997 969 520 780 827 019 587 355 169 257 114 749 756 8;
  • 5) 0.627 416 997 969 520 780 827 019 587 355 169 257 114 749 756 8 × 2 = 1 + 0.254 833 995 939 041 561 654 039 174 710 338 514 229 499 513 6;
  • 6) 0.254 833 995 939 041 561 654 039 174 710 338 514 229 499 513 6 × 2 = 0 + 0.509 667 991 878 083 123 308 078 349 420 677 028 458 999 027 2;
  • 7) 0.509 667 991 878 083 123 308 078 349 420 677 028 458 999 027 2 × 2 = 1 + 0.019 335 983 756 166 246 616 156 698 841 354 056 917 998 054 4;
  • 8) 0.019 335 983 756 166 246 616 156 698 841 354 056 917 998 054 4 × 2 = 0 + 0.038 671 967 512 332 493 232 313 397 682 708 113 835 996 108 8;
  • 9) 0.038 671 967 512 332 493 232 313 397 682 708 113 835 996 108 8 × 2 = 0 + 0.077 343 935 024 664 986 464 626 795 365 416 227 671 992 217 6;
  • 10) 0.077 343 935 024 664 986 464 626 795 365 416 227 671 992 217 6 × 2 = 0 + 0.154 687 870 049 329 972 929 253 590 730 832 455 343 984 435 2;
  • 11) 0.154 687 870 049 329 972 929 253 590 730 832 455 343 984 435 2 × 2 = 0 + 0.309 375 740 098 659 945 858 507 181 461 664 910 687 968 870 4;
  • 12) 0.309 375 740 098 659 945 858 507 181 461 664 910 687 968 870 4 × 2 = 0 + 0.618 751 480 197 319 891 717 014 362 923 329 821 375 937 740 8;
  • 13) 0.618 751 480 197 319 891 717 014 362 923 329 821 375 937 740 8 × 2 = 1 + 0.237 502 960 394 639 783 434 028 725 846 659 642 751 875 481 6;
  • 14) 0.237 502 960 394 639 783 434 028 725 846 659 642 751 875 481 6 × 2 = 0 + 0.475 005 920 789 279 566 868 057 451 693 319 285 503 750 963 2;
  • 15) 0.475 005 920 789 279 566 868 057 451 693 319 285 503 750 963 2 × 2 = 0 + 0.950 011 841 578 559 133 736 114 903 386 638 571 007 501 926 4;
  • 16) 0.950 011 841 578 559 133 736 114 903 386 638 571 007 501 926 4 × 2 = 1 + 0.900 023 683 157 118 267 472 229 806 773 277 142 015 003 852 8;
  • 17) 0.900 023 683 157 118 267 472 229 806 773 277 142 015 003 852 8 × 2 = 1 + 0.800 047 366 314 236 534 944 459 613 546 554 284 030 007 705 6;
  • 18) 0.800 047 366 314 236 534 944 459 613 546 554 284 030 007 705 6 × 2 = 1 + 0.600 094 732 628 473 069 888 919 227 093 108 568 060 015 411 2;
  • 19) 0.600 094 732 628 473 069 888 919 227 093 108 568 060 015 411 2 × 2 = 1 + 0.200 189 465 256 946 139 777 838 454 186 217 136 120 030 822 4;
  • 20) 0.200 189 465 256 946 139 777 838 454 186 217 136 120 030 822 4 × 2 = 0 + 0.400 378 930 513 892 279 555 676 908 372 434 272 240 061 644 8;
  • 21) 0.400 378 930 513 892 279 555 676 908 372 434 272 240 061 644 8 × 2 = 0 + 0.800 757 861 027 784 559 111 353 816 744 868 544 480 123 289 6;
  • 22) 0.800 757 861 027 784 559 111 353 816 744 868 544 480 123 289 6 × 2 = 1 + 0.601 515 722 055 569 118 222 707 633 489 737 088 960 246 579 2;
  • 23) 0.601 515 722 055 569 118 222 707 633 489 737 088 960 246 579 2 × 2 = 1 + 0.203 031 444 111 138 236 445 415 266 979 474 177 920 493 158 4;
  • 24) 0.203 031 444 111 138 236 445 415 266 979 474 177 920 493 158 4 × 2 = 0 + 0.406 062 888 222 276 472 890 830 533 958 948 355 840 986 316 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).


4. 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:


0.414 213 562 373 095 048 801 688 724 209 698 078 569 671 859 8(10) =

0.0110 1010 0000 1001 1110 0110(2)

5. Positive number before normalization:

1.414 213 562 373 095 048 801 688 724 209 698 078 569 671 859 8(10) =

1.0110 1010 0000 1001 1110 0110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.414 213 562 373 095 048 801 688 724 209 698 078 569 671 859 8(10) =

1.0110 1010 0000 1001 1110 0110(2) =

1.0110 1010 0000 1001 1110 0110(2) × 20


7. 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 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.0110 1010 0000 1001 1110 0110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

0 + 2(8-1) - 1 =

(0 + 127)(10) =

127(10)


9. 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;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

127(10) =

0111 1111(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 011 0101 0000 0100 1111 0011 0 =

011 0101 0000 0100 1111 0011


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0111 1111

Mantissa (23 bits) =
011 0101 0000 0100 1111 0011


Decimal number 1.414 213 562 373 095 048 801 688 724 209 698 078 569 671 859 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 011 0101 0000 0100 1111 0011

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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