0.000 000 001 862 645 149 230 957 064 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 001 862 645 149 230 957 064(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
0.000 000 001 862 645 149 230 957 064(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: 0.000 000 001 862 645 149 230 957 064.

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) 0.000 000 001 862 645 149 230 957 064 × 2 = 0 + 0.000 000 003 725 290 298 461 914 128;
  • 2) 0.000 000 003 725 290 298 461 914 128 × 2 = 0 + 0.000 000 007 450 580 596 923 828 256;
  • 3) 0.000 000 007 450 580 596 923 828 256 × 2 = 0 + 0.000 000 014 901 161 193 847 656 512;
  • 4) 0.000 000 014 901 161 193 847 656 512 × 2 = 0 + 0.000 000 029 802 322 387 695 313 024;
  • 5) 0.000 000 029 802 322 387 695 313 024 × 2 = 0 + 0.000 000 059 604 644 775 390 626 048;
  • 6) 0.000 000 059 604 644 775 390 626 048 × 2 = 0 + 0.000 000 119 209 289 550 781 252 096;
  • 7) 0.000 000 119 209 289 550 781 252 096 × 2 = 0 + 0.000 000 238 418 579 101 562 504 192;
  • 8) 0.000 000 238 418 579 101 562 504 192 × 2 = 0 + 0.000 000 476 837 158 203 125 008 384;
  • 9) 0.000 000 476 837 158 203 125 008 384 × 2 = 0 + 0.000 000 953 674 316 406 250 016 768;
  • 10) 0.000 000 953 674 316 406 250 016 768 × 2 = 0 + 0.000 001 907 348 632 812 500 033 536;
  • 11) 0.000 001 907 348 632 812 500 033 536 × 2 = 0 + 0.000 003 814 697 265 625 000 067 072;
  • 12) 0.000 003 814 697 265 625 000 067 072 × 2 = 0 + 0.000 007 629 394 531 250 000 134 144;
  • 13) 0.000 007 629 394 531 250 000 134 144 × 2 = 0 + 0.000 015 258 789 062 500 000 268 288;
  • 14) 0.000 015 258 789 062 500 000 268 288 × 2 = 0 + 0.000 030 517 578 125 000 000 536 576;
  • 15) 0.000 030 517 578 125 000 000 536 576 × 2 = 0 + 0.000 061 035 156 250 000 001 073 152;
  • 16) 0.000 061 035 156 250 000 001 073 152 × 2 = 0 + 0.000 122 070 312 500 000 002 146 304;
  • 17) 0.000 122 070 312 500 000 002 146 304 × 2 = 0 + 0.000 244 140 625 000 000 004 292 608;
  • 18) 0.000 244 140 625 000 000 004 292 608 × 2 = 0 + 0.000 488 281 250 000 000 008 585 216;
  • 19) 0.000 488 281 250 000 000 008 585 216 × 2 = 0 + 0.000 976 562 500 000 000 017 170 432;
  • 20) 0.000 976 562 500 000 000 017 170 432 × 2 = 0 + 0.001 953 125 000 000 000 034 340 864;
  • 21) 0.001 953 125 000 000 000 034 340 864 × 2 = 0 + 0.003 906 250 000 000 000 068 681 728;
  • 22) 0.003 906 250 000 000 000 068 681 728 × 2 = 0 + 0.007 812 500 000 000 000 137 363 456;
  • 23) 0.007 812 500 000 000 000 137 363 456 × 2 = 0 + 0.015 625 000 000 000 000 274 726 912;
  • 24) 0.015 625 000 000 000 000 274 726 912 × 2 = 0 + 0.031 250 000 000 000 000 549 453 824;
  • 25) 0.031 250 000 000 000 000 549 453 824 × 2 = 0 + 0.062 500 000 000 000 001 098 907 648;
  • 26) 0.062 500 000 000 000 001 098 907 648 × 2 = 0 + 0.125 000 000 000 000 002 197 815 296;
  • 27) 0.125 000 000 000 000 002 197 815 296 × 2 = 0 + 0.250 000 000 000 000 004 395 630 592;
  • 28) 0.250 000 000 000 000 004 395 630 592 × 2 = 0 + 0.500 000 000 000 000 008 791 261 184;
  • 29) 0.500 000 000 000 000 008 791 261 184 × 2 = 1 + 0.000 000 000 000 000 017 582 522 368;
  • 30) 0.000 000 000 000 000 017 582 522 368 × 2 = 0 + 0.000 000 000 000 000 035 165 044 736;
  • 31) 0.000 000 000 000 000 035 165 044 736 × 2 = 0 + 0.000 000 000 000 000 070 330 089 472;
  • 32) 0.000 000 000 000 000 070 330 089 472 × 2 = 0 + 0.000 000 000 000 000 140 660 178 944;
  • 33) 0.000 000 000 000 000 140 660 178 944 × 2 = 0 + 0.000 000 000 000 000 281 320 357 888;
  • 34) 0.000 000 000 000 000 281 320 357 888 × 2 = 0 + 0.000 000 000 000 000 562 640 715 776;
  • 35) 0.000 000 000 000 000 562 640 715 776 × 2 = 0 + 0.000 000 000 000 001 125 281 431 552;
  • 36) 0.000 000 000 000 001 125 281 431 552 × 2 = 0 + 0.000 000 000 000 002 250 562 863 104;
  • 37) 0.000 000 000 000 002 250 562 863 104 × 2 = 0 + 0.000 000 000 000 004 501 125 726 208;
  • 38) 0.000 000 000 000 004 501 125 726 208 × 2 = 0 + 0.000 000 000 000 009 002 251 452 416;
  • 39) 0.000 000 000 000 009 002 251 452 416 × 2 = 0 + 0.000 000 000 000 018 004 502 904 832;
  • 40) 0.000 000 000 000 018 004 502 904 832 × 2 = 0 + 0.000 000 000 000 036 009 005 809 664;
  • 41) 0.000 000 000 000 036 009 005 809 664 × 2 = 0 + 0.000 000 000 000 072 018 011 619 328;
  • 42) 0.000 000 000 000 072 018 011 619 328 × 2 = 0 + 0.000 000 000 000 144 036 023 238 656;
  • 43) 0.000 000 000 000 144 036 023 238 656 × 2 = 0 + 0.000 000 000 000 288 072 046 477 312;
  • 44) 0.000 000 000 000 288 072 046 477 312 × 2 = 0 + 0.000 000 000 000 576 144 092 954 624;
  • 45) 0.000 000 000 000 576 144 092 954 624 × 2 = 0 + 0.000 000 000 001 152 288 185 909 248;
  • 46) 0.000 000 000 001 152 288 185 909 248 × 2 = 0 + 0.000 000 000 002 304 576 371 818 496;
  • 47) 0.000 000 000 002 304 576 371 818 496 × 2 = 0 + 0.000 000 000 004 609 152 743 636 992;
  • 48) 0.000 000 000 004 609 152 743 636 992 × 2 = 0 + 0.000 000 000 009 218 305 487 273 984;
  • 49) 0.000 000 000 009 218 305 487 273 984 × 2 = 0 + 0.000 000 000 018 436 610 974 547 968;
  • 50) 0.000 000 000 018 436 610 974 547 968 × 2 = 0 + 0.000 000 000 036 873 221 949 095 936;
  • 51) 0.000 000 000 036 873 221 949 095 936 × 2 = 0 + 0.000 000 000 073 746 443 898 191 872;
  • 52) 0.000 000 000 073 746 443 898 191 872 × 2 = 0 + 0.000 000 000 147 492 887 796 383 744;

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:


0.000 000 001 862 645 149 230 957 064(10) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0000 0000 0000(2)

5. Positive number before normalization:

0.000 000 001 862 645 149 230 957 064(10) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0000 0000 0000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 29 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 001 862 645 149 230 957 064(10) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0000 0000 0000(2) =

0.0000 0000 0000 0000 0000 0000 0000 1000 0000 0000 0000 0000 0000(2) × 20 =

1.0000 0000 0000 0000 0000 000(2) × 2-29


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): -29

Mantissa (not normalized):
1.0000 0000 0000 0000 0000 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-29 + 2(8-1) - 1 =

(-29 + 127)(10) =

98(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;
  • 98 ÷ 2 = 49 + 0;
  • 49 ÷ 2 = 24 + 1;
  • 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) =

98(10) =

0110 0010(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. 000 0000 0000 0000 0000 0000 =

000 0000 0000 0000 0000 0000


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 0010

Mantissa (23 bits) =
000 0000 0000 0000 0000 0000


Decimal number 0.000 000 001 862 645 149 230 957 064 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0010 - 000 0000 0000 0000 0000 0000

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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