0.000 000 001 862 645 149 230 956 96 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 001 862 645 149 230 956 96(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
0.000 000 001 862 645 149 230 956 96(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: 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;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 001 862 645 149 230 956 96.

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) 0.000 000 001 862 645 149 230 956 96 × 2 = 0 + 0.000 000 003 725 290 298 461 913 92;
  • 2) 0.000 000 003 725 290 298 461 913 92 × 2 = 0 + 0.000 000 007 450 580 596 923 827 84;
  • 3) 0.000 000 007 450 580 596 923 827 84 × 2 = 0 + 0.000 000 014 901 161 193 847 655 68;
  • 4) 0.000 000 014 901 161 193 847 655 68 × 2 = 0 + 0.000 000 029 802 322 387 695 311 36;
  • 5) 0.000 000 029 802 322 387 695 311 36 × 2 = 0 + 0.000 000 059 604 644 775 390 622 72;
  • 6) 0.000 000 059 604 644 775 390 622 72 × 2 = 0 + 0.000 000 119 209 289 550 781 245 44;
  • 7) 0.000 000 119 209 289 550 781 245 44 × 2 = 0 + 0.000 000 238 418 579 101 562 490 88;
  • 8) 0.000 000 238 418 579 101 562 490 88 × 2 = 0 + 0.000 000 476 837 158 203 124 981 76;
  • 9) 0.000 000 476 837 158 203 124 981 76 × 2 = 0 + 0.000 000 953 674 316 406 249 963 52;
  • 10) 0.000 000 953 674 316 406 249 963 52 × 2 = 0 + 0.000 001 907 348 632 812 499 927 04;
  • 11) 0.000 001 907 348 632 812 499 927 04 × 2 = 0 + 0.000 003 814 697 265 624 999 854 08;
  • 12) 0.000 003 814 697 265 624 999 854 08 × 2 = 0 + 0.000 007 629 394 531 249 999 708 16;
  • 13) 0.000 007 629 394 531 249 999 708 16 × 2 = 0 + 0.000 015 258 789 062 499 999 416 32;
  • 14) 0.000 015 258 789 062 499 999 416 32 × 2 = 0 + 0.000 030 517 578 124 999 998 832 64;
  • 15) 0.000 030 517 578 124 999 998 832 64 × 2 = 0 + 0.000 061 035 156 249 999 997 665 28;
  • 16) 0.000 061 035 156 249 999 997 665 28 × 2 = 0 + 0.000 122 070 312 499 999 995 330 56;
  • 17) 0.000 122 070 312 499 999 995 330 56 × 2 = 0 + 0.000 244 140 624 999 999 990 661 12;
  • 18) 0.000 244 140 624 999 999 990 661 12 × 2 = 0 + 0.000 488 281 249 999 999 981 322 24;
  • 19) 0.000 488 281 249 999 999 981 322 24 × 2 = 0 + 0.000 976 562 499 999 999 962 644 48;
  • 20) 0.000 976 562 499 999 999 962 644 48 × 2 = 0 + 0.001 953 124 999 999 999 925 288 96;
  • 21) 0.001 953 124 999 999 999 925 288 96 × 2 = 0 + 0.003 906 249 999 999 999 850 577 92;
  • 22) 0.003 906 249 999 999 999 850 577 92 × 2 = 0 + 0.007 812 499 999 999 999 701 155 84;
  • 23) 0.007 812 499 999 999 999 701 155 84 × 2 = 0 + 0.015 624 999 999 999 999 402 311 68;
  • 24) 0.015 624 999 999 999 999 402 311 68 × 2 = 0 + 0.031 249 999 999 999 998 804 623 36;
  • 25) 0.031 249 999 999 999 998 804 623 36 × 2 = 0 + 0.062 499 999 999 999 997 609 246 72;
  • 26) 0.062 499 999 999 999 997 609 246 72 × 2 = 0 + 0.124 999 999 999 999 995 218 493 44;
  • 27) 0.124 999 999 999 999 995 218 493 44 × 2 = 0 + 0.249 999 999 999 999 990 436 986 88;
  • 28) 0.249 999 999 999 999 990 436 986 88 × 2 = 0 + 0.499 999 999 999 999 980 873 973 76;
  • 29) 0.499 999 999 999 999 980 873 973 76 × 2 = 0 + 0.999 999 999 999 999 961 747 947 52;
  • 30) 0.999 999 999 999 999 961 747 947 52 × 2 = 1 + 0.999 999 999 999 999 923 495 895 04;
  • 31) 0.999 999 999 999 999 923 495 895 04 × 2 = 1 + 0.999 999 999 999 999 846 991 790 08;
  • 32) 0.999 999 999 999 999 846 991 790 08 × 2 = 1 + 0.999 999 999 999 999 693 983 580 16;
  • 33) 0.999 999 999 999 999 693 983 580 16 × 2 = 1 + 0.999 999 999 999 999 387 967 160 32;
  • 34) 0.999 999 999 999 999 387 967 160 32 × 2 = 1 + 0.999 999 999 999 998 775 934 320 64;
  • 35) 0.999 999 999 999 998 775 934 320 64 × 2 = 1 + 0.999 999 999 999 997 551 868 641 28;
  • 36) 0.999 999 999 999 997 551 868 641 28 × 2 = 1 + 0.999 999 999 999 995 103 737 282 56;
  • 37) 0.999 999 999 999 995 103 737 282 56 × 2 = 1 + 0.999 999 999 999 990 207 474 565 12;
  • 38) 0.999 999 999 999 990 207 474 565 12 × 2 = 1 + 0.999 999 999 999 980 414 949 130 24;
  • 39) 0.999 999 999 999 980 414 949 130 24 × 2 = 1 + 0.999 999 999 999 960 829 898 260 48;
  • 40) 0.999 999 999 999 960 829 898 260 48 × 2 = 1 + 0.999 999 999 999 921 659 796 520 96;
  • 41) 0.999 999 999 999 921 659 796 520 96 × 2 = 1 + 0.999 999 999 999 843 319 593 041 92;
  • 42) 0.999 999 999 999 843 319 593 041 92 × 2 = 1 + 0.999 999 999 999 686 639 186 083 84;
  • 43) 0.999 999 999 999 686 639 186 083 84 × 2 = 1 + 0.999 999 999 999 373 278 372 167 68;
  • 44) 0.999 999 999 999 373 278 372 167 68 × 2 = 1 + 0.999 999 999 998 746 556 744 335 36;
  • 45) 0.999 999 999 998 746 556 744 335 36 × 2 = 1 + 0.999 999 999 997 493 113 488 670 72;
  • 46) 0.999 999 999 997 493 113 488 670 72 × 2 = 1 + 0.999 999 999 994 986 226 977 341 44;
  • 47) 0.999 999 999 994 986 226 977 341 44 × 2 = 1 + 0.999 999 999 989 972 453 954 682 88;
  • 48) 0.999 999 999 989 972 453 954 682 88 × 2 = 1 + 0.999 999 999 979 944 907 909 365 76;
  • 49) 0.999 999 999 979 944 907 909 365 76 × 2 = 1 + 0.999 999 999 959 889 815 818 731 52;
  • 50) 0.999 999 999 959 889 815 818 731 52 × 2 = 1 + 0.999 999 999 919 779 631 637 463 04;
  • 51) 0.999 999 999 919 779 631 637 463 04 × 2 = 1 + 0.999 999 999 839 559 263 274 926 08;
  • 52) 0.999 999 999 839 559 263 274 926 08 × 2 = 1 + 0.999 999 999 679 118 526 549 852 16;
  • 53) 0.999 999 999 679 118 526 549 852 16 × 2 = 1 + 0.999 999 999 358 237 053 099 704 32;

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:


0.000 000 001 862 645 149 230 956 96(10) =

0.0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1(2)

5. Positive number before normalization:

0.000 000 001 862 645 149 230 956 96(10) =

0.0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 30 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 001 862 645 149 230 956 96(10) =

0.0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1(2) =

0.0000 0000 0000 0000 0000 0000 0000 0111 1111 1111 1111 1111 1111 1(2) × 20 =

1.1111 1111 1111 1111 1111 111(2) × 2-30


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): -30

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-30 + 2(8-1) - 1 =

(-30 + 127)(10) =

97(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;
  • 97 ÷ 2 = 48 + 1;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

97(10) =

0110 0001(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 111 1111 1111 1111 1111 1111 =

111 1111 1111 1111 1111 1111


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) =
0110 0001

Mantissa (23 bits) =
111 1111 1111 1111 1111 1111


Decimal number 0.000 000 001 862 645 149 230 956 96 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0001 - 111 1111 1111 1111 1111 1111

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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