Base ten decimal number 101 001 110 011 001 100 110 011 converted to 32 bit single precision IEEE 754 binary floating point standard

How to convert the decimal number 101 001 110 011 001 100 110 011(10)
to
32 bit single precision IEEE 754 binary floating point
(1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 101 001 110 011 001 100 110 011 ÷ 2 = 50 500 555 005 500 550 055 005 + 1;
  • 50 500 555 005 500 550 055 005 ÷ 2 = 25 250 277 502 750 275 027 502 + 1;
  • 25 250 277 502 750 275 027 502 ÷ 2 = 12 625 138 751 375 137 513 751 + 0;
  • 12 625 138 751 375 137 513 751 ÷ 2 = 6 312 569 375 687 568 756 875 + 1;
  • 6 312 569 375 687 568 756 875 ÷ 2 = 3 156 284 687 843 784 378 437 + 1;
  • 3 156 284 687 843 784 378 437 ÷ 2 = 1 578 142 343 921 892 189 218 + 1;
  • 1 578 142 343 921 892 189 218 ÷ 2 = 789 071 171 960 946 094 609 + 0;
  • 789 071 171 960 946 094 609 ÷ 2 = 394 535 585 980 473 047 304 + 1;
  • 394 535 585 980 473 047 304 ÷ 2 = 197 267 792 990 236 523 652 + 0;
  • 197 267 792 990 236 523 652 ÷ 2 = 98 633 896 495 118 261 826 + 0;
  • 98 633 896 495 118 261 826 ÷ 2 = 49 316 948 247 559 130 913 + 0;
  • 49 316 948 247 559 130 913 ÷ 2 = 24 658 474 123 779 565 456 + 1;
  • 24 658 474 123 779 565 456 ÷ 2 = 12 329 237 061 889 782 728 + 0;
  • 12 329 237 061 889 782 728 ÷ 2 = 6 164 618 530 944 891 364 + 0;
  • 6 164 618 530 944 891 364 ÷ 2 = 3 082 309 265 472 445 682 + 0;
  • 3 082 309 265 472 445 682 ÷ 2 = 1 541 154 632 736 222 841 + 0;
  • 1 541 154 632 736 222 841 ÷ 2 = 770 577 316 368 111 420 + 1;
  • 770 577 316 368 111 420 ÷ 2 = 385 288 658 184 055 710 + 0;
  • 385 288 658 184 055 710 ÷ 2 = 192 644 329 092 027 855 + 0;
  • 192 644 329 092 027 855 ÷ 2 = 96 322 164 546 013 927 + 1;
  • 96 322 164 546 013 927 ÷ 2 = 48 161 082 273 006 963 + 1;
  • 48 161 082 273 006 963 ÷ 2 = 24 080 541 136 503 481 + 1;
  • 24 080 541 136 503 481 ÷ 2 = 12 040 270 568 251 740 + 1;
  • 12 040 270 568 251 740 ÷ 2 = 6 020 135 284 125 870 + 0;
  • 6 020 135 284 125 870 ÷ 2 = 3 010 067 642 062 935 + 0;
  • 3 010 067 642 062 935 ÷ 2 = 1 505 033 821 031 467 + 1;
  • 1 505 033 821 031 467 ÷ 2 = 752 516 910 515 733 + 1;
  • 752 516 910 515 733 ÷ 2 = 376 258 455 257 866 + 1;
  • 376 258 455 257 866 ÷ 2 = 188 129 227 628 933 + 0;
  • 188 129 227 628 933 ÷ 2 = 94 064 613 814 466 + 1;
  • 94 064 613 814 466 ÷ 2 = 47 032 306 907 233 + 0;
  • 47 032 306 907 233 ÷ 2 = 23 516 153 453 616 + 1;
  • 23 516 153 453 616 ÷ 2 = 11 758 076 726 808 + 0;
  • 11 758 076 726 808 ÷ 2 = 5 879 038 363 404 + 0;
  • 5 879 038 363 404 ÷ 2 = 2 939 519 181 702 + 0;
  • 2 939 519 181 702 ÷ 2 = 1 469 759 590 851 + 0;
  • 1 469 759 590 851 ÷ 2 = 734 879 795 425 + 1;
  • 734 879 795 425 ÷ 2 = 367 439 897 712 + 1;
  • 367 439 897 712 ÷ 2 = 183 719 948 856 + 0;
  • 183 719 948 856 ÷ 2 = 91 859 974 428 + 0;
  • 91 859 974 428 ÷ 2 = 45 929 987 214 + 0;
  • 45 929 987 214 ÷ 2 = 22 964 993 607 + 0;
  • 22 964 993 607 ÷ 2 = 11 482 496 803 + 1;
  • 11 482 496 803 ÷ 2 = 5 741 248 401 + 1;
  • 5 741 248 401 ÷ 2 = 2 870 624 200 + 1;
  • 2 870 624 200 ÷ 2 = 1 435 312 100 + 0;
  • 1 435 312 100 ÷ 2 = 717 656 050 + 0;
  • 717 656 050 ÷ 2 = 358 828 025 + 0;
  • 358 828 025 ÷ 2 = 179 414 012 + 1;
  • 179 414 012 ÷ 2 = 89 707 006 + 0;
  • 89 707 006 ÷ 2 = 44 853 503 + 0;
  • 44 853 503 ÷ 2 = 22 426 751 + 1;
  • 22 426 751 ÷ 2 = 11 213 375 + 1;
  • 11 213 375 ÷ 2 = 5 606 687 + 1;
  • 5 606 687 ÷ 2 = 2 803 343 + 1;
  • 2 803 343 ÷ 2 = 1 401 671 + 1;
  • 1 401 671 ÷ 2 = 700 835 + 1;
  • 700 835 ÷ 2 = 350 417 + 1;
  • 350 417 ÷ 2 = 175 208 + 1;
  • 175 208 ÷ 2 = 87 604 + 0;
  • 87 604 ÷ 2 = 43 802 + 0;
  • 43 802 ÷ 2 = 21 901 + 0;
  • 21 901 ÷ 2 = 10 950 + 1;
  • 10 950 ÷ 2 = 5 475 + 0;
  • 5 475 ÷ 2 = 2 737 + 1;
  • 2 737 ÷ 2 = 1 368 + 1;
  • 1 368 ÷ 2 = 684 + 0;
  • 684 ÷ 2 = 342 + 0;
  • 342 ÷ 2 = 171 + 0;
  • 171 ÷ 2 = 85 + 1;
  • 85 ÷ 2 = 42 + 1;
  • 42 ÷ 2 = 21 + 0;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

101 001 110 011 001 100 110 011(10) =


1 0101 0110 0011 0100 0111 1111 1001 0001 1100 0011 0000 1010 1110 0111 1001 0000 1000 1011 1011(2)

3. Normalize the binary representation of the number, shifting the decimal mark 76 positions to the left so that only one non zero digit remains to the left of it:

101 001 110 011 001 100 110 011(10) =


1 0101 0110 0011 0100 0111 1111 1001 0001 1100 0011 0000 1010 1110 0111 1001 0000 1000 1011 1011(2) =


1 0101 0110 0011 0100 0111 1111 1001 0001 1100 0011 0000 1010 1110 0111 1001 0000 1000 1011 1011(2) × 20 =


1.0101 0110 0011 0100 0111 1111 1001 0001 1100 0011 0000 1010 1110 0111 1001 0000 1000 1011 1011(2) × 276

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): 76


Mantissa (not normalized): 1.0101 0110 0011 0100 0111 1111 1001 0001 1100 0011 0000 1010 1110 0111 1001 0000 1000 1011 1011

4. 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:

Exponent (adjusted) =


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


76 + 2(8-1) - 1 =


(76 + 127)(10) =


203(10)


  • division = quotient + remainder;
  • 203 ÷ 2 = 101 + 1;
  • 101 ÷ 2 = 50 + 1;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


203(10) =


1100 1011(2)

5. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then 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. 010 1011 0001 1010 0011 1111 1 1001 0001 1100 0011 0000 1010 1110 0111 1001 0000 1000 1011 1011 =


010 1011 0001 1010 0011 1111

Conclusion:

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) =
1100 1011


Mantissa (23 bits) =
010 1011 0001 1010 0011 1111

Number 101 001 110 011 001 100 110 011, a decimal, converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:


0 - 1100 1011 - 010 1011 0001 1010 0011 1111

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 1

      30
    • 1

      29
    • 0

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 1

      24
    • 1

      23
  • Mantissa (23 bits):

    • 0

      22
    • 1

      21
    • 0

      20
    • 1

      19
    • 0

      18
    • 1

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 1

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 1

      5
    • 1

      4
    • 1

      3
    • 1

      2
    • 1

      1
    • 1

      0

Convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits), mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

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