1.153 216 282 338 999 514 602 459 887 537 951 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.153 216 282 338 999 514 602 459 887 537 951 2(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
1.153 216 282 338 999 514 602 459 887 537 951 2(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: 0.153 216 282 338 999 514 602 459 887 537 951 2.

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) 0.153 216 282 338 999 514 602 459 887 537 951 2 × 2 = 0 + 0.306 432 564 677 999 029 204 919 775 075 902 4;
  • 2) 0.306 432 564 677 999 029 204 919 775 075 902 4 × 2 = 0 + 0.612 865 129 355 998 058 409 839 550 151 804 8;
  • 3) 0.612 865 129 355 998 058 409 839 550 151 804 8 × 2 = 1 + 0.225 730 258 711 996 116 819 679 100 303 609 6;
  • 4) 0.225 730 258 711 996 116 819 679 100 303 609 6 × 2 = 0 + 0.451 460 517 423 992 233 639 358 200 607 219 2;
  • 5) 0.451 460 517 423 992 233 639 358 200 607 219 2 × 2 = 0 + 0.902 921 034 847 984 467 278 716 401 214 438 4;
  • 6) 0.902 921 034 847 984 467 278 716 401 214 438 4 × 2 = 1 + 0.805 842 069 695 968 934 557 432 802 428 876 8;
  • 7) 0.805 842 069 695 968 934 557 432 802 428 876 8 × 2 = 1 + 0.611 684 139 391 937 869 114 865 604 857 753 6;
  • 8) 0.611 684 139 391 937 869 114 865 604 857 753 6 × 2 = 1 + 0.223 368 278 783 875 738 229 731 209 715 507 2;
  • 9) 0.223 368 278 783 875 738 229 731 209 715 507 2 × 2 = 0 + 0.446 736 557 567 751 476 459 462 419 431 014 4;
  • 10) 0.446 736 557 567 751 476 459 462 419 431 014 4 × 2 = 0 + 0.893 473 115 135 502 952 918 924 838 862 028 8;
  • 11) 0.893 473 115 135 502 952 918 924 838 862 028 8 × 2 = 1 + 0.786 946 230 271 005 905 837 849 677 724 057 6;
  • 12) 0.786 946 230 271 005 905 837 849 677 724 057 6 × 2 = 1 + 0.573 892 460 542 011 811 675 699 355 448 115 2;
  • 13) 0.573 892 460 542 011 811 675 699 355 448 115 2 × 2 = 1 + 0.147 784 921 084 023 623 351 398 710 896 230 4;
  • 14) 0.147 784 921 084 023 623 351 398 710 896 230 4 × 2 = 0 + 0.295 569 842 168 047 246 702 797 421 792 460 8;
  • 15) 0.295 569 842 168 047 246 702 797 421 792 460 8 × 2 = 0 + 0.591 139 684 336 094 493 405 594 843 584 921 6;
  • 16) 0.591 139 684 336 094 493 405 594 843 584 921 6 × 2 = 1 + 0.182 279 368 672 188 986 811 189 687 169 843 2;
  • 17) 0.182 279 368 672 188 986 811 189 687 169 843 2 × 2 = 0 + 0.364 558 737 344 377 973 622 379 374 339 686 4;
  • 18) 0.364 558 737 344 377 973 622 379 374 339 686 4 × 2 = 0 + 0.729 117 474 688 755 947 244 758 748 679 372 8;
  • 19) 0.729 117 474 688 755 947 244 758 748 679 372 8 × 2 = 1 + 0.458 234 949 377 511 894 489 517 497 358 745 6;
  • 20) 0.458 234 949 377 511 894 489 517 497 358 745 6 × 2 = 0 + 0.916 469 898 755 023 788 979 034 994 717 491 2;
  • 21) 0.916 469 898 755 023 788 979 034 994 717 491 2 × 2 = 1 + 0.832 939 797 510 047 577 958 069 989 434 982 4;
  • 22) 0.832 939 797 510 047 577 958 069 989 434 982 4 × 2 = 1 + 0.665 879 595 020 095 155 916 139 978 869 964 8;
  • 23) 0.665 879 595 020 095 155 916 139 978 869 964 8 × 2 = 1 + 0.331 759 190 040 190 311 832 279 957 739 929 6;
  • 24) 0.331 759 190 040 190 311 832 279 957 739 929 6 × 2 = 0 + 0.663 518 380 080 380 623 664 559 915 479 859 2;

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:


0.153 216 282 338 999 514 602 459 887 537 951 2(10) =

0.0010 0111 0011 1001 0010 1110(2)

5. Positive number before normalization:

1.153 216 282 338 999 514 602 459 887 537 951 2(10) =

1.0010 0111 0011 1001 0010 1110(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:


1.153 216 282 338 999 514 602 459 887 537 951 2(10) =

1.0010 0111 0011 1001 0010 1110(2) =

1.0010 0111 0011 1001 0010 1110(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):
1.0010 0111 0011 1001 0010 1110


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. 001 0011 1001 1100 1001 0111 0 =

001 0011 1001 1100 1001 0111


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) =
001 0011 1001 1100 1001 0111


Decimal number 1.153 216 282 338 999 514 602 459 887 537 951 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 001 0011 1001 1100 1001 0111

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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